R-循环多个变量group_by

编程问答 2022-06-13

问题描述

我希望能够遍历group_by变量，以便按变量和总数的每种组合进行汇总，并将其组合为一个。

我在这里看到了类似的内容：
dplyr- group by in a for loop r

我尝试通过几种不同的方式修改代码，但似乎无法使其与交叉配合工作。

df <- data.frame(
  location = c(rep("UK",5),rep("USA",5)),industry = c(rep("RETAIL",3),rep("TECH",7)),department = c(rep("SALES",4),rep("MANUFACTURING",6)),pay = rnorm(10),tax = rnorm(10)
)

temp <- crossing(vara = c("location",""),varB = c("industry",varC = c("department",""))

data <- data.frame()
for(i in 1:nrow(temp)){
test <- df %>%
  group_by(!!temp[i,]) %>%
  summarise_at(c("pay","tax"),sum,na.rm = TRUE)

data <- rbind.fill(test,data)
}

解决方法

这是我认为您要寻找的。这是ginput解决方案。

dplyr

如果您想将set.seed(10) df <- data.frame( location = c(rep("UK",5),rep("USA",5)),industry = c(rep("RETAIL",3),rep("TECH",7)),department = c(rep("SALES",4),rep("MANUFACTURING",6)),pay = rnorm(10),tax = rnorm(10) ) temp <- crossing(varA = c("location",""),varB = c("industry",varC = c("department","")) data <- data.frame() for(i in 1:nrow(temp)){ # extracts only non "" values from temp[i,] and unnames them (else group_by will use names) vars <- unname(unlist(temp[i,which(temp[i,] != "")])) test <- df %>% # tells tidyselect to use all columns that match the contents of vars group_by(across(all_of(vars))) %>% summarise_at(c("pay","tax"),sum,na.rm = TRUE) # union_all does what you want rbind.fill to do data <- union_all(test,data) } print(data,n = 20) # A tibble: 20 x 5 # Groups: location,industry [8] location industry department pay tax <chr> <chr> <chr> <dbl> <dbl> 1 UK RETAIL SALES -1.54 1.62 2 UK TECH MANUFACTURING 0.295 0.741 3 UK TECH SALES -0.599 0.987 4 USA TECH MANUFACTURING -3.07 0.348 5 UK RETAIL NA -1.54 1.62 6 UK TECH NA -0.305 1.73 7 USA TECH NA -3.07 0.348 8 UK NA MANUFACTURING 0.295 0.741 9 UK NA SALES -2.14 2.61 10 USA NA MANUFACTURING -3.07 0.348 11 UK NA NA -1.84 3.35 12 USA NA NA -3.07 0.348 13 NA RETAIL SALES -1.54 1.62 14 NA TECH MANUFACTURING -2.77 1.09 15 NA TECH SALES -0.599 0.987 16 NA RETAIL NA -1.54 1.62 17 NA TECH NA -3.37 2.08 18 NA NA MANUFACTURING -2.77 1.09 19 NA NA SALES -2.14 2.61 20 NA NA NA -4.91 3.70 的值替换为NA，可以简单地做到这一点：

"ALL"
,

也许您正在寻找这个。尝试以下带有循环的tidyverse解决方案：

library(tidyverse)
#Data
df <- data.frame(
  location = c(rep("UK",tax = rnorm(10)
)
#Loop
vars <- names(df)[1:3]
List <- list()
#Code df[,i]
for(i in 1:length(vars)){
  test <- df %>%
    group_by(eval(parse(text=vars[i]))) %>%
    summarise_at(c("pay",na.rm = TRUE)
  names(test)[1] <- 'var'
  #Var
  vardf <- data.frame(Mainvar=rep(vars[i],nrow(test)))
  test <- cbind(vardf,test)
  #Save
  List[[i]] <- test
  
}
#Bind all
mydf <- do.call(rbind,List)
rownames(mydf)<-NULL

输出：

     Mainvar           var        pay       tax
1   location            UK -0.8347144 -1.719750
2   location           USA -2.8887471 -4.079747
3   industry        RETAIL  0.1327241 -1.212067
4   industry          TECH -3.8561856 -4.587430
5 department MANUFACTURING -4.5570133 -4.248031
6 department         SALES  0.8335518 -1.551466
, 
vars <- names(df)[1:3]  
vars_subsets <- 0:length(vars) %>%
  map(~combn(vars,.x,simplify = FALSE)) %>%
  unlist(recursive = FALSE)

vars_subsets %>%
  map(~
        df %>% 
          {if(length(.x) > 0) group_by(.,across(all_of(.x))) else .} %>% 
          summarise(pay = sum(pay,na.rm = TRUE),tax = sum(tax,na.rm = TRUE)) 
  ) %>%
  bind_rows() %>%
  select(all_of(vars),pay,tax)

给予：

> head(x)
location industry    department        pay        tax
1     <NA>     <NA>          <NA>  2.7641031  3.2347055
2       UK     <NA>          <NA> -0.2370619  3.5215502
3      USA     <NA>          <NA>  3.0011650 -0.2868447
4     <NA>   RETAIL          <NA>  1.3318324  0.4189127
5     <NA>     TECH          <NA>  1.4322707  2.8157928
6     <NA>     <NA> MANUFACTURING  2.8567654  0.7405478
aggregateaggregateloopsrr