Question

我试图回答这个问题：Create vectors containing 0 or 1 based on subjects in R 我只得到以下代码行，它返回一个名为matrix的{{1}}：

pollution

假设我开始使用pollution <- apply(mydf[,2:4], c(1,2), function(i) { ifelse(grep('POLLUTION|EMISSION|WASTE',i)==1,1,0)} ) pollution # SUBJECT.1 SUBJECT.2 SUBJECT.3 # [1,] 1 Logical,0 Logical,0 # [2,] Logical,0 1 Logical,0 # [3,] Logical,0 Logical,0 1 # [4,] Logical,0 Logical,0 1 matrix，我该如何将其转换为pollution matrix和1的：{/ p>

我已经回答了有关如何将# SUBJECT.1 SUBJECT.2 SUBJECT.3 # [1,] 1 0 0 # [2,] 0 1 0 # [3,] 0 0 1 # [4,] 0 0 1替换为NA等的Stack Overflow的几个问题。但是，我无法弄清楚如何替换这些0的。

例如：

Logical,0

感谢您的任何建议。对不起，如果这是重复的。搜索单词pollution[is.logical(pollution)] = 0 # SUBJECT.1 SUBJECT.2 SUBJECT.3 # [1,] 1 Logical,0 Logical,0 # [2,] Logical,0 1 Logical,0 # [3,] Logical,0 Logical,0 1 # [4,] Logical,0 Logical,0 1 pollution2 <- as.data.frame(pollution) pollution2 # SUBJECT.1 SUBJECT.2 SUBJECT.3 # 1 1 # 2 1 # 3 1 # 4 1会返回大量点击。

Answer 1

这是我能想到的最容易的：

pollution2 <- pollution
pollution2[] <- as.numeric(sapply(pollution, function(x) length(x) > 0))
pollution2
     SUBJECT.1 SUBJECT.2 SUBJECT.3
[1,] 1         0         0        
[2,] 0         1         0        
[3,] 0         0         1        
[4,] 0         0         1

请注意我如何使用[]表示法保留矩阵结构和dimnames。

编辑：对apply的调整非常简单，仅供参考

apply(mydf[,2:4], c(1,2), function(i) {  
  ifelse(length(grep('POLLUTION|EMISSION|WASTE',i)),1,0)} )
     SUBJECT.1 SUBJECT.2 SUBJECT.3
[1,]         1         0         0
[2,]         0         1         0
[3,]         0         0         1
[4,]         0         0         1

Answer 2

可能这也有帮助：

 matrix(sapply(c(pollution),length),4)
    [,1] [,2] [,3]
[1,]    1    0    0
[2,]    0    1    0
[3,]    0    0    1
[4,]    0    0    1

或

matrix((!sapply(pollution, is.logical))+0,4)

Answer 3

我知道我可以这样做：

> new.pollution <- matrix(as.numeric(pollution), nrow=4, byrow=FALSE)
> new.pollution[is.na(new.pollution)] = 0
> new.pollution
     [,1] [,2] [,3]
[1,]    1    0    0
[2,]    0    1    0
[3,]    0    0    1
[4,]    0    0    1

但肯定有一种更简单的方法。

替换Logical，0替换为0

3 个答案: