提问人：Seyma Kalay 提问时间：6/15/2021 更新时间：6/15/2021 访问量：28
如何在 FORLOOP 中运行不同的函数
how to run different functions in forloop
问：
我想在同一个循环上运行两个不同的函数，并使用特定名称和 保存结果。提前非常感谢。df.rescale01df.rescale.02
df <- tibble( a = rnorm(10),b = rnorm(10))
rescale01 <- function(x) {
rng <- range(x, na.rm = TRUE)
(x - rng[1]) / (rng[2] - rng[1])
}
rescale02 <- function(x) {
rng <- range(x, na.rm = TRUE)
(rng[2] - rng[1]) / (x - rng[1])
}
for (i in seq_along(df)) {
df[[i]] <- rescale01(df[[i]])
}
df
预期答案
rescale01
A tibble: 10 x 2
a b
<dbl> <dbl>
1 0.354 0.627
2 0.657 0.545
3 0 0.397
4 0.602 0
5 0.539 0.698
6 0.775 1
7 1 0.216
8 0.949 0.286
9 0.782 0.244
10 0.239 0.523
rescale02
A tibble: 10 x 2
a b
<dbl> <dbl>
1 2.83 1.59
2 1.52 1.83
3 Inf 2.52
4 1.66 Inf
5 1.85 1.43
6 1.29 1
7 1 4.63
8 1.05 3.49
9 1.28 4.10
10 4.18 1.91
答：
1赞
KaptajnKasper
6/15/2021
#1
您可以将函数和结果存储在单独的列表中，如下所示
library(tidyverse)
set.seed(123)
df <- tibble(a = rnorm(10), b = rnorm(10))
rescale01 <- function(x) {
rng <- range(x, na.rm = TRUE)
(x - rng[1]) / (rng[2] - rng[1])
}
rescale02 <- function(x) {
rng <- range(x, na.rm = TRUE)
(rng[2] - rng[1]) / (x - rng[1])
}
rescale = list(rescale01 = rescale01, rescale02 = rescale02)
rslt = list()
for (i in seq_along(rescale)) {
for (j in seq_along(df)) {
df[[j]] <- rescale[[i]](df[[j]])
}
rslt[[names(rescale)[i]]] <- df
}
rslt
这将给出输出
> rslt
$rescale01
# A tibble: 10 x 2
a b
<dbl> <dbl>
1 0.236 0.850
2 0.347 0.620
3 0.948 0.631
4 0.448 0.553
5 0.468 0.376
6 1 1
7 0.579 0.657
8 0 0
9 0.194 0.711
10 0.275 0.398
$rescale02
# A tibble: 10 x 2
a b
<dbl> <dbl>
1 4.23 1.18
2 2.88 1.61
3 1.06 1.59
4 2.23 1.81
5 2.14 2.66
6 1 1
7 1.73 1.52
8 Inf Inf
9 5.15 1.41
10 3.64 2.51
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