在 R 中按组汇总百分比
Summarize Percentage by Group in R
我希望按行业的女性和男性百分比来总结我的数据集。我仍在学习 R 并且无法解决这个问题。
我的数据：
Industry
Male
Female
Art/Entertainment
100
500
Banking
600
100
Healthcare
53
65
Education
20
766
Military
47
96
Medicine
500
400
Law
500
500
Computer
200
144
Sales
420
69
目标：
Industry
Male
Female
F%
M%
Art/Entertainment
100
500
Banking
600
100
Healthcare
53
65
Education
20
766
Military
47
96
Medicine
500
400
Law
500
500
Computer
200
144
Sales
420
69
如果您的数据名为 df，我们可以像这样为男性和女性百分比创建列：
df$Fpct <- df$Female / (df$Male + df$Female)
df$Mpct <- df$Male / (df$Male + df$Female)
注意，不要在变量名中使用 % 符号。
您可以使用 group_by，然后使用 F% 和 M% 创建两个新列
也许你可以使用这个：
library(dplyr)
df %>% group_by(Industry) %>% mutate(F_prec=Female/(Male+Female), M_prec=Male/(Male+Female))
1) proportions 如果您的输入是 df1（在最后的注释中重复显示）然后将列名称更改为所需的名称并将其转换为矩阵 m。最后使用 proportions，边距为 1 表示行比例——2 表示列比例。请注意，我们在第一行转换为矩阵，因为 proportions 需要。
m <- as.matrix(setNames(df1[-1], c("%M", "%F")))
cbind(df1, 100 * proportions(m, 1))
## Industry Male Female %M %F
## 1 Art/Entertainment 100 500 16.666667 83.33333
## 2 Banking 600 100 85.714286 14.28571
## 3 Healthcare 53 65 44.915254 55.08475
## ...snip...
2) rowSums 另一种方法是将 df1[-1] 除以 rowSums 给出相同的结果。
cbind(df1, setNames(100 * df1[-1] / rowSums(df1[-1]), c("%M", "%F")))
## Industry Male Female %M %F
## 1 Art/Entertainment 100 500 16.666667 83.33333
## 2 Banking 600 100 85.714286 14.28571
## 3 Healthcare 53 65 44.915254 55.08475
## ...snip...
3) dplyr 使用 across 复制具有指定名称的列，然后将其乘以 100，然后使用 c_across 除按列的总和
df1 %>%
group_by(Industry) %>%
mutate(100 * across(.names = "%{.col}") / sum(c_across())) %>%
ungroup
## # A tibble: 9 x 5
## Industry Male Female `%Male` `%Female`
## <chr> <int> <int> <dbl> <dbl>
## 1 Art/Entertainment 100 500 16.7 83.3
## 2 Banking 600 100 85.7 14.3
## 3 Healthcare 53 65 44.9 55.1
## ...snip...
4) transform 这个接近另一个答案但它不会覆盖输入：
transform(df1,
"%M" = 100 * Male / (Male + Female),
"%F" = 100 * Female / (Male + Female),
check.names = FALSE)
