如何解决R-循环多个变量group_by
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
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