如何用r中下一列的值替换匹配

Question

我有一个数据框：

structure(list(City = structure(c(4L, 2L, 1L, 3L, 3L, 3L), .Label = c("Gold Cost", 
"Melbourne", "Other", "Sydney"), class = "factor"), Town = structure(c(1L, 
1L, 1L, 3L, 4L, 2L), .Label = c("", "Brighton", "Hurstville", 
"Penhurst"), class = "factor")), .Names = c("City", "Town"), class = "data.frame", row.names = c(NA, 
-6L))

我想替换名为 City 的列中包含名为 Other 的值的所有行，并将其替换为同一行的下一列中的值。

我的输出应该是这样的：

structure(list(City = structure(c(6L, 4L, 2L, 3L, 5L, 1L), .Label = c("Brighton", 
"Gold Cost", "Hurstville", "Melbourne", "Penhurst", "Sydney"), class = "factor"), 
    Town = structure(c(1L, 1L, 1L, 3L, 4L, 2L), .Label = c("", 
    "Brighton", "Hurstville", "Penhurst"), class = "factor")), .Names = c("City", 
"Town"), class = "data.frame", row.names = c(NA, -6L))

我之前没有写过任何函数，但我猜它应该是这样的：

for(data1 in 1:nrow(data1)) {
        if(data1$City[i] == 'Other') {
                data1$city[i] <- data1$Town[i]
        } else {
                break
        }
}

1. 我哪里出错了？
2. 未来解决此类问题的思考过程应该是什么？
3. 我怎样才能得到想要的结果？

Answer 1

有2个错误和2个低效。

错误 1： 你写的是for(data1 而不是for(i。

错误 2： 你的班级是 factor for City，你正在尝试添加新的关卡。而是将此操作更改为character。否则，新的因子水平将转换为NA。还有其他方法可以解决此问题，但效率较低；之后您可以随时将其改回因子。

效率低下 1： 您也不需要else 语句。

效率低下 2：如果没有 for 循环（即以矢量化方式），您可以做到这一点。

data1 <- structure(list(City = structure(c(4L, 2L, 1L, 3L, 3L, 3L), 
.Label = c("Gold Cost", "Melbourne", "Other", "Sydney"), class = "factor"), 
                        Town = structure(c(1L, 1L, 1L, 3L, 4L, 2L),
 .Label = c("", "Brighton", "Hurstville", "Penhurst"), class = "factor")), 
                   .Names = c("City", "Town"), class = "data.frame",
   row.names = c(NA, -6L))

desired_output <- structure(list(City = structure(c(6L, 4L, 2L, 3L, 5L, 1L),
 .Label = c("Brighton", "Gold Cost", "Hurstville", "Melbourne", "Penhurst", "Sydney"),
  class = "factor"), Town = structure(c(1L, 1L, 1L, 3L, 4L, 2L), 
 .Label = c("", "Brighton", "Hurstville", "Penhurst"), class = "factor")), 
  .Names = c("City", "Town"), class = "data.frame",
 row.names = c(NA, -6L))

data1$City <- as.character(data1$City)
data1$Town <- as.character(data1$Town)
for(i in 1:nrow(data1)){
  if(data1$City[i]=='Other'){
    data1$City[i]<- data1$Town[i]
  }
}

data1

        City       Town
1     Sydney           
2  Melbourne           
3  Gold Cost           
4 Hurstville Hurstville
5   Penhurst   Penhurst
6   Brighton   Brighton

data1 == desired_output

     City Town
[1,] TRUE TRUE
[2,] TRUE TRUE
[3,] TRUE TRUE
[4,] TRUE TRUE
[5,] TRUE TRUE
[6,] TRUE TRUE

现在是矢量化解决方案。通过避免使用循环，您的代码运行速度将呈指数级增长，而且您还必须输入更少的代码。

data1$City[data1$City == "Other"] <- data1$Town[data1$City == "Other"]