import itertools import warnings import numpy as np import pandas as pd import pandas_flavor as pf from scipy.stats import t, norm from scipy.linalg import pinvh, lstsq from pingouin.config import options from pingouin.utils import remove_na as rm_na from pingouin.utils import _flatten_list as _fl from pingouin.utils import _postprocess_dataframe __all__ = ["linear_regression", "logistic_regression", "mediation_analysis"] def linear_regression( X, y, add_intercept=True, weights=None, coef_only=False, alpha=0.05, as_dataframe=True, remove_na=False, relimp=False, ): """(Multiple) Linear regression. Parameters ---------- X : array_like Predictor(s), of shape *(n_samples, n_features)* or *(n_samples)*. y : array_like Dependent variable, of shape *(n_samples)*. add_intercept : bool If False, assume that the data are already centered. If True, add a constant term to the model. In this case, the first value in the output dict is the intercept of the model. .. note:: It is generally recommended to include a constant term (intercept) to the model to limit the bias and force the residual mean to equal zero. The intercept coefficient and p-values are however rarely meaningful. weights : array_like An optional vector of sample weights to be used in the fitting process, of shape *(n_samples)*. Missing or negative weights are not allowed. If not null, a weighted least squares is calculated. .. versionadded:: 0.3.5 coef_only : bool If True, return only the regression coefficients. alpha : float Alpha value used for the confidence intervals. :math:`\\text{CI} = [\\alpha / 2 ; 1 - \\alpha / 2]` as_dataframe : bool If True, returns a pandas DataFrame. If False, returns a dictionnary. remove_na : bool If True, apply a listwise deletion of missing values (i.e. the entire row is removed). Default is False, which will raise an error if missing values are present in either the predictor(s) or dependent variable. relimp : bool If True, returns the relative importance (= contribution) of predictors. This is irrelevant when the predictors are uncorrelated: the total :math:`R^2` of the model is simply the sum of each univariate regression :math:`R^2`-values. However, this does not apply when predictors are correlated. Instead, the total :math:`R^2` of the model is partitioned by averaging over all combinations of predictors, as done in the `relaimpo `_ R package (``calc.relimp(type="lmg")``). .. warning:: The computation time roughly doubles for each additional predictor and therefore this can be extremely slow for models with more than 12-15 predictors. .. versionadded:: 0.3.0 Returns ------- stats : :py:class:`pandas.DataFrame` or dict Linear regression summary: * ``'names'``: name of variable(s) in the model (e.g. x1, x2...) * ``'coef'``: regression coefficients * ``'se'``: standard errors * ``'T'``: T-values * ``'pval'``: p-values * ``'r2'``: coefficient of determination (:math:`R^2`) * ``'adj_r2'``: adjusted :math:`R^2` * ``'CI[2.5%]'``: lower confidence intervals * ``'CI[97.5%]'``: upper confidence intervals * ``'relimp'``: relative contribution of each predictor to the final\ :math:`R^2` (only if ``relimp=True``). * ``'relimp_perc'``: percent relative contribution In addition, the output dataframe comes with hidden attributes such as the residuals, and degrees of freedom of the model and residuals, which can be accessed as follow, respectively: >>> lm = pg.linear_regression() # doctest: +SKIP >>> lm.residuals_, lm.df_model_, lm.df_resid_ # doctest: +SKIP Note that to follow scikit-learn convention, these hidden atributes end with an "_". When ``as_dataframe=False`` however, these attributes are no longer hidden and can be accessed as any other keys in the output dictionary. >>> lm = pg.linear_regression() # doctest: +SKIP >>> lm['residuals'], lm['df_model'], lm['df_resid'] # doctest: +SKIP When ``as_dataframe=False`` the dictionary also contains the processed ``X`` and ``y`` arrays (i.e, with NaNs removed if ``remove_na=True``) and the model's predicted values ``pred``. >>> lm['X'], lm['y'], lm['pred'] # doctest: +SKIP For a weighted least squares fit, the weighted ``Xw`` and ``yw`` arrays are included in the dictionary. >>> lm['Xw'], lm['yw'] # doctest: +SKIP See also -------- logistic_regression, mediation_analysis, corr Notes ----- The :math:`\\beta` coefficients are estimated using an ordinary least squares (OLS) regression, as implemented in the :py:func:`scipy.linalg.lstsq` function. The OLS method minimizes the sum of squared residuals, and leads to a closed-form expression for the estimated :math:`\\beta`: .. math:: \\hat{\\beta} = (X^TX)^{-1} X^Ty It is generally recommended to include a constant term (intercept) to the model to limit the bias and force the residual mean to equal zero. Note that intercept coefficient and p-values are however rarely meaningful. The standard error of the estimates is a measure of the accuracy of the prediction defined as: .. math:: \\sigma = \\sqrt{\\text{MSE} \\cdot (X^TX)^{-1}} where :math:`\\text{MSE}` is the mean squared error, .. math:: \\text{MSE} = \\frac{SS_{\\text{resid}}}{n - p - 1} = \\frac{\\sum{(\\text{true} - \\text{pred})^2}}{n - p - 1} :math:`p` is the total number of predictor variables in the model (excluding the intercept) and :math:`n` is the sample size. Using the :math:`\\beta` coefficients and the standard errors, the T-values can be obtained: .. math:: T = \\frac{\\beta}{\\sigma} and the p-values approximated using a T-distribution with :math:`n - p - 1` degrees of freedom. The coefficient of determination (:math:`R^2`) is defined as: .. math:: R^2 = 1 - (\\frac{SS_{\\text{resid}}}{SS_{\\text{total}}}) The adjusted :math:`R^2` is defined as: .. math:: \\overline{R}^2 = 1 - (1 - R^2) \\frac{n - 1}{n - p - 1} The relative importance (``relimp``) column is a partitioning of the total :math:`R^2` of the model into individual :math:`R^2` contribution. This is calculated by taking the average over average contributions in models of different sizes. For more details, please refer to `Groemping et al. 2006 `_ and the R package `relaimpo `_. Note that Pingouin will automatically remove any duplicate columns from :math:`X`, as well as any column with only one unique value (constant), excluding the intercept. Results have been compared against sklearn, R, statsmodels and JASP. Examples -------- 1. Simple linear regression using columns of a pandas dataframe In this first example, we'll use the tips dataset to see how well we can predict the waiter's tip (in dollars) based on the total bill (also in dollars). >>> import numpy as np >>> import pingouin as pg >>> df = pg.read_dataset('tips') >>> # Let's predict the tip ($) based on the total bill (also in $) >>> lm = pg.linear_regression(df['total_bill'], df['tip']) >>> lm.round(2) names coef se T pval r2 adj_r2 CI[2.5%] CI[97.5%] 0 Intercept 0.92 0.16 5.76 0.0 0.46 0.45 0.61 1.23 1 total_bill 0.11 0.01 14.26 0.0 0.46 0.45 0.09 0.12 It comes as no surprise that total bill is indeed a significant predictor of the waiter's tip (T=14.26, p<0.05). The :math:`R^2` of the model is 0.46 and the adjusted :math:`R^2` is 0.45, which means that our model roughly explains ~45% of the total variance in the tip amount. 2. Multiple linear regression We can also have more than one predictor and run a multiple linear regression. Below, we add the party size as a second predictor of tip. >>> # We'll add a second predictor: the party size >>> lm = pg.linear_regression(df[['total_bill', 'size']], df['tip']) >>> lm.round(2) names coef se T pval r2 adj_r2 CI[2.5%] CI[97.5%] 0 Intercept 0.67 0.19 3.46 0.00 0.47 0.46 0.29 1.05 1 total_bill 0.09 0.01 10.17 0.00 0.47 0.46 0.07 0.11 2 size 0.19 0.09 2.26 0.02 0.47 0.46 0.02 0.36 The party size is also a significant predictor of tip (T=2.26, p=0.02). Note that adding this new predictor however only improved the :math:`R^2` of our model by ~1%. This function also works with numpy arrays: >>> X = df[['total_bill', 'size']].to_numpy() >>> y = df['tip'].to_numpy() >>> pg.linear_regression(X, y).round(2) names coef se T pval r2 adj_r2 CI[2.5%] CI[97.5%] 0 Intercept 0.67 0.19 3.46 0.00 0.47 0.46 0.29 1.05 1 x1 0.09 0.01 10.17 0.00 0.47 0.46 0.07 0.11 2 x2 0.19 0.09 2.26 0.02 0.47 0.46 0.02 0.36 3. Get the residuals >>> # For clarity, only display the first 9 values >>> np.round(lm.residuals_, 2)[:9] array([-1.62, -0.55, 0.31, 0.06, -0.11, 0.93, 0.13, -0.81, -0.49]) Using pandas, we can show a summary of the distribution of the residuals: >>> import pandas as pd >>> pd.Series(lm.residuals_).describe().round(2) count 244.00 mean -0.00 std 1.01 min -2.93 25% -0.55 50% -0.09 75% 0.51 max 4.04 dtype: float64 5. No intercept and return only the regression coefficients Sometimes it may be useful to remove the constant term from the regression, or to only return the regression coefficients without calculating the standard errors or p-values. This latter can potentially save you a lot of time if you need to calculate hundreds of regression and only care about the coefficients! >>> pg.linear_regression(X, y, add_intercept=False, coef_only=True) array([0.1007119 , 0.36209717]) 6. Return a dictionnary instead of a dataframe >>> lm_dict = pg.linear_regression(X, y, as_dataframe=False) >>> lm_dict.keys() dict_keys(['names', 'coef', 'se', 'T', 'pval', 'r2', 'adj_r2', 'CI[2.5%]', 'CI[97.5%]', 'df_model', 'df_resid', 'residuals', 'X', 'y', 'pred']) 7. Remove missing values >>> X[4, 1] = np.nan >>> y[7] = np.nan >>> pg.linear_regression(X, y, remove_na=True, coef_only=True) array([0.65749955, 0.09262059, 0.19927529]) 8. Get the relative importance of predictors >>> lm = pg.linear_regression(X, y, remove_na=True, relimp=True) >>> lm[['names', 'relimp', 'relimp_perc']] names relimp relimp_perc 0 Intercept NaN NaN 1 x1 0.342503 73.045583 2 x2 0.126386 26.954417 The ``relimp`` column is a partitioning of the total :math:`R^2` of the model into individual contribution. Therefore, it sums to the :math:`R^2` of the full model. The ``relimp_perc`` is normalized to sum to 100%. See `Groemping 2006 `_ for more details. >>> lm[['relimp', 'relimp_perc']].sum() relimp 0.468889 relimp_perc 100.000000 dtype: float64 9. Weighted linear regression >>> X = [1, 2, 3, 4, 5, 6] >>> y = [10, 22, 11, 13, 13, 16] >>> w = [1, 0.1, 1, 1, 0.5, 1] # Array of weights. Must be >= 0. >>> lm = pg.linear_regression(X, y, weights=w) >>> lm.round(2) names coef se T pval r2 adj_r2 CI[2.5%] CI[97.5%] 0 Intercept 9.00 2.03 4.42 0.01 0.51 0.39 3.35 14.64 1 x1 1.04 0.50 2.06 0.11 0.51 0.39 -0.36 2.44 """ # Extract names if X is a Dataframe or Series if isinstance(X, pd.DataFrame): names = X.keys().tolist() elif isinstance(X, pd.Series): names = [X.name] else: names = [] # Convert input to numpy array X = np.asarray(X) y = np.asarray(y) assert y.ndim == 1, "y must be one-dimensional." assert 0 < alpha < 1 if X.ndim == 