## Industry Male Female %M %F
## 1 Art/Entertainment 100 500 16.666667 83.33333
## 2 Banking 600 100 85.714286 14.28571
## 3 Healthcare 53 65 44.915254 55.08475
## ...snip...
备注
以可复制的形式输入：
df1 <- structure(list(Industry = c("Art/Entertainment", "Banking", "Healthcare",
"Education", "Military", "Medicine", "Law", "Computer", "Sales"
), Male = c(100L, 600L, 53L, 20L, 47L, 500L, 500L, 200L, 420L
), Female = c(500L, 100L, 65L, 766L, 96L, 400L, 500L, 144L, 69L
)), class = "data.frame", row.names = c(NA, -9L))
有一个简单的解决方案，使用用于交叉制表目的的库 janitor
library(janitor)
data %>%
adorn_totals(where = c("row","col")) %>%
adorn_percentages(denominator = "row") %>%
adorn_pct_formatting(digits = 0) %>%
adorn_ns(position = "front")
Industry Male Female Total
Art/Entertainment 100 (17%) 500 (83%) 600 (100%)
Banking 600 (86%) 100 (14%) 700 (100%)
Healthcare 53 (45%) 65 (55%) 118 (100%)
Education 20 (3%) 766 (97%) 786 (100%)
Military 47 (33%) 96 (67%) 143 (100%)
Medicine 500 (56%) 400 (44%) 900 (100%)
Law 500 (50%) 500 (50%) 1000 (100%)
Computer 200 (58%) 144 (42%) 344 (100%)
Sales 420 (86%) 69 (14%) 489 (100%)
Total 2440 (48%) 2640 (52%) 5080 (100%)
#OR
data %>%
adorn_percentages(denominator = "row") %>%
adorn_pct_formatting(digits = 2) %>%
adorn_ns(position = "front")
Industry Male Female
Art/Entertainment 100 (16.67%) 500 (83.33%)
Banking 600 (85.71%) 100 (14.29%)
Healthcare 53 (44.92%) 65 (55.08%)
Education 20 (2.54%) 766 (97.46%)
Military 47 (32.87%) 96 (67.13%)
Medicine 500 (55.56%) 400 (44.44%)
Law 500 (50.00%) 500 (50.00%)
Computer 200 (58.14%) 144 (41.86%)
Sales 420 (85.89%) 69 (14.11%)
使用的数据
> data
Industry Male Female
1 Art/Entertainment 100 500
2 Banking 600 100
3 Healthcare 53 65
4 Education 20 766
5 Military 47 96
6 Medicine 500 400
7 Law 500 500
8 Computer 200 144
9 Sales 420 69
我希望按行业的女性和男性百分比来总结我的数据集。我仍在学习 R 并且无法解决这个问题。
我的数据：
| Industry | Male | Female |
|---|---|---|
| Art/Entertainment | 100 | 500 |
| Banking | 600 | 100 |
| Healthcare | 53 | 65 |
| Education | 20 | 766 |
| Military | 47 | 96 |
| Medicine | 500 | 400 |
| Law | 500 | 500 |
| Computer | 200 | 144 |
| Sales | 420 | 69 |
目标：
| Industry | Male | Female | F% | M% |
|---|---|---|---|---|
| Art/Entertainment | 100 | 500 | ||
| Banking | 600 | 100 | ||
| Healthcare | 53 | 65 | ||
| Education | 20 | 766 | ||
| Military | 47 | 96 | ||
| Medicine | 500 | 400 | ||
| Law | 500 | 500 | ||
| Computer | 200 | 144 | ||
| Sales | 420 | 69 |
如果您的数据名为 df，我们可以像这样为男性和女性百分比创建列：
df$Fpct <- df$Female / (df$Male + df$Female)
df$Mpct <- df$Male / (df$Male + df$Female)
注意，不要在变量名中使用 % 符号。
您可以使用 group_by，然后使用 F% 和 M% 创建两个新列
也许你可以使用这个：
library(dplyr)
df %>% group_by(Industry) %>% mutate(F_prec=Female/(Male+Female), M_prec=Male/(Male+Female))
1) proportions 如果您的输入是 df1（在最后的注释中重复显示）然后将列名称更改为所需的名称并将其转换为矩阵 m。最后使用 proportions，边距为 1 表示行比例——2 表示列比例。请注意，我们在第一行转换为矩阵，因为 proportions 需要。
m <- as.matrix(setNames(df1[-1], c("%M", "%F")))
cbind(df1, 100 * proportions(m, 1))
