我有这个示例矩阵，我想根据条件if语句用"YES"或"NO"更改矩阵的条目。
a<-c(5,1,0,3,2,0.6,1.6,7,9,0)
b<-c(11,0,1,18,11,11,0,13,20,10)
c<-c(10,20,0.7,0.8,0.3,0.4,0,0.9,1,1)
MAT<-cbind(a,b,c)
MAT
for (i in 1:nrow(MAT)){
for (j in 1:ncol(MAT)){
if (MAT[i,j]>5){
MAT[i,j]="YES"
} else {
MAT[i,j]="NO"
}
}
}
print(MAT)
我得到的输出是这样的，它是错误的。请帮忙告诉我出了什么问题，如何解决?
a b c
[1,] "NO" "NO" "NO"
[2,] "NO" "NO" "NO"
[3,] "NO" "NO" "NO"
[4,] "NO" "NO" "NO"
[5,] "NO" "NO" "NO"
[6,] "NO" "NO" "NO"
[7,] "NO" "NO" "NO"
[8,] "YES" "NO" "NO"
[9,] "YES" "NO" "NO"
[10,] "NO" "NO" "NO"
这里不需要循环。只需在调用x>5时使用整个矩阵
ifelse(MAT>5, "YES", "NO")
这将对整个矩阵进行逻辑运算，并输出一个逻辑矩阵。
您可以使用空括号[]从ifelse()的输出中重新分配VALUES，同时保留MAT的STRUCTURE，如:
MAT[]<-ifelse(MAT>5, "YES", "NO")
故障原因
你尝试失败的原因来自这一部分:
if (MAT[i,j]>5){
MAT[i,j]="YES"
} else {
MAT[i,j]="NO"
}
}
您应该知道MAT是数字，但是您在if...else...语句中为MAT分配字符，这将使MAT转换为字符矩阵。在这种情况下，当你运行MAT[i,j] > 5时，你正在将一个字符与一个数值进行比较，例如，"18" > 5，它返回一个不希望的FALSE。
解决方案
一个解决方法是使用另一个变量来存储if...else...之后的值，而不是替换MAT中的值:
a <- c(5, 1, 0, 3, 2, 0.6, 1.6, 7, 9, 0)
b <- c(11, 0, 1, 18, 11, 11, 0, 13, 20, 10)
c <- c(10, 20, 0.7, 0.8, 0.3, 0.4, 0, 0.9, 1, 1)
MAT <- cbind(a, b, c)
out <- MAT
for (i in 1:nrow(MAT)) {
for (j in 1:ncol(MAT)) {
if (MAT[i, j] > 5) {
out[i, j] <- "YES"
} else {
out[i, j] <- "NO"
}
}
}
,
> out
a b c
[1,] "NO" "YES" "YES"
[2,] "NO" "NO" "YES"
[3,] "NO" "NO" "NO"
[4,] "NO" "YES" "NO"
[5,] "NO" "YES" "NO"
[6,] "NO" "YES" "NO"
[7,] "NO" "NO" "NO"
[8,] "YES" "YES" "NO"
[9,] "YES" "YES" "NO"
[10,] "NO" "YES" "NO"
选择
这个问题已经有很多答案了，下面是另一个基本的R选项
> `dim<-`(as.character(factor(MAT > 5, labels = c("NO", "YES"))), dim(MAT))
[,1] [,2] [,3]
[1,] "NO" "YES" "YES"
[2,] "NO" "NO" "YES"
[3,] "NO" "NO" "NO"
[4,] "NO" "YES" "NO"
[5,] "NO" "YES" "NO"
[6,] "NO" "YES" "NO"
[7,] "NO" "NO" "NO"
[8,] "YES" "YES" "NO"
[9,] "YES" "YES" "NO"
[10,] "NO" "YES" "NO"
仅使用逻辑矩阵转换为数字索引
MAT[] <- c("NO", "YES")[1 + (MAT > 5)]
-ouptut
> MAT
a b c
[1,] "NO" "YES" "YES"
[2,] "NO" "NO" "YES"
[3,] "NO" "NO" "NO"
[4,] "NO" "YES" "NO"
[5,] "NO" "YES" "NO"
[6,] "NO" "YES" "NO"
[7,] "NO" "NO" "NO"
[8,] "YES" "YES" "NO"
[9,] "YES" "YES" "NO"
[10,] "NO" "YES" "NO"
试试这个:
apply(MAT, 2, function(x) ifelse(x > 5, "YES", "NO"))
a b c
[1,] "NO" "YES" "YES"
[2,] "NO" "NO" "YES"
[3,] "NO" "NO" "NO"
[4,] "NO" "YES" "NO"
[5,] "NO" "YES" "NO"
[6,] "NO" "YES" "NO"
[7,] "NO" "NO" "NO"
[8,] "YES" "YES" "NO"
[9,] "YES" "YES" "NO"
[10,] "NO" "YES" "NO"
更新:在Jean-Claude Arbaut, ThomasIsCoding和GuedesBF的帮助说明之后，请注意第一个答案是错误的，这里是dplyr的替代方案:
我们可以在将matrix更改为tibble类后使用across，并在操作后重新更改为matrix类:
library(tibble)
library(dplyr)
MAT <- MAT %>%
as_tibble() %>%
mutate(across(everything(), ~ifelse(. > 5, "YES", "NO"))) %>%
as.matrix()
第一个答案:警告!
不要使用这个代码
MAT[MAT>5] <- "yes"
MAT[MAT<=5] <- "no"
正如Jean-Claude Arbaut, ThomasIsCoding和GuedesBF所指出的，它将在第一次赋值后强制字符，这可能导致下游操作中意想不到的结果。