英文:
How to calculate percentage for each group depends on the different variable?
问题
以下是代码的翻译部分：
这是虚拟数据集的R代码：
c <- c(10, 20, 30, 40, 50, 40, 2, 40, 10, 50)
b <- c(40, 2, 40, 10, 50, 10, 20, 30, 40, 50)
a <- c(10, 50, 3, 60, 100,40, 2, 40, 10, 50)
id <- c("a", "b", "b", "a", "c", "a", "b", "b", "a", "c")
variation <- c("a3", "a3", "b1", "a2", "b1","a3", "a1", "b1", "a1", "b1")
data <- data.frame(id, a, b, c, variation)
head(data)
# id a b c variation
# 1 a 10 40 10 a3
# 2 b 50 2 20 a3
# 3 b 3 40 30 b1
# 4 a 60 10 40 a2
# 5 c 100 50 50 b1
# 6 a 40 10 40 a3
# 7 b 2 20 2 a1
# 8 b 40 30 40 b1
# 9 a 10 40 10 a1
# 10 c 50 50 50 b1
我可以为每个单独的id进行百分比计算，过滤后如下：
data_filter <- data %>% filter(id == "a")
data_filter
# id a b c variation
# 1 a 10 40 10 a3
# 2 a 60 10 40 a2
# 3 a 40 10 40 a3
# 4 a 10 40 10 a1
# 数据转换
data_filter_percentage <- data_filter %>%
group_by(variation) %>%
count() %>%
ungroup() %>%
mutate(perc = `n` / sum(`n`)) %>%
arrange(perc) %>%
mutate(labels = scales::percent(perc))
head(data_filter_percentage)
# A tibble: 3 x 4
# variation n perc labels
# <chr> <int> <dbl> <chr>
# 1 a1 1 0.25 25%
# 2 a2 1 0.25 25%
# 3 a3 2 0.5 50%
然而，我的问题是，是否可以对所有"id"执行上述管道而无需单独过滤？
英文:
This the dummy dataset R code:
c <- c(10, 20, 30, 40, 50, 40, 2, 40, 10, 50)
b <- c(40, 2, 40, 10, 50, 10, 20, 30, 40, 50)
a <- c(10, 50, 3, 60, 100,40, 2, 40, 10, 50)
id <- c("a", "b", "b", "a", "c", "a", "b", "b", "a", "c")
variation <- c("a3", "a3", "b1", "a2", "b1","a3", "a1", "b1", "a1", "b1" )
data <- data.frame(id, a, b, c, variation)
head(data)
# id a b c variation
# 1 a 10 40 10 a3
# 2 b 50 2 20 a3
# 3 b 3 40 30 b1
# 4 a 60 10 40 a2
# 5 c 100 50 50 b1
# 6 a 40 10 40 a3
# 7 b 2 20 2 a1
# 8 b 40 30 40 b1
# 9 a 10 40 10 a1
# 10 c 50 50 50 b1
I can calculate percentages for individual id after filtering:
data_filter <- data %>% filter(id == "a")
data_filter
# id a b c variation
# 1 a 10 40 10 a3
# 2 a 60 10 40 a2
# 3 a 40 10 40 a3
# 4 a 10 40 10 a1
# Data transformation
data_filter_percentage <- data_filter %>%
group_by(variation) %>% # Variable to be transformed
count() %>%
ungroup() %>%
mutate(perc = `n` / sum(`n`)) %>%
arrange(perc) %>%
mutate(labels = scales::percent(perc))
head(data_filter_percentage)
# A tibble: 3 x 4
# variation n perc labels
# <chr> <int> <dbl> <chr>
# 1 a1 1 0.25 25%
# 2 a2 1 0.25 25%
# 3 a3 2 0.5 50%
However, my question is, Is it possible to perform above pipeline for all "id" without filtering individually?
答案1
得分: 1
以下是翻译好的代码部分：
library(dplyr)
data %>%
group_by(id) %>%
count(variation) %>%
mutate(perc = n / sum(n), labels = scales::percent(perc)) %>%
ungroup()
Briefly,
data %>%
count(id, variation) %>%
mutate(perc = n / sum(n), labels = scales::percent(perc), .by = id)
# # A tibble: 7 × 5
# id variation n perc labels
# <chr> <chr> <int> <dbl> <chr>
# 1 a a1 1 0.25 25%
# 2 a a2 1 0.25 25%
# 3 a a3 2 0.5 50%
# 4 b a1 1 0.25 25%
# 5 b a3 1 0.25 25%
# 6 b b1 2 0.5 50%
# 7 c b1 2 1 100%
英文:
You can try the following workflow:
library(dplyr)
data %>%
group_by(id) %>%
count(variation) %>%
mutate(perc = n / sum(n), labels = scales::percent(perc)) %>%
ungroup()
Briefly,
data %>%
count(id, variation) %>%
mutate(perc = n / sum(n), labels = scales::percent(perc), .by = id)
# # A tibble: 7 × 5
# id variation n perc labels
# <chr> <chr> <int> <dbl> <chr>
# 1 a a1 1 0.25 25%
# 2 a a2 1 0.25 25%
# 3 a a3 2 0.5 50%
# 4 b a1 1 0.25 25%
# 5 b a3 1 0.25 25%
# 6 b b1 2 0.5 50%
# 7 c b1 2 1 100%
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