英文:
Mutate across pairs of columns with dplyr
问题
我试图在dplyr中对列对应的函数进行应用。在下面的简单示例中，有列a和b，以及a_exp和b_exp。我想创建一个新列叫做a_mul = a*a_exp，以及另一个用于b（以及c，d，e ...）。我知道可以使用循环和函数、purrr，或pivot_longer来解决这个问题，但我想知道是否可以使用mutate(across())来解决它。这是一个简单的示例。实际问题更复杂，所以我无法一次性创建两列，因为在实际情况中，a_exp不是a的函数。我正在尝试以下代码，并且得到以下错误信息：
"promise already under evaluation: recursive default argument reference or earlier problems?"
我不明白为什么或如何修复它。
test <- tibble(a = runif(10), b = runif(10)) %>%mutate(across(c(a,b), exp, .names = '{.col}_exp'))test <- test %>%mutate(across(c(a,b),~ .x * .data[[paste0(cur_column(), '_exp')]],.names = '{.col}_mult'))
我已经翻译了你的代码示例，如你所要求，没有包括其他内容。
英文:
I am trying to apply a function to pairs of columns in dplyr. In the simple example below, there are columns a and b, as well as a_exp and b_exp. I want to create a new column called a_mul = a*a_exp and another one for b (and c, d, e ...). I know I can solve this problem with a loop and a function, with purrr, or with pivot_longer, but I want to know if I can solve it with mutate(across()). This is a minimal example. The real problem is more complicated, so I can't create both columns in one go because in the real case a_exp is not a function of a. I am trying the following code, and I get
>"promise already under evaluation: recursive default argument reference or earlier problems?"
I don't understand why or how to fix it.
test <- tibble(a = runif(10), b = runif(10)) %>%mutate(across(c(a,b), exp, .names = '{.col}_exp'))test <- test %>%mutate(across(c(a,b),~ .x * .data[[paste0(cur_column(), '_exp')]],.names = '{.col}_mult'))
答案1
得分: 1
你可以从列中获取值。
library(tidyverse)set.seed(12)test <- tibble(a = runif(10), b = runif(10)) %>%mutate(across(c(a, b), exp, .names = "{.col}_exp"))test %>% mutate(across(c(a, b), ~.x * get(paste0(cur_column(), "_exp")), .names = "{.col}_mult"))
如果你可以决定先进行哪个算术操作，也许先进行乘法，然后再进行exp会更容易。
test <- tibble(a = runif(10), b = runif(10)) %>%mutate(across(c(a, b), ~.x * exp(.x), .names = "{.col}_mult"),across(c(a, b), exp, .names = "{.col}_exp"))
输出
# A tibble: 10 × 6a b a_exp b_exp a_mult b_mult<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>1 0.0694 0.393 1.07 1.48 0.0743 0.5822 0.818 0.814 2.27 2.26 1.85 1.843 0.943 0.376 2.57 1.46 2.42 0.5484 0.269 0.381 1.31 1.46 0.353 0.5575 0.169 0.265 1.18 1.30 0.201 0.3456 0.0339 0.439 1.03 1.55 0.0351 0.6827 0.179 0.458 1.20 1.58 0.214 0.7238 0.642 0.541 1.90 1.72 1.22 0.9299 0.0229 0.666 1.02 1.95 0.0234 1.3010 0.00832 0.113 1.01 1.12 0.00839 0.126
英文:
You can get the values from the column.
library(tidyverse)set.seed(12)test <- tibble(a = runif(10), b = runif(10)) %>%mutate(across(c(a, b), exp, .names = "{.col}_exp"))test %>% mutate(across(c(a, b), ~.x * get(paste0(cur_column(), "_exp")), .names = "{.col}_mult"))
If you can decide which arithmetic operation to go first, it maybe easier to do multiplication first, then exp.