1: # Convert to (n_samples, n_features) shape X = X[..., np.newaxis] # Check for NaN / Inf if remove_na: X, y = rm_na(X, y[..., np.newaxis], paired=True, axis="rows") y = np.squeeze(y) y_gd = np.isfinite(y).all() X_gd = np.isfinite(X).all() assert y_gd, ( "Target (y) contains NaN or Inf. Please remove them " "manually or use remove_na=True." ) assert X_gd, ( "Predictors (X) contain NaN or Inf. Please remove them " "manually or use remove_na=True." ) # Check that X and y have same length assert y.shape[0] == X.shape[0], "X and y must have same number of samples" if not names: names = ["x" + str(i + 1) for i in range(X.shape[1])] if add_intercept: # Add intercept X = np.column_stack((np.ones(X.shape[0]), X)) names.insert(0, "Intercept") # FINAL CHECKS BEFORE RUNNING LEAST SQUARES REGRESSION # 1. Let's remove column(s) with only zero, otherwise the regression fails n_nonzero = np.count_nonzero(X, axis=0) idx_zero = np.flatnonzero(n_nonzero == 0) # Find columns that are only 0 if len(idx_zero): X = np.delete(X, idx_zero, 1) names = np.delete(names, idx_zero) # 2. We also want to make sure that there is no more than one constant # column (= intercept), otherwise the regression fails # This is equivalent, but much faster, to pd.DataFrame(X).nunique() idx_unique = np.where(np.all(X == X[0, :], axis=0))[0] if len(idx_unique) > 1: # We remove all but the first "Intercept" column. X = np.delete(X, idx_unique[1:], 1) names = np.delete(names, idx_unique[1:]) # Is there a constant in our predictor matrix? Useful for dof and R^2. constant = 1 if len(idx_unique) > 0 else 0 # 3. Finally, we want to remove duplicate columns if X.shape[1] > 1: idx_duplicate = [] for pair in itertools.combinations(range(X.shape[1]), 2): if np.array_equal(X[:, pair[0]], X[:, pair[1]]): idx_duplicate.append(pair[1]) if len(idx_duplicate): X = np.delete(X, idx_duplicate, 1) names = np.delete(names, idx_duplicate) # 4. Check that we have enough samples / features n, p = X.shape[0], X.shape[1] assert n >= 3, "At least three valid samples are required in X." assert p >= 1, "X must have at least one valid column." # 5. Handle weights if weights is not None: if relimp: raise ValueError("relimp = True is not supported when using " "weights.") w = np.asarray(weights) assert w.ndim == 1, "weights must be a 1D array." assert w.size == n, "weights must be of shape n_samples." assert not np.isnan(w).any(), "Missing weights are not accepted." assert not (w < 0).any(), "Negative weights are not accepted." # Do not count weights == 0 in dof # This gives similar results as R lm() but different from statsmodels n = np.count_nonzero(w) # Rescale (whitening) wts = np.diag(np.sqrt(w)) Xw = wts @ X yw = wts @ y else: # Set all weights to one, [1, 1, 1, ...] w = np.ones(n) Xw = X yw = y # FIT (WEIGHTED) LEAST SQUARES REGRESSION coef, ss_res, rank, _ = lstsq(Xw, yw, cond=None) ss_res = ss_res[0] if ss_res.shape == (1,) else ss_res if coef_only: return coef calc_ss_res = False if rank < Xw.shape[1]: # in this case, ss_res is of shape (0,), i.e., an empty array warnings.warn( "Design matrix supplied with `X` parameter is rank " f"deficient (rank {rank} with {Xw.shape[1]} columns). " "That means that one or more of the columns in `X` " "are a linear combination of one of more of the " "other columns." ) calc_ss_res = True # Degrees of freedom df_model = rank - constant df_resid = n - rank # Calculate predicted values and (weighted) residuals pred = Xw @ coef resid = yw - pred if calc_ss_res: # In case we did not get ss_res from lstsq due to rank deficiency ss_res = (resid**2).sum() # Calculate total (weighted) sums of squares and R^2 ss_tot = yw @ yw ss_wtot = np.sum(w * (y - np.average(y, weights=w)) ** 2) if constant: r2 = 1 - ss_res / ss_wtot else: r2 = 1 - ss_res / ss_tot adj_r2 = 1 - (1 - r2) * (n - constant) / df_resid # Compute mean squared error, variance and SE mse = ss_res / df_resid beta_var = mse * (np.linalg.pinv(Xw.T @ Xw).diagonal()) beta_se = np.sqrt(beta_var) # Compute T and p-values T = coef / beta_se pval = 2 * t.sf(np.fabs(T), df_resid) # Compute confidence intervals crit = t.ppf(1 - alpha / 2, df_resid) marg_error = crit * beta_se ll = coef - marg_error ul = coef + marg_error # Rename CI ll_name = "CI[%.1f%%]" % (100 * alpha / 2) ul_name = "CI[%.1f%%]" % (100 * (1 - alpha / 2)) # Create dict stats = { "names": names, "coef": coef, "se": beta_se, "T": T, "pval": pval, "r2": r2, "adj_r2": adj_r2, ll_name: ll, ul_name: ul, } # Relative importance if relimp: data = pd.concat( [pd.DataFrame(y, columns=["y"]), pd.DataFrame(X, columns=names)], sort=False, axis=1 ) if "Intercept" in names: # Intercept is the first column reli = _relimp(data.drop(columns=["Intercept"]).cov(numeric_only=True)) reli["names"] = ["Intercept"] + reli["names"] reli["relimp"] = np.insert(reli["relimp"], 0, np.nan) reli["relimp_perc"] = np.insert(reli["relimp_perc"], 0, np.nan) else: reli = _relimp(data.cov(numeric_only=True)) stats.update(reli) if as_dataframe: stats = _postprocess_dataframe(pd.DataFrame(stats)) stats.df_model_ = df_model stats.df_resid_ = df_resid stats.residuals_ = 0 # Trick to avoid Pandas warning stats.residuals_ = resid # Residuals is a hidden attribute else: stats["df_model"] = df_model stats["df_resid"] = df_resid stats["residuals"] = resid stats["X"] = X stats["y"] = y stats["pred"] = pred if weights is not None: stats["yw"] = yw stats["Xw"] = Xw return stats def _relimp(S): """Relative importance of predictors in multiple regression. This is an internal function. This function should only be used with a low number of predictors. Indeed, the computation time roughly doubles for each additional predictor. Parameters ---------- S : pd.DataFrame Covariance matrix. The target variable MUST be the FIRST column, followed by the predictors (excluding the intercept). """ assert isinstance(S, pd.DataFrame) cols = S.columns.tolist() # Define indices of columns: .iloc is faster than .loc predictors = cols[1:] npred = len(predictors) target_int = 0 predictors_int = np.arange(1, npred + 1) # Calculate total sum of squares and beta coefficients # Note that the R^2 that we calculate below is always the R^2 of the model # INCLUDING the intercept! ss_tot = S.iat[target_int, target_int] betas = ( np.linalg.pinv(S.iloc[predictors_int, predictors_int]) @ S.iloc[predictors_int, target_int] ) r2_full = betas @ S.iloc[target_int, predictors_int] / ss_tot # Pre-computed SSreg dictionnary ss_reg_precomp = {} # Start looping over predictors all_preds = [] for pred in predictors_int: loo = np.setdiff1d(predictors_int, pred) r2_seq_mean = [] # Loop over number of predictors for k in np.arange(0, npred - 1): r2_seq = [] # Loop over combinations of predictors for p in itertools.combinations(loo, int(k)): p = list(p) p_with = p + [pred] # To avoid calculating several times the same values # we use a trick here: we save the first calculation # to a dictionnary where the key is the sorted string # (hence the order does not matter) if str(sorted(p)) in ss_reg_precomp.keys(): ss_reg_without = ss_reg_precomp[str(sorted(p))] else: S_without = S.iloc[p, target_int] ss_reg_without = np.linalg.pinv(S.iloc[p, p]) @ S_without @ S_without ss_reg_precomp[str(sorted(p))] = ss_reg_without S_with = S.iloc[p_with, target_int] ss_reg_with = pinvh(S.iloc[p_with, p_with]) @ S_with @ S_with ss_reg_precomp[str(sorted(p_with))] = ss_reg_with # Calculate R^2 r2_diff = (ss_reg_with - ss_reg_without) / ss_tot # Append the difference r2_seq.append(r2_diff) # First averaging r2_seq_mean.append(np.mean(r2_seq)) # When Sk(r) = S S_without = S.iloc[loo, target_int] ss_reg = np.linalg.pinv(S.iloc[loo, loo]) @ S_without @ S_without r2_without = ss_reg / ss_tot r2_seq = r2_full - r2_without r2_seq_mean.append(r2_seq) all_preds.append(np.mean(r2_seq_mean)) stats_relimp = { "names": predictors, "relimp": all_preds, "relimp_perc": all_preds / sum(all_preds) * 100, } return stats_relimp def logistic_regression( X, y, coef_only=False, alpha=0.05, as_dataframe=True, remove_na=False, **kwargs ): """(Multiple) Binary logistic regression. Parameters ---------- X : array_like Predictor(s), of shape *(n_samples, n_features)* or *(n_samples)*. y : array_like Dependent variable, of shape *(n_samples)*. ``y`` must be binary, i.e. only contains 0 or 1. Multinomial logistic regression is not supported. coef_only : bool If True, return only the regression coefficients. alpha : float Alpha value used for the confidence intervals. :math:`\\text{CI} = [\\alpha / 2 ; 1 - \\alpha / 2]` as_dataframe : bool If True, returns a pandas DataFrame. If False, returns a dictionnary. remove_na : bool If True, apply a listwise deletion of missing values (i.e. the entire row is removed). Default is False, which will raise an error if missing values are present in either the predictor(s) or dependent variable. **kwargs : optional Optional arguments passed to :py:class:`sklearn.linear_model.LogisticRegression` (see Notes). Returns ------- stats : :py:class:`pandas.DataFrame` or dict Logistic regression summary: * ``'names'``: name of variable(s) in the model (e.g. x1, x2...) * ``'coef'``: regression coefficients (log-odds) * ``'se'``: standard error * ``'z'``: z-scores * ``'pval'``: two-tailed p-values * ``'CI[2.5%]'``: lower confidence interval * ``'CI[97.5%]'``: upper confidence interval See also -------- linear_regression Notes ----- .. caution:: This function is a wrapper around the :py:class:`sklearn.linear_model.LogisticRegression` class. However, Pingouin internally disables the L2 regularization and changes the default solver to 'newton-cg' to obtain results that are similar to R and statsmodels. Logistic regression assumes that the log-odds (the logarithm of the odds) for the value labeled "1" in the response variable is a linear combination of the predictor variables. The log-odds are given by the `logit `_ function, which map a probability :math:`p` of the response variable being "1" from :math:`[0, 1)` to :math:`(-\\infty, +\\infty)`. .. math:: \\text{logit}(p) = \\ln \\frac{p}{1 - p} = \\beta_0 + \\beta X The odds of the response variable being "1" can be obtained by exponentiating the log-odds: .. math:: \\frac{p}{1 - p} = e^{\\beta_0 + \\beta X} and the probability of the response variable being "1" is given by the `logistic function `_: .. math:: p = \\frac{1}{1 + e^{-(\\beta_0 + \\beta X})} The first coefficient is always the constant term (intercept) of the model. Pingouin will automatically add the intercept to your predictor(s) matrix, therefore, :math:`X` should not include a constant term. Pingouin will remove any constant term (e.g column with only one unique value), or duplicate columns from :math:`X`. The calculation of the p-values and confidence interval is adapted from a `code by Rob Speare `_. Results have been compared against statsmodels, R, and JASP. Examples -------- 1. Simple