问用于合并拟合模型的For循环
EN

Stack Overflow用户

提问于 2020-03-31 22:58:31

回答 2查看 40关注 0票数 1

我是R的初学者，所以我想寻求一些帮助。

我正在尝试使用for循环来迭代我估算的拟合模型，以便在合并模型并随后计算模型的R平方时增加一些效率。

# Model with all Trust variables

fits_mod1 <- lm.mids(Trust ~  Tr_Family + Tr_Neighborhood + Tr_Personally
                     + Tr_Initial + Tr_Nationality, data = miceOut3)

# Model with all Trust + Discriminatory attitudes variables

fits_mod2 <- lm.mids(Trust ~  Tr_Family + Tr_Neighborhood + Tr_Personally
                     + Tr_Initial + Tr_Nationality
                     + Racism_neighborhood + Homosexuality, data = miceOut3)

# Model with all Trust + Police variables

fits_mod3 <- lm.mids(Trust ~  Tr_Family + Tr_Neighborhood + Tr_Personally
                     + Tr_Initial + Tr_Nationality 
                     + Confidence_police + Interfere_police, data = miceOut3)

# Model with all Trust + Happiness variables

fits_mod4 <- lm.mids(Trust ~ Tr_Family + Tr_Neighborhood + Tr_Personally
                     + Tr_Initial + Tr_Nationality
                     + Satisfaction + Feeling_happy, data = miceOut3)

# Model with all Trust + Danger variables

fits_mod5 <- lm.mids(Trust ~ Tr_Family + Tr_Neighborhood + Tr_Personally
                     + Tr_Initial + Tr_Nationality
                     + Violence + Avoid_danger, data = miceOut3)

# Model with all Trust + Control and Advantage variables

fits_mod6 <- lm.mids(Trust ~ Tr_Family + Tr_Neighborhood + Tr_Personally
                     + Tr_Initial + Tr_Nationality
                     + Adv_Taken + Control_life 
                     + Wealth_accumulation, data = miceOut3)

## Pool the fitted models:
poolFit1 <- pool(fits_mod1)
poolFit2 <- pool(fits_mod2)
poolFit3 <- pool(fits_mod3)
poolFit4 <- pool(fits_mod4)
poolFit5 <- pool(fits_mod5)
poolFit6 <- pool(fits_mod6)

## Compute the pooled R^2:
pool.r.squared(fits_mod1)
pool.r.squared(fits_mod2)
pool.r.squared(fits_mod3)
pool.r.squared(fits_mod4)
pool.r.squared(fits_mod5)
pool.r.squared(fits_mod6)

# select the model with highest rsquared 
pool.r.squared(fits_mod2)[1] - pool.r.squared(fits_mod1)[1]

我的意图是让每个“fits_model”的“poolFit”在1:6的范围内(对于6个型号)，而不是必须手动制作它们。

谢谢！！

for-loop

Stack Overflow用户

发布于 2020-03-31 23:54:17

一种更像R的方法是这样做的。将不同的公式存储在列表中，然后在列表上使用lapply来拟合和总结您的模型。

models <- list(
  Trust ~  Tr_Family + Tr_Neighborhood + Tr_Personally 
                     + Tr_Initial + Tr_Nationality,
  Trust ~  Tr_Family + Tr_Neighborhood + Tr_Personally
                     + Tr_Initial + Tr_Nationality
                     + Racism_neighborhood + Homosexuality,
  Trust ~  Tr_Family + Tr_Neighborhood + Tr_Personally
                     + Tr_Initial + Tr_Nationality 
                     + Confidence_police + Interfere_police,
  Trust ~ Tr_Family + Tr_Neighborhood + Tr_Personally
                     + Tr_Initial + Tr_Nationality
                     + Satisfaction + Feeling_happy,
  Trust ~ Tr_Family + Tr_Neighborhood + Tr_Personally
                     + Tr_Initial + Tr_Nationality
                     + Violence + Avoid_danger,
  Trust ~ Tr_Family + Tr_Neighborhood + Tr_Personally
                     + Tr_Initial + Tr_Nationality
                     + Adv_Taken + Control_life 
                     + Wealth_accumulation)

fits <- lapply(models, lm.mids, data=miceOut3)
pools <- lapply(fits, pool)
poolR2 <- lapply(fits, pool.r.squared)

您是否可以通过fits[[1]]、fits[[2]]等获得个人合身

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原文链接：

https://stackoverflow.com/questions/60953660

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