test <- tibble(a = runif(10), b = runif(10)) %>%mutate(across(c(a, b), ~.x * exp(.x), .names = "{.col}_mult"),across(c(a, b), exp, .names = "{.col}_exp"))
Output
# A tibble: 10 × 6a b a_exp b_exp a_mult b_mult<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>1 0.0694 0.393 1.07 1.48 0.0743 0.5822 0.818 0.814 2.27 2.26 1.85 1.843 0.943 0.376 2.57 1.46 2.42 0.5484 0.269 0.381 1.31 1.46 0.353 0.5575 0.169 0.265 1.18 1.30 0.201 0.3456 0.0339 0.439 1.03 1.55 0.0351 0.6827 0.179 0.458 1.20 1.58 0.214 0.7238 0.642 0.541 1.90 1.72 1.22 0.9299 0.0229 0.666 1.02 1.95 0.0234 1.3010 0.00832 0.113 1.01 1.12 0.00839 0.126
答案2
得分: 1
以下是翻译好的代码部分：
1. 使用 purrr 包中的 reduce 方法的方法：
library(dplyr)library(purrr)library(stringr)set.seed(123)test %>%split.default(str_remove(names(.), "_.*")) %>%map_dfr(reduce, `*`) %>%rename_with(~ paste0(., "_mult"), everything()) %>%bind_cols(test)a_mult b_mult a b a_exp b_exp<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>1 1.86 1.14 0.819 0.616 2.27 1.852 2.53 0.861 0.964 0.515 2.62 1.673 2.33 2.31 0.924 0.920 2.52 2.514 0.194 0.718 0.164 0.455 1.18 1.585 0.500 1.98 0.351 0.847 1.42 2.336 1.14 2.08 0.617 0.871 1.85 2.397 1.04 1.35 0.580 0.683 1.79 1.988 2.42 1.71 0.942 0.781 2.56 2.189 0.259 0.323 0.210 0.251 1.23 1.2910 0.0404 0.0487 0.0388 0.0465 1.04 1.05
2. 使用 across 方法需要使用 rename_with 修改列名以获得合适的列名对：
library(dplyr)library(stringr)test %>%rename_with(., ~ifelse(!str_detect(., "\\_"), paste0(., "_start"), .)) %>%mutate(across(ends_with('exp'), ~ . *get(str_replace(cur_column(), "exp$", "start")), .names = "mult_{.col}")) %>%rename_at(vars(starts_with('mult')), ~ str_remove(., "\\_exp"))a_start b_start a_exp b_exp mult_a mult_b<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>1 0.288 0.957 1.33 2.60 0.383 2.492 0.788 0.453 2.20 1.57 1.73 0.7133 0.409 0.678 1.51 1.97 0.616 1.334 0.883 0.573 2.42 1.77 2.14 1.025 0.940 0.103 2.56 1.11 2.41 0.1146 0.0456 0.900 1.05 2.46 0.0477 2.217 0.528 0.246 1.70 1.28 0.896 0.3158 0.892 0.0421 2.44 1.04 2.18 0.04399 0.551 0.328 1.74 1.39 0.957 0.45510 0.457 0.955 1.58 2.60 0.721 2.48
英文:
We have at least two ways to do this:
1.An approach with reduce from purrr package:
library(dplyr)library(purrr)library(stringr)set.seed(123)test %>%split.default(str_remove(names(.), "_.*")) %>%map_dfr(reduce, `*`) %>%rename_with(~ paste0(., "_mult"), everything()) %>%bind_cols(test)a_mult b_mult a b a_exp b_exp<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>1 1.86 1.14 0.819 0.616 2.27 1.852 2.53 0.861 0.964 0.515 2.62 1.673 2.33 2.31 0.924 0.920 2.52 2.514 0.194 0.718 0.164 0.455 1.18 1.585 0.500 1.98 0.351 0.847 1.42 2.336 1.14 2.08 0.617 0.871 1.85 2.397 1.04 1.35 0.580 0.683 1.79 1.988 2.42 1.71 0.942 0.781 2.56 2.189 0.259 0.323 0.210 0.251 1.23 1.2910 0.0404 0.0487 0.0388 0.0465 1.04 1.05
