scipy.stats.pointbiserialr¶

scipy.stats.
pointbiserialr
(x, y)[source]¶ Calculate a point biserial correlation coefficient and its pvalue.
The point biserial correlation is used to measure the relationship between a binary variable, x, and a continuous variable, y. Like other correlation coefficients, this one varies between 1 and +1 with 0 implying no correlation. Correlations of 1 or +1 imply a determinative relationship.
This function uses a shortcut formula but produces the same result as
pearsonr
.Parameters:  x : array_like of bools
Input array.
 y : array_like
Input array.
Returns:  correlation : float
R value
 pvalue : float
2tailed pvalue
Notes
pointbiserialr
uses a ttest withn1
degrees of freedom. It is equivalent to pearsonr.The value of the pointbiserial correlation can be calculated from:
\[r_{pb} = \frac{\overline{Y_{1}}  \overline{Y_{0}}}{s_{y}}\sqrt{\frac{N_{1} N_{2}}{N (N  1))}}\]Where \(Y_{0}\) and \(Y_{1}\) are means of the metric observations coded 0 and 1 respectively; \(N_{0}\) and \(N_{1}\) are number of observations coded 0 and 1 respectively; \(N\) is the total number of observations and \(s_{y}\) is the standard deviation of all the metric observations.
A value of \(r_{pb}\) that is significantly different from zero is completely equivalent to a significant difference in means between the two groups. Thus, an independent groups t Test with \(N2\) degrees of freedom may be used to test whether \(r_{pb}\) is nonzero. The relation between the tstatistic for comparing two independent groups and \(r_{pb}\) is given by:
\[t = \sqrt{N  2}\frac{r_{pb}}{\sqrt{1  r^{2}_{pb}}}\]References
[1] J. Lev, “The Point Biserial Coefficient of Correlation”, Ann. Math. Statist., Vol. 20, no.1, pp. 125126, 1949. [2] R.F. Tate, “Correlation Between a Discrete and a Continuous Variable. PointBiserial Correlation.”, Ann. Math. Statist., Vol. 25, np. 3, pp. 603607, 1954. [3] http://onlinelibrary.wiley.com/doi/10.1002/9781118445112.stat06227/full Examples
>>> from scipy import stats >>> a = np.array([0, 0, 0, 1, 1, 1, 1]) >>> b = np.arange(7) >>> stats.pointbiserialr(a, b) (0.8660254037844386, 0.011724811003954652) >>> stats.pearsonr(a, b) (0.86602540378443871, 0.011724811003954626) >>> np.corrcoef(a, b) array([[ 1. , 0.8660254], [ 0.8660254, 1. ]])