binary logistic regression. In this first example, we'll use the `penguins dataset `_ to see how well we can predict the sex of penguins based on their bodies mass. >>> import numpy as np >>> import pandas as pd >>> import pingouin as pg >>> df = pg.read_dataset('penguins') >>> # Let's first convert the target variable from string to boolean: >>> df['male'] = (df['sex'] == 'male').astype(int) # male: 1, female: 0 >>> # Since there are missing values in our outcome variable, we need to >>> # set `remove_na=True` otherwise regression will fail. >>> lom = pg.logistic_regression(df['body_mass_g'], df['male'], ... remove_na=True) >>> lom.round(2) names coef se z pval CI[2.5%] CI[97.5%] 0 Intercept -5.16 0.71 -7.24 0.0 -6.56 -3.77 1 body_mass_g 0.00 0.00 7.24 0.0 0.00 0.00 Body mass is a significant predictor of sex (p<0.001). Here, it could be useful to rescale our predictor variable from *g* to *kg* (e.g divide by 1000) in order to get more intuitive coefficients and confidence intervals: >>> df['body_mass_kg'] = df['body_mass_g'] / 1000 >>> lom = pg.logistic_regression(df['body_mass_kg'], df['male'], ... remove_na=True) >>> lom.round(2) names coef se z pval CI[2.5%] CI[97.5%] 0 Intercept -5.16 0.71 -7.24 0.0 -6.56 -3.77 1 body_mass_kg 1.23 0.17 7.24 0.0 0.89 1.56 2. Multiple binary logistic regression We'll now add the species as a categorical predictor in our model. To do so, we first need to dummy-code our categorical variable, dropping the first level of our categorical variable (species = Adelie) which will be used as the reference level: >>> df = pd.get_dummies(df, columns=['species'], dtype=float, drop_first=True) >>> X = df[['body_mass_kg', 'species_Chinstrap', 'species_Gentoo']] >>> y = df['male'] >>> lom = pg.logistic_regression(X, y, remove_na=True) >>> lom.round(2) names coef se z pval CI[2.5%] CI[97.5%] 0 Intercept -26.24 2.84 -9.24 0.00 -31.81 -20.67 1 body_mass_kg 7.10 0.77 9.23 0.00 5.59 8.61 2 species_Chinstrap -0.13 0.42 -0.31 0.75 -0.96 0.69 3 species_Gentoo -9.72 1.12 -8.65 0.00 -11.92 -7.52 3. Using NumPy aray and returning only the coefficients >>> pg.logistic_regression(X.to_numpy(), y.to_numpy(), coef_only=True, ... remove_na=True) array([-26.23906892, 7.09826571, -0.13180626, -9.71718529]) 4. Passing custom parameters to sklearn >>> lom = pg.logistic_regression(X, y, solver='sag', max_iter=10000, ... random_state=42, remove_na=True) >>> print(lom['coef'].to_numpy()) [-25.98248153 7.02881472 -0.13119779 -9.62247569] **How to interpret the log-odds coefficients?** We'll use the `Wikipedia example `_ of the probability of passing an exam versus the hours of study: *A group of 20 students spends between 0 and 6 hours studying for an exam. How does the number of hours spent studying affect the probability of the student passing the exam?* >>> # First, let's create the dataframe >>> Hours = [0.50, 0.75, 1.00, 1.25, 1.50, 1.75, 1.75, 2.00, 2.25, 2.50, ... 2.75, 3.00, 3.25, 3.50, 4.00, 4.25, 4.50, 4.75, 5.00, 5.50] >>> Pass = [0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1] >>> df = pd.DataFrame({'HoursStudy': Hours, 'PassExam': Pass}) >>> # And then run the logistic regression >>> lr = pg.logistic_regression(df['HoursStudy'], df['PassExam']).round(3) >>> lr names coef se z pval CI[2.5%] CI[97.5%] 0 Intercept -4.078 1.761 -2.316 0.021 -7.529 -0.626 1 HoursStudy 1.505 0.629 2.393 0.017 0.272 2.737 The ``Intercept`` coefficient (-4.078) is the log-odds of ``PassExam=1`` when ``HoursStudy=0``. The odds ratio can be obtained by exponentiating the log-odds: >>> np.exp(-4.078) 0.016941314421496552 i.e. :math:`0.017:1`. Conversely the odds of failing the exam are :math:`(1/0.017) \\approx 59:1`. The probability can then be obtained with the following equation .. math:: p = \\frac{1}{1 + e^{-(-4.078 + 0 * 1.505)}} >>> 1 / (1 + np.exp(-(-4.078))) 0.016659087580814722 The ``HoursStudy`` coefficient (1.505) means that for each additional hour of study, the log-odds of passing the exam increase by 1.505, and the odds are multipled by :math:`e^{1.505} \\approx 4.50`. For example, a student who studies 2 hours has a probability of passing the exam of 25%: >>> 1 / (1 + np.exp(-(-4.078 + 2 * 1.505))) 0.2557836148964987 The table below shows the probability of passing the exam for several values of ``HoursStudy``: +----------------+----------+----------------+------------------+ | Hours of Study | Log-odds | Odds | Probability | +================+==========+================+==================+ | 0 | −4.08 | 0.017 ≈ 1:59 | 0.017 | +----------------+----------+----------------+------------------+ | 1 | −2.57 | 0.076 ≈ 1:13 | 0.07 | +----------------+----------+----------------+------------------+ | 2 | −1.07 | 0.34 ≈ 1:3 | 0.26 | +----------------+----------+----------------+------------------+ | 3 | 0.44 | 1.55 | 0.61 | +----------------+----------+----------------+------------------+ | 4 | 1.94 | 6.96 | 0.87 | +----------------+----------+----------------+------------------+ | 5 | 3.45 | 31.4 | 0.97 | +----------------+----------+----------------+------------------+ | 6 | 4.96 | 141.4 | 0.99 | +----------------+----------+----------------+------------------+ """ # Check that sklearn is installed from pingouin.utils import _is_sklearn_installed _is_sklearn_installed(raise_error=True) from sklearn.linear_model import LogisticRegression # Extract names if X is a Dataframe or Series if isinstance(X, pd.DataFrame): names = X.keys().tolist() elif isinstance(X, pd.Series): names = [X.name] else: names = [] # Convert to numpy array X = np.asarray(X) y = np.asarray(y) assert y.ndim == 1, "y must be one-dimensional." assert 0 < alpha < 1, "alpha must be between 0 and 1." # Add axis if only one-dimensional array if X.ndim == 1: X = X[..., np.newaxis] # Check for NaN / Inf if remove_na: X, y = rm_na(X, y[..., np.newaxis], paired=True, axis="rows") y = np.squeeze(y) y_gd = np.isfinite(y).all() X_gd = np.isfinite(X).all() assert y_gd, ( "Target (y) contains NaN or Inf. Please remove them " "manually or use remove_na=True." ) assert X_gd, ( "Predictors (X) contain NaN or Inf. Please remove them " "manually or use remove_na=True." ) # Check that X and y have same length assert y.shape[0] == X.shape[0], "X and y must have same number of samples" # Check that y is binary if np.unique(y).size != 2: raise ValueError("Dependent variable must be binary.") if not names: names = ["x" + str(i + 1) for i in range(X.shape[1])] # We also want to make sure that there is no column # with only one unique value, otherwise the regression fails # This is equivalent, but much faster, to pd.DataFrame(X).nunique() idx_unique = np.where(np.all(X == X[0, :], axis=0))[0] if len(idx_unique): X = np.delete(X, idx_unique, 1) names = np.delete(names, idx_unique).tolist() # Finally, we want to remove duplicate columns if X.shape[1] > 1: idx_duplicate = [] for pair in itertools.combinations(range(X.shape[1]), 2): if np.array_equal(X[:, pair[0]], X[:, pair[1]]): idx_duplicate.append(pair[1]) if len(idx_duplicate): X = np.delete(X, idx_duplicate, 1) names = np.delete(names, idx_duplicate).tolist() # Initialize and fit if "solver" not in kwargs: # https://stats.stackexchange.com/a/204324/253579 # Updated in Pingouin > 0.3.6 to be consistent with R kwargs["solver"] = "newton-cg" if "penalty" not in kwargs: kwargs["penalty"] = "none" lom = LogisticRegression(**kwargs) lom.fit(X, y) if lom.get_params()["fit_intercept"]: names.insert(0, "Intercept") X_design = np.column_stack((np.ones(X.shape[0]), X)) coef = np.append(lom.intercept_, lom.coef_) else: coef = lom.coef_ X_design = X if coef_only: return coef # Fisher Information Matrix n, p = X_design.shape denom = 2 * (1 + np.cosh(lom.decision_function(X))) denom = np.tile(denom, (p, 1)).T fim = (X_design / denom).T @ X_design crao = np.linalg.pinv(fim) # Standard error and Z-scores se = np.sqrt(np.diag(crao)) z_scores = coef / se # Two-tailed p-values pval = 2 * norm.sf(np.fabs(z_scores)) # Wald Confidence intervals # In R: this is equivalent to confint.default(model) # Note that confint(model) will however return the profile CI crit = norm.ppf(1 - alpha / 2) ll = coef - crit * se ul = coef + crit * se # Rename CI ll_name = "CI[%.1f%%]" % (100 * alpha / 2) ul_name = "CI[%.1f%%]" % (100 * (1 - alpha / 2)) # Create dict stats = { "names": names, "coef": coef, "se": se, "z": z_scores, "pval": pval, ll_name: ll, ul_name: ul, } if as_dataframe: return _postprocess_dataframe(pd.DataFrame(stats)) else: return stats def _point_estimate(X_val, XM_val, M_val, y_val, idx, n_mediator, mtype="linear", **logreg_kwargs): """Point estimate of indirect effect based on bootstrap sample.""" # Mediator(s) model (M(j) ~ X + covar) beta_m = [] for j in range(n_mediator): if mtype == "linear": beta_m.append(linear_regression(X_val[idx], M_val[idx, j], coef_only=True)[1]) else: beta_m.append( logistic_regression(X_val[idx], M_val[idx, j], coef_only=True, **logreg_kwargs)[1] ) # Full model (Y ~ X + M + covar) beta_y = linear_regression(XM_val[idx], y_val[idx], coef_only=True)[2 : (2 + n_mediator)] # Point estimate return beta_m * beta_y def _bias_corrected_ci(bootdist, sample_point, alpha=0.05): """Bias-corrected confidence intervals Parameters ---------- bootdist : array-like Array with bootstrap estimates for each sample. sample_point : float Point estimate based on full sample. alpha : float Alpha for confidence interval. Returns ------- CI : 1d array-like Lower and upper bias-corrected confidence interval estimates. Notes ----- This is what's used in the "cper" method implemented in :py:func:`pingouin.compute_bootci`. This differs from the bias-corrected and accelerated method (BCa, default in Matlab and SciPy) because it does not correct for skewness. Indeed, the acceleration parameter, a, is proportional to the skewness of the bootstrap distribution. The bias-correction parameter, z0, is related to the proportion of bootstrap estimates that are less than the observed statistic. """ # Bias of bootstrap estimates # In Matlab bootci they also count when bootdist == sample_point # >>> z0 = norminv(mean(bstat < stat,1) + mean(bstat == stat,1)/2); z0 = norm.ppf(np.mean(bootdist < sample_point)) # Adjusted intervals adjusted_ll = norm.cdf(2 * z0 + norm.ppf(alpha / 2)) * 100 adjusted_ul = norm.cdf(2 * z0 + norm.ppf(1 - alpha / 2)) * 100 # Adjusted percentiles ci = np.percentile(bootdist, q=[adjusted_ll, adjusted_ul]) return ci def _pval_from_bootci(boot, estimate): """Compute p-value from bootstrap distribution. Similar to the pval function in the R package mediation. Note that this is less accurate than a permutation test because the bootstrap distribution is not conditioned on a true null hypothesis. """ if estimate == 0: out = 1 else: out = 2 * min(sum(boot > 0), sum(boot < 0)) / len(boot) return min(out, 1) @pf.register_dataframe_method def