英文:
How can I subtract values of each column of my dataframe from all other columns?
问题
Here is the translated code portion:
我有一个简单的数据框
df <- data.frame(
col1 = c(1, 2, 3),
col2 = c(4, 5, 6),
col3 = c(7, 8, 9)
)
我想要一个新的数据框，其中包含每一列与其他列的减法结果，依次计算。我认为这会完成任务
result_df <- data.frame()
# 从所有其他列中减去每一列
for (i in 1:ncol(df)) {
column_diff <- apply(df[, -i], 2, function(x) df[, i] - x)
result_df <- cbind(result_df, column_diff)
}
但问题是什么？
英文:
I have a simple datafrme
df <- data.frame(
col1 = c(1, 2, 3),
col2 = c(4, 5, 6),
col3 = c(7, 8, 9)
)
And I want to have a new dataframe that will contain the results of subtraction of each column from all others, one by one. I assume that will do the job
result_df <- data.frame()
# Subtract each column from all the other columns
for (i in 1:ncol(df)) {
column_diff <- apply(df[, -i], 2, function(x) df[, i] - x)
result_df <- cbind(result_df, column_diff)
}
but what is the issue?
答案1
得分: 1
如果您想要使用 cbind()，您需要将具有相同列数的元素组合在一起。我认为您可以像这样设置 result_df：
result_df <- data.frame(matrix(nrow=3, ncol=0))
或者创建一个列差异的列表，然后将它们组合起来可能更好：
cdiff <- function(i) apply(df[, -i], 2, function(x) df[, i] - x)
newcols <- lapply(1:ncol(df), cdiff)
data.frame(newcols)
英文:
If you want to cbind() you need to combine elements with the same number of columns. I think you could set up result_df like this:
result_df <- data.frame(matrix(nrow=3, ncol = 0))
Or it might be better to create a list of column-differences, then combine them:
cdiff <- function(i) apply(df[, -i], 2, function(x) df[, i] - x)
newcols <- lapply(1:ncol(df), cdiff)
data.frame(newcols)
答案2
得分: 1
简单
在 {tidyverse} 中，可以使用 purrr::map_dfc() 来执行类似于您所使用的函数：
library(tidyverse)
d <- data.frame(col1 = 1:3, col2 = 4:6, col3 = 7:9)
seq_along(d) %>% map_dfc(.f = \(x){d[,x] - d[,-x]})
#> col2...1 col3...2 col1...3 col3...4 col1...5 col2...6
#> 1 -3 -6 3 -3 6 3
#> 2 -3 -6 3 -3 6 3
#> 3 -3 -6 3 -3 6 3
创建于 2023-06-04，使用 reprex v2.0.2
更新
purrr::map_dfc() 在 {purrr} 1.0.0 中已弃用，现在可以使用 map() 与 list_cbind() 来进行正确的操作，这允许您指定 names_repair 参数以控制命名。
seq_along(d) %>%
map(.f = \(x){d[,x] - d[,-x]}) %>%
list_cbind(name_repair = "minimal")
#> col2 col3 col1 col3 col1 col2
#> 1 -3 -6 3 -3 6 3
#> 2 -3 -6 3 -3 6 3
#> 3 -3 -6 3 -3 6 3
创建于 2023-06-05，使用 reprex v2.0.2
高级
如果您想要更高级的操作，并且希望新列反映生成它们的计算过程，您可以使用 dplyr::rename_with()。
seq_along(d) %>%
map(.f = \(x) {
d[, x] - d[,-x] %>%
rename_with(\(y) {paste0(names(d)[x], "-", y)})
}) %>%
list_cbind()
#> col1-col2 col1-col3 col2-col1 col2-col3 col3-col1 col3-col2
#> 1 -3 -6 3 -3 6 3
#> 2 -3 -6 3 -3 6 3
#> 3 -3 -6 3 -3 6 3
创建于 2023-06-05，使用 reprex v2.0.2
英文:
Simple
In the {tidyverse} this can be done with purrr::map_dfc() using a function similar to what you had:
library(tidyverse)
d <- data.frame(col1 = 1:3, col2 = 4:6, col3 = 7:9)
seq_along(d) %>% map_dfc(.f = \(x){d[,x] - d[,-x]})
#> col2...1 col3...2 col1...3 col3...4 col1...5 col2...6
#> 1 -3 -6 3 -3 6 3
#> 2 -3 -6 3 -3 6 3
#> 3 -3 -6 3 -3 6 3
<sup>Created on 2023-06-04 with reprex v2.0.2</sup>
Updated
purrr::map_dfc() was deprecated in {purrr} 1.0.0 so it would properly be done now using map() with list_cbind() which enables one to specify the names_repair argument to control the naming.
seq_along(d) %>%
map(.f = \(x){d[,x] - d[,-x]}) %>%
list_cbind(name_repair = "minimal")
#> col2 col3 col1 col3 col1 col2
#> 1 -3 -6 3 -3 6 3
#> 2 -3 -6 3 -3 6 3
#> 3 -3 -6 3 -3 6 3
<sup>Created on 2023-06-05 with reprex v2.0.2</sup>
Fancy
If you you want to get fancy and have the new columns reflect the calculation that produced them, you could use dplyr::rename_with().
seq_along(d) %>%
map(.f = \(x) {
d[, x] - d[,-x] %>%
rename_with(\(y) {paste0(names(d)[x], "-", y)})
}) %>%
list_cbind()
#> col1-col2 col1-col3 col2-col1 col2-col3 col3-col1 col3-col2
#> 1 -3 -6 3 -3 6 3
#> 2 -3 -6 3 -3 6 3
#> 3 -3 -6 3 -3 6 3
<sup>Created on 2023-06-05 with reprex v2.0.2</sup>
答案3
得分: 0
以下是您要翻译的内容：
"While I prefer the solution by @Ben Bolker, here is a fun one:
library(dplyr)
library(purrr)
cdiff <- function(x) {
df %>%
select(-{{x}}) %>%
transmute(!!paste0("col", x, ".diff") := df[[x]] - .)
}
map_dfc(1:ncol(df), cdiff)
col1.diff.col2 col1.diff.col3 col2.diff.col1 col2.diff.col3 col3.diff.col1 col3.diff.col2
1 -3 -6 3 -3 6 3
2 -3 -6 3 -3 6 3
3 -3 -6 3 -3 6 3
```"
<details>
<summary>英文:</summary>
While I prefer the solution by @Ben Bolker, here is a fun one:
library(dplyr)
library(purrr)
cdiff <- function(x) {
df %>%
select(-{{x}}) %>%
transmute(!!paste0("col", x, ".diff") := df[[x]] - .)
}
map_dfc(1:ncol(df), cdiff)
col1.diff.col2 col1.diff.col3 col2.diff.col1 col2.diff.col3 col3.diff.col1 col3.diff.col2
1 -3 -6 3 -3 6 3
2 -3 -6 3 -3 6 3
3 -3 -6 3 -3 6 3
</details>
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