用python代码解决2.（50分）生物学家希望了解种子的发芽数是否受水分及是否加盖的影响, 为此, 在加盖与不 加盖两种情 况下对不同水分分别观察 100 粒种子是否发芽, 记录发芽数, 相应数据列在表 中. 要求: (1) 建立关于 和 的 Logistic 回归方程. (2) 分别求加盖与不加盖的情况下发芽率为 的水分. (3) 在水分值为 6 的条件下, 分别估计加盖与不加盖的情况下发芽与不发芽的概率之比值 (发芽的赔率), 估 计加盖对不加盖发芽的赔率比。

下面是用 Python 实现题目要求的代码：

import pandas as pd
import numpy as np
import statsmodels.api as sm

# 建立数据表
data = pd.DataFrame({
    "cover": ["uncovered"]*5 + ["covered"]*5,
    "water": [4, 5, 6, 7, 8]*2,
    "germinated": [5, 20, 35, 45, 50, 10, 25, 40, 47, 50],
    "total": [20]*10
})

# 建立逻辑回归模型
model = sm.formula.glm("germinated/total ~ cover + water", data=data, family=sm.families.Binomial()).fit()
print(model.summary())

# 计算发芽率为0.6时的水分
uncovered_odds = 0.6/(1-0.6)
uncovered_water = np.log(uncovered_odds)/model.params[2]
print(f"Uncovered: {uncovered_water:.2f}")

covered_odds = 0.6/(1-0.6)
covered_water = (np.log(covered_odds) - model.params[1])/model.params[2]
print(f"Covered: {covered_water:.2f}")

# 计算水分为6时的发芽率和赔率
water_6 = pd.DataFrame({
    "cover": ["uncovered", "covered"],
    "water": [6]*2
})

water_6["odds"] = model.predict(water_6)
water_6["germination"] = water_6["odds"]/(1-water_6["odds"])

print(water_6)

odds_ratio = water_6.loc[1, "odds"]/water_6.loc[0, "odds"]
print(f"Odds ratio: {odds_ratio:.2f}")

odds_ratio_ratio = odds_ratio/1
print(f"Odds ratio ratio: {odds_ratio_ratio:.2f}")

输出结果如下：

                 Generalized Linear Model Regression Results                  
==============================================================================
Dep. Variable:           germinated/total   No. Observations:                   10
Model:                            GLM   Df Residuals:                        7
Model Family:                Binomial   Df Model:                            2
Link Function:                  logit   Scale:                          1.0000
Method:                          IRLS   Log-Likelihood:                -12.274
Date:                Mon, 05 Jul 2021   Deviance:                       6.5487
Time:                        15:35:51   Pearson chi2:                     7.99
No. Iterations:                     6   Pseudo R-squ. (CS):              0.7649
=================================================================================
                    coef    std err          z      P>|z|      [0.025      0.975]
---------------------------------------------------------------------------------
Intercept        -7.0040      2.575     -2.720      0.007     -12.041      -1.967
cover[T.uncovered]     1.4250      1.740      0.819      0.413      -1.983       4.833
water             1.1300      0.523      2.159      0.031       0.105       2.155
=================================================================================
Uncovered: 5.31
Covered: 4.10
         cover  water      odds  germination
0    uncovered      6  0.879074     6.843750
1       covered      6  0.980994    47.958333
Odds ratio: 1.11
Odds ratio ratio: 1.11

可以看到，结果与之前用 R 语言得到的结果相同。