Python基本统计分析
常见的统计分析方法
import numpy as np
import scipy.stats as spss
import pandas as pd
鸢尾花数据集
https://github.com/mwaskom/seaborn-data
df = pd.read_csv("iris.csv",index_col="species")
v1 = df.loc["versicolor",:].petal_length.values
v2 = df.loc["virginica",:].petal_length.values
1.组间差异的参数检验
数据是否服从正态分布
符合正态分布（p>0.05）
# Shapiro-Wilk test
stat, p_value = spss.shapiro(v1)
stat, p_value = spss.shapiro(v2)
方差齐性检验
方差齐，即v1和v2的方差没有显著性差异，即p>0.05
# 非参数检验，对于数据的分布没有要求
stat, p_value = spss.levene(v1,v2)
# 要求数据服从正态分布
stat, p_value = spss.bartlett(v1,v2)
两独立样本的 t 检验
stat, p_value = spss.ttest_ind(v1,v2)
非独立样本的 t 检验
配对 Paired Student’s t-test（本例中v1，v2并不是配对样本，这里仅用于演示）
stat, p_value = spss.ttest_rel(v1,v2)
one-way ANOVA
检查是否符合正态分布
df.petal_length.groupby(df.index).apply(spss.shapiro)
# species
# setosa (0.971718966960907, 0.27151283621788025)
# versicolor (0.9741330742835999, 0.3379890024662018)
# virginica (0.9673907160758972, 0.1808987259864807)
# Name: sepal_width, dtype: object
方差齐性检验
p_value > 0.05方差齐
v1 = df.loc["versicolor",:].sepal_width.values
v2 = df.loc["virginica",:].sepal_width.values
v3 = df.loc["setosa",:].sepal_width.values
stat, p_value = spss.bartlett(v1,v2,v3)
单因素方差分析
p_value < 0.05三个物种间的sepal_width有差异
stat, p_value = spss.f_oneway(v1, v2, v3)
也可以使用statsmodels中的函数，结果一致
from statsmodels.formula.api import ols
from statsmodels.stats.anova import anova_lm
df.loc[:,'species'] = df.index
aov_results = anova_lm(ols('sepal_width ~ species', data = df).fit())
aov_results
# df sum_sq mean_sq F PR(>F)
# species 2.0 11.344933 5.672467 49.16004 4.492017e-17
# Residual 147.0 16.962000 0.115388 NaN NaN
两两比较找出哪些组之间存在显著差异
3个物种两两之间的sepal_width都有显著性差异
from statsmodels.stats.multicomp import pairwise_tukeyhsd
tukey = pairwise_tukeyhsd(df.sepal_width, df.index)
print(tukey)
# Multiple Comparison of Means - Tukey HSD, FWER=0.05
# ============================================================
# group1 group2 meandiff p-adj lower upper reject
# ------------------------------------------------------------
# setosa versicolor -0.658 0.0 -0.8189 -0.4971 True
# setosa virginica -0.454 0.0 -0.6149 -0.2931 True
# versicolor virginica 0.204 0.0088 0.0431 0.3649 True
# ------------------------------------------------------------
2.组间差异的非参数检验
两组样本
独立样本秩和检验
stat, p_value = spss.ranksums(v1, v2)
非独立样本秩和检验
stat, p_value = spss.wilcoxon(v1, v2)
多组样本
stat, p_value = spss.kruskal(v1, v2, v3)
3.连续型变量之间的相关性
Pearson’s Correlation Coefficient
v1,v2符合正态分布
r, p_value = spss.pearsonr(v1,v2)
spearman
v1,v2的分布没有特定的要求
r, p_value = spss.spearmanr(v1,v2)
kendalltau
v1,v2的分布没有特定的要求
r, p_value = spss.kendalltau(v1,v2)
多个变量之间的相关性
协方差矩阵
df.cov(numeric_only=True)
# sepal_length sepal_width petal_length petal_width
# sepal_length 0.685694 -0.042434 1.274315 0.516271
# sepal_width -0.042434 0.189979 -0.329656 -0.121639
# petal_length 1.274315 -0.329656 3.116278 1.295609
# petal_width 0.516271 -0.121639 1.295609 0.581006
相关系数矩阵
df.corr(numeric_only=True)
# sepal_length sepal_width petal_length petal_width
# sepal_length 1.000000 -0.117570 0.871754 0.817941
# sepal_width -0.117570 1.000000 -0.428440 -0.366126
# petal_length 0.871754 -0.428440 1.000000 0.962865
# petal_width 0.817941 -0.366126 0.962865 1.000000
3.分类变量
汽车耗油量数据集https://github.com/mwaskom/seaborn-data
mpg = pd.read_csv("mpg.csv")
频数
pd.value_counts(mpg.origin)
# usa 249
# japan 79
# europe 70
# Name: origin, dtype: int64
# 百分比
pd.value_counts(mpg.origin,normalize=True)
# usa 0.625628
# japan 0.198492
# europe 0.175879
# Name: origin, dtype: float64
列联表
两个以上的变量交叉分类的频数分布表
pd.crosstab(mpg.cylinders, mpg.origin)
# origin europe japan usa
# cylinders
# 3 0 4 0
# 4 63 69 72
# 5 3 0 0
# 6 4 6 74
# 8 0 0 103
pd.crosstab(mpg.cylinders, mpg.origin, margins = True)
# origin europe japan usa All
# cylinders
# 3 0 4 0 4
# 4 63 69 72 204
# 5 3 0 0 3
# 6 4 6 74 84
# 8 0 0 103 103
# All 70 79 249 398
每个单元格占总数的比例
pd.crosstab(mpg.cylinders, mpg.origin, normalize = True)
# origin europe japan usa
# cylinders
# 3 0.000000 0.010050 0.000000
# 4 0.158291 0.173367 0.180905
# 5 0.007538 0.000000 0.000000
# 6 0.010050 0.015075 0.185930
# 8 0.000000 0.000000 0.258794
按行求比例
pd.crosstab(mpg.cylinders, mpg.origin, normalize = 0)
# origin europe japan usa
# cylinders
# 3 0.000000 1.000000 0.000000
# 4 0.308824 0.338235 0.352941
# 5 1.000000 0.000000 0.000000
# 6 0.047619 0.071429 0.880952
# 8 0.000000 0.000000 1.000000
按列求比例
pd.crosstab(mpg.cylinders, mpg.origin, normalize = 1)
# origin europe japan usa
# cylinders
# 3 0.000000 0.050633 0.000000
# 4 0.900000 0.873418 0.289157
# 5 0.042857 0.000000 0.000000
# 6 0.057143 0.075949 0.297189
# 8 0.000000 0.000000 0.413655
列联表独立性检验
χ2 独立性检验
在该函数中，参数“correction”用于设置是否进行连续性校正，默认为 True。对于大样本，且频数表中每个单元格的期望频数都比较大（一般要求大于 5），可以不进行连续性校正。
tb = pd.crosstab(mpg.cylinders, mpg.origin)
# χ2 值、 P 值、自由度、期望频数表
chi2, p_value, df, expected = spss.chi2_contingency(tb)
p_value
# 9.800693325588298e-35
expected
# array([[ 0.70351759, 0.79396985, 2.50251256],
# [ 35.87939698, 40.49246231, 127.6281407 ],
# [ 0.52763819, 0.59547739, 1.87688442],
# [ 14.77386935, 16.67336683, 52.55276382],
# [ 18.11557789, 20.44472362, 64.43969849]])
Fisher 精确概率检验
R语言中fisher.test的故事以及示例
Agresti (1990, p. 61f; 2002, p. 91) Fisher's Tea Drinker A British woman claimed to be able to distinguish whether milk or tea was added to the cup first. To test, she was given 8 cups of tea, in four of which milk was added first. The null hypothesis is that there is no association between the true order of pouring and the woman's guess, the alternative that there is a positive association (that the odds ratio is greater than 1).
如果观察总例数 n 小于 40，或者频数表里的某个期望频数很小（小于 1），则需要使用 Fisher 精确概率检验
spss.fisher_exact这个函数的输入只能是2X2的二维列联表,R中的fisher.test输入可以不是2X2列联表。
OR(0,+inf)如果 OR 值大于 1，则说明该因素更容易导致结果事件发生
alternative可以选two-sided（默认，OR可能>1,也可能<1）, less（OR<1）, greater（OR>1）
tea_tasting = pd.DataFrame({"Milk":[3,1],"Tea":[1,3]},index=["Milk", "Tea"])
tea_tasting
# Milk Tea
# Milk 3 1
# Tea 1 3
OR, p_value = spss.fisher_exact(tea_tasting,alternative="greater")
OR, p_value
# (9.0, 0.24285714285714283)
# p > 0.05, association could not be established
配对列联表的Mcnemar 检验
对每个对象分别用两种方法处理
exact：True（样本量小，使用二项分布）；False（样本较大，使用 χ2 分布）
correction：在样本量较大，且不一致的结果总数小于 40 时，需要进行连续性校正
from statsmodels.sandbox.stats.runs import mcnemar
tb = np.array([[11, 12],[2, 33]])
stat, p_value = mcnemar(tb, exact = False, correction = True)
p_value
# 0.016156931261181305
Reference
https://www.heywhale.com/mw/notebook/61e3d3c7ddda3c0017b4658f
https://www.statsmodels.org/stable/generated/statsmodels.sandbox.stats.runs.mcnemar.html
