我认为这很容易，但不适用。我试图按照这个例子来改变我创建的矩阵列表中每个矩阵的列名：
Assign column names to list of dataframes
当我运行下面的代码时，我得到一个非常奇怪的回报，看起来我只是在每个矩阵中设置每个元素的名称，而不仅仅是列名。
#create a list of matrices containing random numbers
randoms<-lapply(1:1000, function(x) matrix(rnorm(1440), ncol=10))
trial<-lapply(randoms, setNames , nm = letters[1:10])
head(trial[[1]])
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 0.89032453 1.02459736 0.7141343 -0.47405630 -2.0719943 -1.5087669
[2,] -0.74866047 0.44086093 -1.7540066 -2.04227094 -0.4875453 1.4207707
[3,] -0.04565454 -1.52336294 -0.1941370 -1.36252338 1.7338307 -1.3536725
[4,] 0.13242099 -0.09157545 -0.6156536 -1.34546174 -0.3279853 0.9663668
[5,] 2.09173141 0.41592339 0.7422889 -0.05991624 0.5319697 0.6413341
[6,] -0.32129540 2.11206231 0.1722047 -0.54404820 1.2685971 -0.0784607
[,7] [,8] [,9] [,10]
[1,] -0.4849624 -1.2590439 -1.5066718 -0.6758746
[2,] -2.5010320 -2.3469163 0.5221117 0.9186142
[3,] -1.3763468 -0.5551194 -0.2304872 -1.6087508
[4,] -2.0282231 -0.1949064 0.9329241 1.0196325
[5,] 1.6429999 1.8176161 -0.6549447 -1.8833887
[6,] 1.0044023 1.5895154 0.3660308 -0.1883819
head(attr(trial[[1]], "names"))
[1] "a" "b" "c" "d" "e" "f"
答案 0 :(得分：0)
我们可以使用for循环
for(i in seq_along(randoms)) {
colnames(randoms[[i]]) <- letters[1:10]
}