[英]How do I replace empty dataframe/data.tables from a list with a placeholder data.table?
這篇文章如何刪除列表中的空白數據框? 談到刪除空的數據框。 如何從列表中刪除空數據框(行= 0)並將其替換為1行占位符數據框/data.tables?
M1 <- data.frame(matrix(1:4, nrow = 2, ncol = 2))
M2 <- data.frame(matrix(nrow = 0, ncol = 0))
M3 <- data.frame(matrix(9:12, nrow = 2, ncol = 2))
mlist <- list(M1, M2, M3)
placeholder = data.table(a=1,b=1)
我試過了:
lapply(mlist, function(x) ifelse(nrow(fundslist[[x]]) == 0, placeholder, x))
一種選擇是使用lengths
mlist[!lengths(mlist)] <- list(placeholder)
str(mlist)
#List of 3
# $ :'data.frame': 2 obs. of 2 variables:
# ..$ X1: int [1:2] 1 2
# ..$ X2: int [1:2] 3 4
# $ :Classes ‘data.table’ and 'data.frame': 1 obs. of 2 variables:
# ..$ a: num 1
# ..$ b: num 1
# ..- attr(*, ".internal.selfref")=<externalptr>
# $ :'data.frame': 2 obs. of 2 variables:
# ..$ X1: int [1:2] 9 10
# ..$ X2: int [1:2] 11 12
這個怎么樣? 由於您的占位符很小,因此將其乘以n倍不是問題。
library(data.table)
M1 <- data.frame(matrix(1:4, nrow = 2, ncol = 2))
M2 <- data.frame(matrix(nrow = 0, ncol = 0))
M3 <- data.frame(matrix(9:12, nrow = 2, ncol = 2))
mlist <- list(M1, M2, M3)
placeholder = data.table(a=1,b=1)
num.rows <- unlist(lapply(mlist, nrow))
num.zeros <- length(num.rows[num.rows == 0])
replacement <- replicate(num.zeros, {placeholder}, simplify = FALSE)
mlist[num.rows == 0] <- replacement
str(mlist)
List of 3
$ :'data.frame': 2 obs. of 2 variables:
..$ X1: int [1:2] 1 2
..$ X2: int [1:2] 3 4
$ :Classes ‘data.table’ and 'data.frame': 1 obs. of 2 variables:
..$ a: num 1
..$ b: num 1
..- attr(*, ".internal.selfref")=<externalptr>
$ :'data.frame': 2 obs. of 2 variables:
..$ X1: int [1:2] 9 10
..$ X2: int [1:2] 11 12
只是說明您如何使用自己的方法來完成它!
M1 <- data.frame(matrix(1:4, nrow = 2, ncol = 2))
M2 <- data.frame(matrix(nrow = 0, ncol = 0))
M3 <- data.frame(matrix(9:12, nrow = 2, ncol = 2))
mlist <- list(M1, M2, M3)
placeholder = data.frame(matrix(c(1,1), nrow=1))
abc <- function(x){
if(sum(dim(x))==0)
return(data.frame(placeholder))
else
return(x)
}
lapply(mlist, abc)
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