## Industry Male Female %M %F
## 1 Art/Entertainment 100 500 16.666667 83.33333
## 2 Banking 600 100 85.714286 14.28571
## 3 Healthcare 53 65 44.915254 55.08475
## ...snip...
2) rowSums 另一种方法是将 df1[-1] 除以 rowSums 给出相同的结果。
cbind(df1, setNames(100 * df1[-1] / rowSums(df1[-1]), c("%M", "%F")))
## Industry Male Female %M %F
## 1 Art/Entertainment 100 500 16.666667 83.33333
## 2 Banking 600 100 85.714286 14.28571
## 3 Healthcare 53 65 44.915254 55.08475
## ...snip...
3) dplyr 使用 across 复制具有指定名称的列，然后将其乘以 100，然后使用 c_across 除按列的总和
df1 %>%
group_by(Industry) %>%
mutate(100 * across(.names = "%{.col}") / sum(c_across())) %>%
ungroup
## # A tibble: 9 x 5
## Industry Male Female `%Male` `%Female`
## <chr> <int> <int> <dbl> <dbl>
## 1 Art/Entertainment 100 500 16.7 83.3
## 2 Banking 600 100 85.7 14.3
## 3 Healthcare 53 65 44.9 55.1
## ...snip...
4) transform 这个接近另一个答案但它不会覆盖输入：
transform(df1,
"%M" = 100 * Male / (Male + Female),
"%F" = 100 * Female / (Male + Female),
check.names = FALSE)
## Industry Male Female %M %F
## 1 Art/Entertainment 100 500 16.666667 83.33333
## 2 Banking 600 100 85.714286 14.28571
## 3 Healthcare 53 65 44.915254 55.08475
## ...snip...
备注
以可复制的形式输入：
df1 <- structure(list(Industry = c("Art/Entertainment", "Banking", "Healthcare",
"Education", "Military", "Medicine", "Law", "Computer", "Sales"
), Male = c(100L, 600L, 53L, 20L, 47L, 500L, 500L, 200L, 420L
), Female = c(500L, 100L, 65L, 766L, 96L, 400L, 500L, 144L, 69L
)), class = "data.frame", row.names = c(NA, -9L))
有一个简单的解决方案，使用用于交叉制表目的的库 janitor
library(janitor)
data %>%
adorn_totals(where = c("row","col")) %>%
adorn_percentages(denominator = "row") %>%
adorn_pct_formatting(digits = 0) %>%
adorn_ns(position = "front")
Industry Male Female Total
Art/Entertainment 100 (17%) 500 (83%) 600 (100%)
Banking 600 (86%) 100 (14%) 700 (100%)
Healthcare 53 (45%) 65 (55%) 118 (100%)
Education 20 (3%) 766 (97%) 786 (100%)
Military 47 (33%) 96 (67%) 143 (100%)
Medicine 500 (56%) 400 (44%) 900 (100%)
Law 500 (50%) 500 (50%) 1000 (100%)
Computer 200 (58%) 144 (42%) 344 (100%)
Sales 420 (86%) 69 (14%) 489 (100%)
Total 2440 (48%) 2640 (52%) 5080 (100%)
#OR
data %>%
adorn_percentages(denominator = "row") %>%
adorn_pct_formatting(digits = 2) %>%
adorn_ns(position = "front")
Industry Male Female
Art/Entertainment 100 (16.67%) 500 (83.33%)
Banking 600 (85.71%) 100 (14.29%)
Healthcare 53 (44.92%) 65 (55.08%)
Education 20 (2.54%) 766 (97.46%)
Military 47 (32.87%) 96 (67.13%)
Medicine 500 (55.56%) 400 (44.44%)
Law 500 (50.00%) 500 (50.00%)
Computer 200 (58.14%) 144 (41.86%)
Sales 420 (85.89%) 69 (14.11%)
使用的数据
> data
Industry Male Female
1 Art/Entertainment 100 500
2 Banking 600 100
3 Healthcare 53 65
4 Education 20 766
5 Military 47 96
6 Medicine 500 400
7 Law 500 500
8 Computer 200 144
9 Sales 420 69