2. For using across we have to modify the names with rename_with to get adequate pairs of names:
library(dplyr)library(stringr)test %>%rename_with(., ~ifelse(!str_detect(., "\\_"), paste0(., "_start"), .)) %>%mutate(across(ends_with('_exp'), ~ . *get(str_replace(cur_column(), "exp$", "start")), .names = "mult_{.col}")) %>%rename_at(vars(starts_with('mult')), ~ str_remove(., "\\_exp"))a_start b_start a_exp b_exp mult_a mult_b<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>1 0.288 0.957 1.33 2.60 0.383 2.492 0.788 0.453 2.20 1.57 1.73 0.7133 0.409 0.678 1.51 1.97 0.616 1.334 0.883 0.573 2.42 1.77 2.14 1.025 0.940 0.103 2.56 1.11 2.41 0.1146 0.0456 0.900 1.05 2.46 0.0477 2.217 0.528 0.246 1.70 1.28 0.896 0.3158 0.892 0.0421 2.44 1.04 2.18 0.04399 0.551 0.328 1.74 1.39 0.957 0.45510 0.457 0.955 1.58 2.60 0.721 2.48
答案3
得分: 1
You can multiply across(a:b) by across(a_exp:b_exp) directly if they are in pairs.
test %>%mutate(across(a:b, .names = '{.col}_mul') * across(a_exp:b_exp))
# A tibble: 10 × 6
a b a_exp b_exp a_mul b_mul
1 0.288 0.957 1.33 2.60 0.383 2.49
2 0.788 0.453 2.20 1.57 1.73 0.713
3 0.409 0.678 1.51 1.97 0.616 1.33
4 0.883 0.573 2.42 1.77 2.14 1.02
5 0.940 0.103 2.56 1.11 2.41 0.114
6 0.0456 0.900 1.05 2.46 0.0477 2.21
7 0.528 0.246 1.70 1.28 0.896 0.315
8 0.892 0.0421 2.44 1.04 2.18 0.0439
9 0.551 0.328 1.74 1.39 0.957 0.455
10 0.457 0.955 1.58 2.60 0.721 2.48
---##### Data```rlibrary(dplyr)set seed(123)test <- tibble(a = runif(10), b = runif(10)) %>%mutate(across(c(a,b), exp, .names = '{.col}_exp'))
英文:
You can multiply across(a:b) by across(a_exp:b_exp) directly if they are in pairs.
test %>%mutate(across(a:b, .names = '{.col}_mul') * across(a_exp:b_exp))# # A tibble: 10 × 6# a b a_exp b_exp a_mul b_mul# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl># 1 0.288 0.957 1.33 2.60 0.383 2.49# 2 0.788 0.453 2.20 1.57 1.73 0.713# 3 0.409 0.678 1.51 1.97 0.616 1.33# 4 0.883 0.573 2.42 1.77 2.14 1.02# 5 0.940 0.103 2.56 1.11 2.41 0.114# 6 0.0456 0.900 1.05 2.46 0.0477 2.21# 7 0.528 0.246 1.70 1.28 0.896 0.315# 8 0.892 0.0421 2.44 1.04 2.18 0.0439# 9 0.551 0.328 1.74 1.39 0.957 0.455# 10 0.457 0.955 1.58 2.60 0.721 2.48
Data
library(dplyr)set.seed(123)test <- tibble(a = runif(10), b = runif(10)) %>%mutate(across(c(a,b), exp, .names = '{.col}_exp'))
答案4
得分: 0
test[, 1:2] * test[, 3:4]
英文:
If the data is organised as shown in your example data, this should do:
test[, 1:2] * test[, 3:4]
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