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R循环矩阵比较第一列是否与另一列相同

[英]R loop matrix comparing whether the first column is the same as the other column

I wanted would like to check whether in each column of a matrix the values are the same as in the first column: 我想检查矩阵的每一列中的值是否与第一列中的值相同:

 x1  <- c("x", "y", "x")
 x2  <- c("x", "y", "y")
 x3  <- c("y", "y","x")

 dat <- data.frame(x1,x2,x3)

   x1 x2 x3
1  x  x  y
2  y  y  y
3  x  y  x

So if x2 != 2, there should be a "1", otherwise a NA. 因此,如果x2!= 2,则应为“ 1”,否则为NA。 How can I do that? 我怎样才能做到这一点? The result in this case would be: 在这种情况下,结果将是:

x2     x3 
NA     1
NA     NA
1      NA

My (not working) solution was: 我(不起作用)的解决方案是:

fun <- function (x) {
for(j in 2: ncol(x)){
  ifelse(x[,1]== j, NA,1)
}
}

fun(dat) 乐趣(日期)

I would need a function to perform this with lapply. 我需要一个函数来用lapply执行此操作。 How can I do that? 我怎样才能做到这一点? Thanks a lot! 非常感谢!

Different approach using the fact that column-operations are often easy because of argument recycling: 不同的方法利用了由于参数循环而导致列操作通常很容易的事实:

 dat <- data.frame(x1,x2,x3, stringsAsFactors=FALSE) # keeps as character 

 ne.dat1 <-  (dat != dat[,1])[ , -1]  # True/False rather than 1,NA
 is.na(ne.dat1) <- !ne.dat1
 ne.dat1
       x2   x3
[1,]   NA TRUE
[2,]   NA   NA
[3,] TRUE   NA

You could do: 您可以这样做:

 m1 <- (!(as.character(dat[,1])==dat[,-1])) +1
 m1[] <- c(NA,1)[m1]
 m1
 #    x2 x3
 #[1,] NA  1
 #[2,] NA NA
 #[3,]  1 NA

This is easier with the wrapper sapply as it applies applies the names to the result: 这是与包装物容易sapply ,因为它适用于应用名称的结果:

sapply(dat[-1], 
       function(x) 
          ifelse(as.character(x)==as.character(dat[[1]]), NA, 1)
)
##      x2 x3
## [1,] NA  1
## [2,] NA NA
## [3,]  1 NA

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