英文:
Add columns in certain position to a set of data.frames contained in a list, preferably with mapply
问题
以下是翻译好的部分：
我有一个包含数据框的列表，我想在每个数据框的第4和第17位置添加两个空列，以便稍后与其他数据框合并它们。我需要按位置进行操作，因为列名不匹配。
df1 <- data.frame(x = 1:3, y=letters[1:3])
df2 <- data.frame(x = 4:6, y=letters[4:6])
data_list <- list(df1,df2)
我认为我可以使用类似于以下的代码作为基础，这段代码我从另一个问题中得到。
data_list2 <- mapply(`[<-`, data_list, 'emptyColumn', value = 'NA', SIMPLIFY = FALSE)
但是，我需要在特定位置添加列，比如在第2位。
此外，关于[<-是如何工作的解释将非常有帮助。
谢谢！
英文:
I have a set of data.frames on a list, and I would like to add two empty columns in the 4th and 17th position to each of the data frames, so I can later rbind them with others dfs. I need to do it using position because the names are not matching.
df1 <- data.frame(x = 1:3, y=letters[1:3])
df2 <- data.frame(x = 4:6, y=letters[4:6])
data_list <- list(df1,df2)
I think I could use something like this as a base, which I took from another question.
data_list2 <- mapply(`[<-`, data_list, 'emptyColumn', value = 'NA', SIMPLIFY = FALSE)
But instead of just adding it at the end, I need to add it on a certain position, let's say on the 2nd, for example.
Also an explanation of what [<- does would be great.
Thanks!
答案1
得分: 2
我提出了以下内容：
data_list2 <- mapply(add_column, data_list, 'emptyColumn'='NA',.after=4, SIMPLIFY = FALSE)
英文:
I came out with this:
data_list2 <- mapply(add_column, data_list, 'emptyColumn'='NA',.after=4, SIMPLIFY = FALSE)
答案2
得分: 1
你可以使用lapply和append来实现类似这样的操作。只有当emptyColumn具有不同值向量时，mapply才有用。
lapply(data_list, \(x) data.frame(append(x, list("emptyColumn" = rep(NA, nrow(df1))), after = 1)))
[[1]]
x emptyColumn y
1 1 NA a
2 2 NA b
3 3 NA c
[[2]]
x emptyColumn y
1 4 NA d
2 5 NA e
3 6 NA f
<details>
<summary>英文:</summary>
You can do something like this with `lapply` and `append`. `mapply` will only be useful if you have different vectors of values for `emptyColumn`.
```r
lapply(data_list, \(x) data.frame(append(x, list("emptyColumn" = rep(NA, nrow(df1))), after = 1)))
[[1]]
x emptyColumn y
1 1 NA a
2 2 NA b
3 3 NA c
[[2]]
x emptyColumn y
1 4 NA d
2 5 NA e
3 6 NA f
答案3
得分: 1
可以的，这是代码的翻译部分：
Alternatively using `purrr` and `tibble::add_column()`:
df1 <- data.frame(x = 1:3, y=letters[1:3], z = 1:3, a = 1:3)
df2 <- data.frame(x = 4:6, y=letters[4:6], z = 1:3, a = 1:3)
data_list <- list(df1,df2)
library(purrr)
library(tibble)
positions <- c(4,1)
map2(data_list, positions, ~.x |>
add_column(empty_column = NA, .before = .y))
Output:
[[1]]
x y z empty_column a
1 1 a 1 NA 1
2 2 b 2 NA 2
3 3 c 3 NA 3
[[2]]
empty_column x y z a
1 NA 4 d 1 1
2 NA 5 e 2 2
3 NA 6 f 3 3
英文:
Alternatively using purrr and tibble::add_column():
df1 <- data.frame(x = 1:3, y=letters[1:3], z = 1:3, a = 1:3)
df2 <- data.frame(x = 4:6, y=letters[4:6], z = 1:3, a = 1:3)
data_list <- list(df1,df2)
library(purrr)
library(tibble)
positions <- c(4,1)
map2(data_list, positions, ~.x |>
add_column(empty_column = NA, .before = .y))
Output:
[[1]]
x y z empty_column a
1 1 a 1 NA 1
2 2 b 2 NA 2
3 3 c 3 NA 3
[[2]]
empty_column x y z a
1 NA 4 d 1 1
2 NA 5 e 2 2
3 NA 6 f 3 3
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