R k折交叉验证后调用stepAIC逐步回归提示行数变化错误如何解决？

这是R语言惰性求值的典型坑，和训练集本身的行数差异无关：
你在循环中构建lm模型时，mod <- lm(z_mara ~ . -1, data = trainCV_sets[[i]]) 这句的调用记录里存储的是trainCV_sets[[i]]这个表达式，而非i对应的当时的实际子集数据。循环结束后全局环境的i值固定为4，你后续调用stepwise/stepAIC时，函数会解析模型调用记录里的data参数，所有模型的训练集都会被解析为trainCV_sets[[4]]（12行），但前3个模型初始拟合用的是11行的训练集，拟合子模型时用的是12行数据，因此触发行数不一致的报错。

两种方案任选即可：

方案1：构建模型时直接固化拟合数据，避免后续动态解析i #

将构建lm的代码修改为：

mod <- lm(z_mara ~ . -1, data = trainCV_sets[[i]],
          # 显式将当前子集的自变量、因变量直接存入模型对象
          x = TRUE, y = TRUE)

x=TRUE和y=TRUE会把拟合用到的变量直接存在lm对象里，stepAIC会优先用这部分数据拟合子模型，不需要再解析调用记录里的data参数。

方案2：调用stepwise时显式传入对应训练集 #

修改遍历代码，同时传入模型和对应的训练集：

# 用map2同时遍历模型列表和训练集列表
map2(modList, trainCV_sets, ~ stepwise(.x, direction = "backward", criterion = "AIC", data = .y))

library(tidyverse)
library(modelr)
library(MASS) # 若使用car包的stepwise，替换加载包即可，用法一致

training_pca <- tibble(z_mara = c(0.875112399579427, 1.13894237770963, 
0.981180086218575, 1.8299894270253, -0.0977104233331723, -0.86241558837047, 
-1.40078945237221, 0.030856469562766, -0.0969068802525719, -0.223063143906711, 
-0.339309042900122, -0.760633464827852, -1.01830294600679, -1.29686454728133, 
-1.39007554463088), PC1 = c(2.50796906215996, 2.68002216940135, 
-3.71526007848766, -4.1866795459591, 0.0486434530855795, 0.938304238296873, 
0.615097559709769, 4.51822579066407, 0.0667126218105262, -0.287378553280471, 
4.56603423346247, -0.933301952142784, -0.216058021757341, 0.0694218190496208, 
9.2292656007433), PC2 = c(2.17078441684463, -0.199721476838284, 
2.11735035612631, 3.03544498625801, -2.40626489140099, 2.35791076587204, 
-1.25582416436226, 1.05994935645006, -0.527916055713799, 0.070729476106444, 
-0.132485078109658, 0.949419002097709, -5.09998831832504, 1.0619915074115, 
2.3100848754419), PC3 = c(-0.678144405424994, 1.09058577380406, 
-1.0176297169444, -1.88552787304444, -1.25338902691145, 0.210499981706639, 
0.04477690001186, -1.38969813324165, -0.804920509008458, 0.879138015029564, 
1.29938967213637, 0.138188073442918, -5.26695126718185, -2.14797795807358, 
1.62335052575001), PC4 = c(-2.3887668089524, 3.97903517088607, 
0.867800635964586, -2.03442689569174, 0.13869948487601, -0.0664645238981098, 
0.670700717048399, 3.11260554952338, 0.408133658723152, 3.44132931097882, 
1.10549707214258, 1.0954984094734, -0.49754510472686, -2.85815085785707, 
2.04309500230981), PC5 = c(0.821714377740757, 3.78554017369849, 
0.230169453679477, -1.9039365507055, 0.0828762592235243, -1.83817906200712, 
-2.4669972669382, 1.88139312797292, -1.28458814157848, -1.74987295529079, 
0.342015969239523, -0.819381168596516, 1.64847329083668, -3.95311876523071, 
-0.306975624219868))
set.seed(3654)
trainingCV_pca <- crossv_kfold(training_pca, k = 4)
trainCV_sets <- vector(mode = "list", length = nrow(trainingCV_pca))
testCV_sets <- vector(mode = "list", length = nrow(trainingCV_pca))
for(i in 1:nrow(trainingCV_pca)) {
  trainCV_idx <- trainingCV_pca[[1]][[i]][[2]]
  trainCV_sets[[i]] <- training_pca[trainCV_idx,]
  testCV_sets[[i]] <- training_pca[-trainCV_idx,]
}
modList <- vector(mode = "list", length = nrow(trainingCV_pca))
for(i in 1:nrow(trainingCV_pca)) {
  # 新增x=TRUE y=TRUE固化拟合数据
  mod <- lm(z_mara ~ . -1, data = trainCV_sets[[i]], x = TRUE, y = TRUE)
  modList[[i]] <- mod
}
# 直接调用即可正常运行，trace=FALSE可关闭逐步过程输出
map(modList, ~ stepAIC(., direction = "backward", criterion = "AIC", trace = FALSE))

内容的提问来源于stack exchange，提问作者a.hesse