英文:
Perform a specific Mathematical Function on each column dynamically in R
问题
我想要在数据框中对每个唯一项目执行数学函数。
通常，要执行数学函数，我们使用 mutate 语句并创建一个列，然后在每个 mutate 语句之后手动执行数学函数。
这在几列上是可行的。但如果我有100列，并且需要执行2-5个数学函数，例如：一个是初始数字的增加20%，另一个是在每列上将初始数字除以2并保持原始列不变。
除了为每个特定项目编写 mutate 语句之外，R中是否有可能实现这一点？
我正在使用的数据框是：
structure(list(`Row Labels` = c("2023-03-01", "2023-04-01", "2023-05-01","2023-06-01", "2023-07-01", "2023-08-01", "2023-09-01", "2023-10-01"), X6 = c(14, 16, 14, 11, 9, 9, 11, 11), X7 = c(50, 50, 50, 50,50, 50, 50, 50), X8 = c(75, 75, 75, 75, 75, 75, 75, 75), X9 = c(100,100, 100, 100, 100, 100, 100, 100), X11 = c(25, 25, 50, 75, 125,200, 325, 525), X12 = c(50, 50, 100, 150, 250, 400, 650, 1050)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,-8L))
对于单独的情况，以下代码就足够了：
library(readxl)library(dplyr)Book1 <- read_excel("C:/X/X/X- X/X/Book1.xlsx", sheet = "Sheet6")dput(Book1)Book1 <- Book1 %>%mutate(`X6 20%` = X6*1.20) %>%mutate(`X6 by 2` = X6/2)
我考虑过通过循环来运行这个代码，但是选择要进行乘法的列会成为一个问题，因为我们必须在 mutate 语句中指定列名，我认为这在这里可能不可行。
有没有人能告诉我是否可以用一种简单的方法实现这个目标？
期望的输出如下：
英文:
I wanted to perform a mathematical function on each unique item in a data frame dynamically.
Normally to perform a mathematical function, we use mutate statement and create a column and perform the mathematical function manually by writing mutate statement after mutate statement.
Which is feasible on a few columns. But what if I have 100 columns and I have to perform 2-5 mathematical function, For example: one would be 20% increase on the initial number, The other one would be to divide the initial number by 2 on each column and keep the original column as is.
Is this possible in R other than writing mutate statement for each specific item?
The data frame I am working with is:
structure(list(`Row Labels` = c("2023-03-01", "2023-04-01", "2023-05-01","2023-06-01", "2023-07-01", "2023-08-01", "2023-09-01", "2023-10-01"), X6 = c(14, 16, 14, 11, 9, 9, 11, 11), X7 = c(50, 50, 50, 50,50, 50, 50, 50), X8 = c(75, 75, 75, 75, 75, 75, 75, 75), X9 = c(100,100, 100, 100, 100, 100, 100, 100), X11 = c(25, 25, 50, 75, 125,200, 325, 525), X12 = c(50, 50, 100, 150, 250, 400, 650, 1050)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,-8L))
For individual cases this code would suffice:
library(readxl)library(dplyr)Book1 <- read_excel("C:/X/X/X- X/X/Book1.xlsx",sheet = "Sheet6")dput(Book1)Book1 <- Book1 %>%mutate(`X6 20%` = X6*1.20) %>%mutate(`X6 by 2`= X6/2)
I was thinking of running this through a loop but then selection of columns to multiple becomes a problem as we have to specify the column name in mutate statement, which I believe would not be possible here right.
Can anyone let me know if this can be achieved in a simple approach?
The expected output is given below:
答案1
得分: 2
We could use across()
update: shorter:
library(dplyr)df %>%mutate(across(2:7, list("20" = ~. * 1.20,"By_2" = ~. / 2), .names = "{col}_{fn}"))
first answer:
library(dplyr)df %>%mutate(across(2:7, ~. * 1.20, .names = "{.col}_20%"),across(2:7, ~. /2, .names = "{.col}_By 2"))`Row Labels` X6 X7 X8 X9 X11 X12 `X6_20%` `X7_20%` `X8_20%` `X9_20%` `X11_20%` `X12_20%` `X6_By 2` `X7_By 2` `X8_By 2` `X9_By 2` `X11_By 2` `X12_By 2`<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>1 2023-03-01 14 50 75 100 25 50 16.8 60 90 120 30 60 7 25 37.5 50 12.5 252 2023-04-01 16 50 75 100 25 50 19.2 60 90 120 30 60 8 25 37.5 50 12.5 253 2023-05-01 14 50 75 100 50 100 16.8 60 90 120 60 120 7 25 37.5 50 25 504 2023-06-01 11 50 75 100 75 150 13.2 60 90 120 90 180 5.5 25 37.5 50 37.5 755 2023-07-01 9 50 75 100 125 250 10.8 60 90 120 150 300 4.5 25 37.5 50 62.5 1256 2023-08-01 9 50 75 100 200 400 10.8 60 90 120 240 480 4.5 25 37.5 50 100 2007 2023-09-01 11 50 75 100 325 650 13.2 60 90 120 390 780 5.5 25 37.5 50 162. 3258 2023-10-01 11 50 75 100 525 1050 13.2 60 90 120 630 1260 5.5 25 37.5 50 262. 525
英文:
We could use across()
update: shorter:
library(dplyr)df %>%mutate(across(2:7, list("20" = ~. * 1.20,"By_2" = ~. / 2), .names = "{col}_{fn}"))
first answer:
library(dplyr)df %>%mutate(across(2:7, ~. * 1.20, .names = "{.col}_20%"),across(2:7, ~. /2, .names = "{.col}_By 2"))`Row Labels` X6 X7 X8 X9 X11 X12 `X6_20%` `X7_20%` `X8_20%` `X9_20%` `X11_20%` `X12_20%` `X6_By 2` `X7_By 2` `X8_By 2` `X9_By 2` `X11_By 2` `X12_By 2`<chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>1 2023-03-01 14 50 75 100 25 50 16.8 60 90 120 30 60 7 25 37.5 50 12.5 252 2023-04-01 16 50 75 100 25 50 19.2 60 90 120 30 60 8 25 37.5 50 12.5 253 2023-05-01 14 50 75 100 50 100 16.8 60 90 120 60 120 7 25 37.5 50 25 504 2023-06-01 11 50 75 100 75 150 13.2 60 90 120 90 180 5.5 25 37.5 50 37.5 755 2023-07-01 9 50 75 100 125 250 10.8 60 90 120 150 300 4.5 25 37.5 50 62.5 1256 2023-08-01 9 50 75 100 200 400 10.8 60 90 120 240 480 4.5 25 37.5 50 100 2007 2023-09-01 11 50 75 100 325 650 13.2 60 90 120 390 780 5.5 25 37.5 50 162. 3258 2023-10-01 11 50 75 100 525 1050 13.2 60 90 120 630 1260 5.5 25 37.5 50 262. 525
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