[英]How to calculate rowMeans for dataframes in a list?
我有一個看起來像這樣的列表(my.list):
> my.list
$S1
A B C D
1 5 2 3 2
2 6 3 4 3
3 7 5 5 5
4 2 3 6 7
5 6 6 7 6
$S2
A B C D
1 5 2 3 2
2 6 3 4 3
3 7 5 5 5
4 2 3 6 7
5 6 6 7 6
我想通過采用 AD 列的行均值來創建一個新變量“E”。
我試過使用兩個循環:
test_list<-list()
for(i in 1:5){
test_data$E <- list.append(test_list, mylist[[i]] %>% rowMeans(mylist[,c("A","B","C","D")]))
}
並應用:
merged_data <- lapply(merged_data, transform, E = rowMeans(mylist[,c("A","B","C","D")]))
兩者似乎都不起作用。 我收到錯誤消息:
Error in mylist[c("A","B","C","D"), :
incorrect number of dimensions
我該怎么做呢?
可重現的數據:
my.list <- structure(list(S1 = structure(list(A = c(5, 6, 7, 2, 6), B = c(2,3,5,3,6), C = c(3,4,5,6,7), D = c(2,3,5,7,6)),.Names = c("A", "B", "C", "D"), class = "data.frame", row.names = c("1","2", "3", "4", "5")), S2 = structure(list(A = c(5, 6, 7, 2, 6), B = c(2,3,5,3,6), C = c(3,4,5,6,7), D = c(2,3,5,7,6)), .Names = c("A", "B", "C",
"D"), class = "data.frame", row.names = c("1", "2", "3", "4","5"))), .Names = c("S1", "S2"))
使用lapply
:
lapply(my.list, \(x)transform(x, E = rowMeans(x)))
甚至
lapply(my.list, \(x)cbind(x, E = rowMeans(x)))
兩者都導致:
$S1
A B C D E
1 5 2 3 2 3.00
2 6 3 4 3 4.00
3 7 5 5 5 5.50
4 2 3 6 7 4.50
5 6 6 7 6 6.25
$S2
A B C D E
1 5 2 3 2 3.00
2 6 3 4 3 4.00
3 7 5 5 5 5.50
4 2 3 6 7 4.50
5 6 6 7 6 6.25
附加選項
library(tidyverse)
map(my.list, ~mutate(.x, E = rowMeans(.x)))
$S1
A B C D E
1 5 2 3 2 3.00
2 6 3 4 3 4.00
3 7 5 5 5 5.50
4 2 3 6 7 4.50
5 6 6 7 6 6.25
$S2
A B C D E
1 5 2 3 2 3.00
2 6 3 4 3 4.00
3 7 5 5 5 5.50
4 2 3 6 7 4.50
5 6 6 7 6 6.25
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