1

我有这个示例脚本:

library(tidyverse)

placeholder <- c("foo", "bar", "baz", "bash")

x <- 1
for (name in placeholder) {
    iris <- iris %>% add_column(name = rep("example", nrow(iris)),
                                .after = x)
    x <- x + 2

}

我想做的是在每个花卉特征之后插入一列,其中的列从placeholder.

预期输出:

  Sepal.Length     foo Sepal.Width     bar Petal.Length     baz Petal.Width    bash Species
1          5.1 example         3.5 example          1.4 example         0.2 example  setosa
2          4.9 example         3.0 example          1.4 example         0.2 example  setosa
3          4.7 example         3.2 example          1.3 example         0.2 example  setosa
4          4.6 example         3.1 example          1.5 example         0.2 example  setosa
5          5.0 example         3.6 example          1.4 example         0.2 example  setosa
6          5.4 example         3.9 example          1.7 example         0.4 example  setosa

但是我的 for 循环标记了 for 循环中的第一列name,然后以错误提示终止Error: Column name already exists

我得到什么:

  Sepal.Length    name Sepal.Width Petal.Length Petal.Width Species
1          5.1 example         3.5          1.4         0.2  setosa
2          4.9 example         3.0          1.4         0.2  setosa
3          4.7 example         3.2          1.3         0.2  setosa
4          4.6 example         3.1          1.5         0.2  setosa
5          5.0 example         3.6          1.4         0.2  setosa
6          5.4 example         3.9          1.7         0.4  setosa
4

1 回答 1

2

我们可以在 lhs 上使用赋值运算符 ( :=) 和name评估 ( !!) 来将占位符的值设置为列名

x <- 1
for (name in placeholder) {
    iris <- iris %>% add_column(!! name := rep("example", nrow(iris)),
                            .after = x)
     x <- x + 2

 }
head(iris)
#  Sepal.Length     foo Sepal.Width     bar Petal.Length     baz Petal.Width    bash Species
#1          5.1 example         3.5 example          1.4 example         0.2 example  setosa
#2          4.9 example         3.0 example          1.4 example         0.2 example  setosa
#3          4.7 example         3.2 example          1.3 example         0.2 example  setosa
#4          4.6 example         3.1 example          1.5 example         0.2 example  setosa
#5          5.0 example         3.6 example          1.4 example         0.2 example  setosa
#6          5.4 example         3.9 example          1.7 example         0.4 example  setosa
于 2019-01-22T10:49:46.473 回答