R data.table-删除与给定边距相对应的行

Posted 2021-04-13

我有以下问题。我有一个data.table和一列M的子集。我在x上定义了矢量M。

library(data.table)
data <- matrix(c(0,0,NA,1,0,1,NA,1,0,0,1,0,1,1,NA,NA,1,0,0,1,0,0,1,1,1,0,0,1,NA,0,1,1,0,1,1,1), byrow = T, ncol = 6, dimnames = LETTERS[1:6])
dt <- data.table(data)
dt
%     A B  C  D  E F
% 1:  0 0 NA  1  0 1
% 2: NA 1  0  0  1 0
% 3:  1 1 NA NA  1 0
% 4:  0 1  0  0  1 1
% 5:  1 0  0  1 NA 0
% 6:  1 1  0  1  1 1

M = LETTERS[2:5]
x <- dt[2,..M]
x
%    B C D E
% 1: 1 0 0 1

我想从dt中删除所有行，其中M上的边距等于x。即行号2和4。M和x都在编程期间改变。给定M和x的结果将是：

   A B  C  D  E F
1: 0 0 NA  1  0 1
2: 1 1 NA NA  1 0
3: 1 0  0  1 NA 0
4: 1 1  0  1  1 1

答案

不是真正的data.table选项，但可以使用base R：

data[rowSums(sweep(data[, M], 2, FUN = `==`, x), na.rm = TRUE) != length(x), ]

     A B  C  D  E F
[1,] 0 0 NA  1  0 1
[2,] 1 1 NA NA  1 0
[3,] 1 0  0  1 NA 0
[4,] 1 1  0  1  1 1

另一答案

数据表防联接

dt[!x, on = M]

#    A B  C  D  E F
# 1: 0 0 NA  1  0 1
# 2: 1 1 NA NA  1 0
# 3: 1 0  0  1 NA 0
# 4: 1 1  0  1  1 1

Base R

eq2 <- Reduce('&', lapply(dt[, ..M], function(x) x == x[2]))

dt[-which(eq2),]
#    A B  C  D  E F
# 1: 0 0 NA  1  0 1
# 2: 1 1 NA NA  1 0
# 3: 1 0  0  1 NA 0
# 4: 1 1  0  1  1 1

另一答案

另一个R基础解决方案

> subset(dt,!data.frame(t(dt[,..M])) %in% data.frame(t(x)))
   A B  C  D  E F
1: 0 0 NA  1  0 1
2: 1 1 NA NA  1 0
3: 1 0  0  1 NA 0
4: 1 1  0  1  1 1