Question

我有一个以下数据框，我想检查二进制列并将非空值更改为1.

a <- c("","a","a","","a")
b <- c("","b","b","b","b")
c <- c("c","","","","c")
d <- c("b","a","","c","d")

dt <- data.frame(a,b,c,d)

我能够通过循环遍历每一列来获得解决方案。但是，我想要一些有效的解决方案，因为我的数据框确实非常大，而下面的解决方案要慢得多。

我的解决方案 -

for(i in 1:length(colnames(dt)))
{
  if(length(table(dt[,i]))==2){
  dt[which(dt[,i]!=""),i] <- 1
  }
}

Expected Output:

 a b c d
     1 b
 1 1   a
 1 1    
   1   c
 1 1 1 d

有没有办法提高效率。

Answer 1

inds = lengths(lapply(dt, unique)) == 2
dt[inds] = lapply(dt[inds], function(x) as.numeric(as.character(x) != ""))
dt
#  a b c d
#1 0 0 1 b
#2 1 1 0 a
#3 1 1 0  
#4 0 1 0 c
#5 1 1 1 d

如果您想要""而不是0

dt[inds] = lapply(dt[inds], function(x) c("", 1)[(as.character(x) != "") + 1])
dt
#  a b c d
#1     1 b
#2 1 1   a
#3 1 1    
#4   1   c
#5 1 1 1 d

Answer 2

由于您的问题似乎很有效，您可能需要查看dplyr或data.table

等软件包

library(dplyr)
mutate_all(dt, .funs = quo(if_else(n_distinct(.) <= 2L & . != "", "1", .)))

library(data.table)
setDT(dt)
dt[ , lapply(.SD, function(x) ifelse(uniqueN(x) <= 2L & x != "", 1, x))]

用R中的1替换二进制列中的值的有效方法

2 个答案: