我有以下数字0和1的矩阵，每列始终包含相同数量的字符串。一列中的最小字符串数为2。当它们同时满足这两个条件时，我想删除它们。
10和01），01只发生一两次。但是我想保留所有其他列： r1 <- c("10","001","0001","01","100","10")
r2 <- c("01","001","0001","10","100","10")
r3 <- c("10","100","1000","10","010","01")
r4 <- c("10","010","0100","10","001","10")
r5<- c("01","010","0010","10","001","10")
r6<- c("01","010","0010","10","001","01")
n.mat <- rbind(r1,r2,r3,r4,r5,r6)
输出：
r1 <- c("10","001","0001","100")
r2 <- c("01","001","0001","100")
r3 <- c("10","100","1000","010")
r4 <- c("10","010","0100","001")
r5<- c("01","010","0010","001")
r6<- c("01","010","0010","001")
n.mat <- rbind(r1,r2,r3,r4,r5,r6)
卸下第4列和第6列。
到目前为止，我的代码是：
del_two<- function(x){
length(unique(x)) != 2
}
msa_protein.mat_1<-msa_protein.mat[, apply(msa_protein.mat, 2, del_two)]
但是我不确定如何添加if函数。
您可以添加&将逻辑选择与“ AND”逻辑结合在一起。尽管在这种情况下，我认为您想删除这些值而不是保留它们，所以您需要取消!最终选择：
n.mat[, apply(n.mat, 2, FUN=function(x) !(length(unique(x)) == 2 & sum(x == '01') <= 2))]
甚至：
n.mat[, !apply(n.mat, 2, FUN=function(x) length(unique(x)) == 2 & sum(x == '01') <= 2)]
您也可以将其表示为逻辑条件失败，再加上|“ OR”逻辑：
n.mat[, apply(n.mat, 2, FUN=function(x) length(unique(x)) != 2 | sum(x == '01') > 2)]
全部给予：
# [,1] [,2] [,3] [,4]
#r1 "10" "001" "0001" "100"
#r2 "01" "001" "0001" "100"
#r3 "10" "100" "1000" "010"
#r4 "10" "010" "0100" "001"
#r5 "01" "010" "0010" "001"
#r6 "01" "010" "0010" "001"
也可能有一些棘手的方法使用列总和来完成此操作，如果您拥有大量数据，则可能会更快，例如：
n.mat[, !(
(colSums(n.mat == "01") <= 2) &
colSums(matrix(n.mat %in% c("10","01"), nrow=nrow(n.mat), ncol=ncol(n.mat))) == nrow(n.mat)
)]
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