15

我有这个示例矩阵，我想使用"YES"或"NO"基于条件if语句更改矩阵的条目。

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。请帮助告诉我出了什么问题以及如何解决？

      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"
4

5 回答 5

17

这里不需要循环。只需在调用中使用整个矩阵x>5

ifelse(MAT>5, "YES", "NO")

这将对整个矩阵进行逻辑运算。但是，输出将是一个向量，它基本上是生成的逻辑矩阵，但从其属性中剥离dim。

您可以使用空括号重新分配VALUES的输出，ifelse()同时保留 的STRUCTURE，如下所示：MAT[]

MAT[]<-ifelse(MAT>5, "YES", "NO")
于 2021-08-15T11:21:23.340 回答
11

失败原因

您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"
    }
  }

您应该知道这MAT是数字，但是您正在将字符分配给MATwith inif...else...语句，这将MAT转换为字符矩阵。在这种情况下，当您运行 时MAT[i,j] > 5，您正在将一个字符与一个数值进行比较，例如"18" > 5，它返回一个不想要的FALSE。

解决方法

一种解决方法是使用另一个变量来存储 之后的值if...else...，而不是替换 中的值MAT：

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"
    }
  }
}

这样

> 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base 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"
于 2021-08-15T20:12:32.240 回答
5

仅使用转换为数字索引的逻辑矩阵

MAT[] <- c("NO", "YES")[1 + (MAT > 5)]

-输出

> 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"
于 2021-08-15T18:46:10.827 回答
3

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"
于 2021-08-15T11:21:31.207 回答
2

更新： 在 Jean-Claude Arbaut、ThomasIsCoding 和 GuedesBF 的有用说明之后，请注意第一个答案是错误的，这里有一个替代方案dplyr：

我们可以across在更改matrix为tibble类后使用，并matrix在操作后重新更改为：

library(tibble)
library(dplyr)

MAT <- MAT %>% 
  as_tibble() %>% 
  mutate(across(everything(), ~ifelse(. > 5, "YES", "NO"))) %>% 
  as.matrix()

第一个答案：警告！

不要使用此代码

MAT[MAT>5] <- "yes"
MAT[MAT<=5] <- "no"

正如 Jean-Claude Arbaut、ThomasIsCoding 和 GuedesBF 所指出的，它将在第一次分配后强制转换字符，这可能会导致下游操作出现意外结果。

于 2021-08-15T11:21:38.163 回答