python中R data.chisq$residuals的等价物是什么？

我有以下数据：

array([[33, 250, 196, 136, 32],
       [55, 293, 190,  71, 13]])

我可以从stats.chi2_或有数据中获得p值

---

有没有类似于R object-data.chisq$残差的东西来获得皮尔逊残差和标准化残差

如果您不介意依赖关系，那么它有一个用于的模块。比如说,

In [2]: import numpy as np                                                                                   

In [3]: import statsmodels.api as sm                                                                         

In [4]: F = np.array([[33, 250, 196, 136, 32], [55, 293, 190,  71, 13]])                                     

In [5]: table = sm.stats.Table(F)                                                                            

In [6]: table.resid_pearson  # Pearson's residuals
Out[6]: 
array([[-1.77162519, -1.61362277, -0.05718356,  2.96508777,  1.89079393],
       [ 1.80687785,  1.64573143,  0.05832142, -3.02408853, -1.92841787]])

In [7]: table.standardized_resids  # Standardized residuals
Out[7]: 
array([[-2.62309082, -3.0471942 , -0.09791681,  4.6295814 ,  2.74991911],
       [ 2.62309082,  3.0471942 ,  0.09791681, -4.6295814 , -2.74991911]])

如果您不想依赖statsmodels，可以使用scipy.stats.chi2_的结果在几行中实现这些计算。下面是一个简短的模块，它定义了这些残差的函数。他们采用chi2_偶然事件返回的观测频率和预期频率。请注意，尽管chi2_列联和以下残差函数适用于n维阵列，但此处实现的STDRE仅适用于2D阵列

from __future__ import division

import numpy as np
from scipy.stats.contingency import margins


def residuals(observed, expected):
    return (observed - expected) / np.sqrt(expected)

def stdres(observed, expected):
    n = observed.sum()
    rsum, csum = margins(observed)
    # With integers, the calculation
    #     csum * rsum * (n - rsum) * (n - csum)
    # might overflow, so convert rsum and csum to floating point.
    rsum = rsum.astype(np.float64)
    csum = csum.astype(np.float64)
    v = csum * rsum * (n - rsum) * (n - csum) / n**3
    return (observed - expected) / np.sqrt(v)

根据您的数据，我们可以：

>>> F = np.array([[33, 250, 196, 136, 32], [55, 293, 190, 71, 13]])

>>> chi2, p, dof, expected = chi2_contingency(F)

>>> residuals(F, expected)
array([[-1.77162519, -1.61362277, -0.05718356,  2.96508777,  1.89079393],
       [ 1.80687785,  1.64573143,  0.05832142, -3.02408853, -1.92841787]])

>>> stdres(F, expected)
array([[-2.62309082, -3.0471942 , -0.09791681,  4.6295814 ,  2.74991911],
       [ 2.62309082,  3.0471942 ,  0.09791681, -4.6295814 , -2.74991911]])

以下是R中的计算值，以供比较：

> F <- as.table(rbind(c(33, 250, 196, 136, 32), c(55, 293, 190, 71, 13)))

> result <- chisq.test(F)

> result$residuals
            A           B           C           D           E
A -1.77162519 -1.61362277 -0.05718356  2.96508777  1.89079393
B  1.80687785  1.64573143  0.05832142 -3.02408853 -1.92841787

> result$stdres
            A           B           C           D           E
A -2.62309082 -3.04719420 -0.09791681  4.62958140  2.74991911
B  2.62309082  3.04719420  0.09791681 -4.62958140 -2.74991911

---

如果您不介意依赖关系，那么它有一个用于的模块。比如说,

In [2]: import numpy as np                                                                                   

In [3]: import statsmodels.api as sm                                                                         

In [4]: F = np.array([[33, 250, 196, 136, 32], [55, 293, 190,  71, 13]])                                     

In [5]: table = sm.stats.Table(F)                                                                            

In [6]: table.resid_pearson  # Pearson's residuals
Out[6]: 
array([[-1.77162519, -1.61362277, -0.05718356,  2.96508777,  1.89079393],
       [ 1.80687785,  1.64573143,  0.05832142, -3.02408853, -1.92841787]])

In [7]: table.standardized_resids  # Standardized residuals
Out[7]: 
array([[-2.62309082, -3.0471942 , -0.09791681,  4.6295814 ,  2.74991911],
       [ 2.62309082,  3.0471942 ,  0.09791681, -4.6295814 , -2.74991911]])

如果您不想依赖statsmodels，可以使用scipy.stats.chi2_的结果在几行中实现这些计算。下面是一个简短的模块，它定义了这些残差的函数。他们采用chi2_偶然事件返回的观测频率和预期频率。请注意，尽管chi2_列联和以下残差函数适用于n维阵列，但此处实现的STDRE仅适用于2D阵列

from __future__ import division

import numpy as np
from scipy.stats.contingency import margins


def residuals(observed, expected):
    return (observed - expected) / np.sqrt(expected)

def stdres(observed, expected):
    n = observed.sum()
    rsum, csum = margins(observed)
    # With integers, the calculation
    #     csum * rsum * (n - rsum) * (n - csum)
    # might overflow, so convert rsum and csum to floating point.
    rsum = rsum.astype(np.float64)
    csum = csum.astype(np.float64)
    v = csum * rsum * (n - rsum) * (n - csum) / n**3
    return (observed - expected) / np.sqrt(v)

根据您的数据，我们可以：

>>> F = np.array([[33, 250, 196, 136, 32], [55, 293, 190, 71, 13]])

>>> chi2, p, dof, expected = chi2_contingency(F)

>>> residuals(F, expected)
array([[-1.77162519, -1.61362277, -0.05718356,  2.96508777,  1.89079393],
       [ 1.80687785,  1.64573143,  0.05832142, -3.02408853, -1.92841787]])

>>> stdres(F, expected)
array([[-2.62309082, -3.0471942 , -0.09791681,  4.6295814 ,  2.74991911],
       [ 2.62309082,  3.0471942 ,  0.09791681, -4.6295814 , -2.74991911]])

以下是R中的计算值，以供比较：

> F <- as.table(rbind(c(33, 250, 196, 136, 32), c(55, 293, 190, 71, 13)))

> result <- chisq.test(F)

> result$residuals
            A           B           C           D           E
A -1.77162519 -1.61362277 -0.05718356  2.96508777  1.89079393
B  1.80687785  1.64573143  0.05832142 -3.02408853 -1.92841787

> result$stdres
            A           B           C           D           E
A -2.62309082 -3.04719420 -0.09791681  4.62958140  2.74991911
B  2.62309082  3.04719420  0.09791681 -4.62958140 -2.74991911

---

谢谢@Warren！我以为有一些我不知道的内置功能。谢谢@Warren！我以为有一些我不知道的内置功能。