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)