mediation_analysis( data=None, x=None, m=None, y=None, covar=None, alpha=0.05, n_boot=500, seed=None, return_dist=False, logreg_kwargs=None, ): """Mediation analysis using a bias-correct non-parametric bootstrap method. Parameters ---------- data : :py:class:`pandas.DataFrame` Dataframe. x : str Column name in data containing the predictor variable. The predictor variable must be continuous. m : str or list of str Column name(s) in data containing the mediator variable(s). The mediator(s) can be continuous or binary (e.g. 0 or 1). This function supports multiple parallel mediators. y : str Column name in data containing the outcome variable. The outcome variable must be continuous. covar : None, str, or list Covariate(s). If not None, the specified covariate(s) will be included in all regressions. alpha : float Significance threshold. Used to determine the confidence interval, :math:`\\text{CI} = [\\alpha / 2 ; 1 - \\alpha / 2]`. n_boot : int Number of bootstrap iterations for confidence intervals and p-values estimation. The greater, the slower. seed : int or None Random state seed. logreg_kwargs : dict or None Dictionary with optional arguments passed to :py:func:`pingouin.logistic_regression` return_dist : bool If True, the function also returns the indirect bootstrapped beta samples (size = n_boot). Can be plotted for instance using :py:func:`seaborn.distplot()` or :py:func:`seaborn.kdeplot()` functions. Returns ------- stats : :py:class:`pandas.DataFrame` Mediation summary: * ``'path'``: regression model * ``'coef'``: regression estimates * ``'se'``: standard error * ``'CI[2.5%]'``: lower confidence interval * ``'CI[97.5%]'``: upper confidence interval * ``'pval'``: two-sided p-values * ``'sig'``: statistical significance See also -------- linear_regression, logistic_regression Notes ----- Mediation analysis [1]_ is a *"statistical procedure to test whether the effect of an independent variable X on a dependent variable Y (i.e., X → Y) is at least partly explained by a chain of effects of the independent variable on an intervening mediator variable M and of the intervening variable on the dependent variable (i.e., X → M → Y)"* [2]_. The **indirect effect** (also referred to as average causal mediation effect or ACME) of X on Y through mediator M quantifies the estimated difference in Y resulting from a one-unit change in X through a sequence of causal steps in which X affects M, which in turn affects Y. It is considered significant if the specified confidence interval does not include 0. The path 'X --> Y' is the sum of both the indirect and direct effect. It is sometimes referred to as total effect. A linear regression is used if the mediator variable is continuous and a logistic regression if the mediator variable is dichotomous (binary). Multiple parallel mediators are also supported. This function will only work well if the outcome variable is continuous. It does not support binary or ordinal outcome variable. For more advanced mediation models, please refer to the `lavaan `_ or `mediation `_ R packages, or the `PROCESS macro `_ for SPSS. The two-sided p-value of the indirect effect is computed using the bootstrap distribution, as in the mediation R package. However, the p-value should be interpreted with caution since it is not constructed conditioned on a true null hypothesis [3]_ and varies depending on the number of bootstrap samples and the random seed. Note that rows with missing values are automatically removed. Results have been tested against the R mediation package and this tutorial https://data.library.virginia.edu/introduction-to-mediation-analysis/ References ---------- .. [1] Baron, R. M. & Kenny, D. A. The moderator–mediator variable distinction in social psychological research: Conceptual, strategic, and statistical considerations. J. Pers. Soc. Psychol. 51, 1173–1182 (1986). .. [2] Fiedler, K., Schott, M. & Meiser, T. What mediation analysis can (not) do. J. Exp. Soc. Psychol. 47, 1231–1236 (2011). .. [3] Hayes, A. F. & Rockwood, N. J. Regression-based statistical mediation and moderation analysis in clinical research: Observations, recommendations, and implementation. Behav. Res. Ther. 98, 39–57 (2017). Code originally adapted from https://github.com/rmill040/pymediation. Examples -------- 1. Simple mediation analysis >>> from pingouin import mediation_analysis, read_dataset >>> df = read_dataset('mediation') >>> mediation_analysis(data=df, x='X', m='M', y='Y', alpha=0.05, ... seed=42) path coef se pval CI[2.5%] CI[97.5%] sig 0 M ~ X 0.561015 0.094480 4.391362e-08 0.373522 0.748509 Yes 1 Y ~ M 0.654173 0.085831 1.612674e-11 0.483844 0.824501 Yes 2 Total 0.396126 0.111160 5.671128e-04 0.175533 0.616719 Yes 3 Direct 0.039604 0.109648 7.187429e-01 -0.178018 0.257226 No 4 Indirect 0.356522 0.083313 0.000000e+00 0.219818 0.537654 Yes 2. Return the indirect bootstrapped beta coefficients >>> stats, dist = mediation_analysis(data=df, x='X', m='M', y='Y', ... return_dist=True) >>> print(dist.shape) (500,) 3. Mediation analysis with a binary mediator variable >>> mediation_analysis(data=df, x='X', m='Mbin', y='Y', seed=42).round(3) path coef se pval CI[2.5%] CI[97.5%] sig 0 Mbin ~ X -0.021 0.116 0.857 -0.248 0.206 No 1 Y ~ Mbin -0.135 0.412 0.743 -0.952 0.682 No 2 Total 0.396 0.111 0.001 0.176 0.617 Yes 3 Direct 0.396 0.112 0.001 0.174 0.617 Yes 4 Indirect 0.002 0.050 0.960 -0.072 0.146 No 4. Mediation analysis with covariates >>> mediation_analysis(data=df, x='X', m='M', y='Y', ... covar=['Mbin', 'Ybin'], seed=42).round(3) path coef se pval CI[2.5%] CI[97.5%] sig 0 M ~ X 0.559 0.097 0.000 0.367 0.752 Yes 1 Y ~ M 0.666 0.086 0.000 0.495 0.837 Yes 2 Total 0.420 0.113 0.000 0.196 0.645 Yes 3 Direct 0.064 0.110 0.561 -0.155 0.284 No 4 Indirect 0.356 0.086 0.000 0.209 0.553 Yes 5. Mediation analysis with multiple parallel mediators >>> mediation_analysis(data=df, x='X', m=['M', 'Mbin'], y='Y', ... seed=42).round(3) path coef se pval CI[2.5%] CI[97.5%] sig 0 M ~ X 0.561 0.094 0.000 0.374 0.749 Yes 1 Mbin ~ X -0.005 0.029 0.859 -0.063 0.052 No 2 Y ~ M 0.654 0.086 0.000 0.482 0.825 Yes 3 Y ~ Mbin -0.064 0.328 0.846 -0.715 0.587 No 4 Total 0.396 0.111 0.001 0.176 0.617 Yes 5 Direct 0.040 0.110 0.721 -0.179 0.258 No 6 Indirect M 0.356 0.085 0.000 0.215 0.538 Yes 7 Indirect Mbin 0.000 0.010 0.952 -0.017 0.025 No """ # Sanity check assert isinstance(x, (str, int)), "y must be a string or int." assert isinstance(y, (str, int)), "y must be a string or int." assert isinstance(m, (list, str, int)), "Mediator(s) must be a list, string or int." assert isinstance(covar, (type(None), str, list, int)) if isinstance(m, (str, int)): m = [m] n_mediator = len(m) assert isinstance(data, pd.DataFrame), "Data must be a DataFrame." # Check for duplicates assert n_mediator == len(set(m)), "Cannot have duplicates mediators." if isinstance(covar, (str, int)): covar = [covar] if isinstance(covar, list): assert len(covar) == len(set(covar)), "Cannot have duplicates covar." assert set(m).isdisjoint(covar), "Mediator cannot be in covar." # Check that columns are in dataframe columns = _fl([x, m, y, covar]) keys = data.columns assert all([c in keys for c in columns]), "Column(s) are not in DataFrame." # Check that columns are numeric err_msg = "Columns must be numeric or boolean." assert all([data[c].dtype.kind in "bfiu" for c in columns]), err_msg # Drop rows with NAN Values data = data[columns].dropna() n = data.shape[0] assert n > 5, "DataFrame must have at least 5 samples (rows)." # Check if mediator is binary mtype = "logistic" if all(data[m].nunique() == 2) else "linear" # Check if a dict with kwargs for logistic_regression has been passed logreg_kwargs = {} if logreg_kwargs is None else logreg_kwargs # Name of CI ll_name = "CI[%.1f%%]" % (100 * alpha / 2) ul_name = "CI[%.1f%%]" % (100 * (1 - alpha / 2)) # Compute regressions cols = ["names", "coef", "se", "pval", ll_name, ul_name] # For speed, we pass np.array instead of pandas DataFrame X_val = data[_fl([x, covar])].to_numpy() # X + covar as predictors XM_val = data[_fl([x, m, covar])].to_numpy() # X + M + covar as predictors M_val = data[m].to_numpy() # M as target (no covariates) y_val = data[y].to_numpy() # y as target (no covariates) # For max precision, make sure rounding is disabled old_options = options.copy() options["round"] = None # M(j) ~ X + covar sxm = {} for idx, j in enumerate(m): if mtype == "linear": sxm[j] = linear_regression(X_val, M_val[:, idx], alpha=alpha).loc[[1], cols] else: sxm[j] = logistic_regression(X_val, M_val[:, idx], alpha=alpha, **logreg_kwargs).loc[ [1], cols ] sxm[j].at[1, "names"] = "%s ~ X" % j sxm = pd.concat(sxm, ignore_index=True) # Y ~ M + covar smy = linear_regression(data[_fl([m, covar])], y_val, alpha=alpha).loc[1:n_mediator, cols] # Average Total Effects (Y ~ X + covar) sxy = linear_regression(X_val, y_val, alpha=alpha).loc[[1], cols] # Average Direct Effects (Y ~ X + M + covar) direct = linear_regression(XM_val, y_val, alpha=alpha).loc[[1], cols] # Rename paths smy["names"] = smy["names"].apply(lambda x: "Y ~ %s" % x) direct.at[1, "names"] = "Direct" sxy.at[1, "names"] = "Total" # Concatenate and create sig column stats = pd.concat((sxm, smy, sxy, direct), ignore_index=True) stats["sig"] = np.where(stats["pval"] < alpha, "Yes", "No") # Bootstrap confidence intervals rng = np.random.RandomState(seed) idx = rng.choice(np.arange(n), replace=True, size=(n_boot, n)) ab_estimates = np.zeros(shape=(n_boot, n_mediator)) for i in range(n_boot): ab_estimates[i, :] = _point_estimate( X_val, XM_val, M_val, y_val, idx[i, :], n_mediator, mtype, **logreg_kwargs ) ab = _point_estimate( X_val, XM_val, M_val, y_val, np.arange(n), n_mediator, mtype, **logreg_kwargs ) indirect = { "names": m, "coef": ab, "se": ab_estimates.std(ddof=1, axis=0), "pval": [], ll_name: [], ul_name: [], "sig": [], } for j in range(n_mediator): ci_j = _bias_corrected_ci(ab_estimates[:, j], indirect["coef"][j], alpha=alpha) indirect[ll_name].append(min(ci_j)) indirect[ul_name].append(max(ci_j)) # Bootstrapped p-value of indirect effect # Note that this is less accurate than a permutation test because the # bootstrap distribution is not conditioned on a true null hypothesis. # For more details see Hayes and Rockwood 2017 indirect["pval"].append(_pval_from_bootci(ab_estimates[:, j], indirect["coef"][j])) indirect["sig"].append("Yes" if indirect["pval"][j] < alpha else "No") # Create output dataframe indirect = pd.DataFrame.from_dict(indirect) if n_mediator == 1: indirect["names"] = "Indirect" else: indirect["names"] = indirect["names"].apply(lambda x: "Indirect %s" % x) stats = pd.concat([stats, indirect], axis=0, ignore_index=True, sort=False) stats = stats.rename(columns={"names": "path"}) # Restore options options.update(old_options) if return_dist: return _postprocess_dataframe(stats), np.squeeze(ab_estimates) else: return _postprocess_dataframe(stats)