按两列分组并汇总多列答案

【问题标题】：Group by two column and summarize multiple columns按两列分组并汇总多列
【发布时间】：2020-09-10 20:20:38
【问题描述】：

我有一个数据框，我想按“状态”和“日期”列进行分组，然后像这样总结其他列的值。

State  Female  Male   Date
------------------------------
Texas  2       2     01/01/04
Texas  3        1     01/01/04
Texas  5        4     02/01/04
Cali   1        1     05/06/05
Cali   2        1     05/06/05
Cali   3         1    10/06/05
Cali   1         2     10/06/05
NY    10         5    11/06/05
NY    11         6    12/06/05

预期结果

State  Female  Male   Date
------------------------------
Texas  5       3     01/01/04
Texas  5        4     02/01/04
Cali   3        2     05/06/05
Cali   4         3    10/06/05
NY    10         5    11/06/05
NY    11         6    12/06/05

我尝试使用 group by 和 summarise，但我不知道我如何为 2 列做同样的事情

我的尝试

df <- df_homicides %>% 
        group_by(state) %>% 
        summarise(Female = sum(Female))

``
Thanks for your help!

【问题讨论】：

标签： r dataframe dplyr tidyverse

【解决方案1】：

我们可以使用summarise 和across 的dplyr 版本> = 1.00

library(dplyr)
df %>%
   group_by(State, Date) %>%
   summarise(across(everything(), sum, na.rm = TRUE), .groups = 'drop')
# A tibble: 6 x 4
#  State Date       Female  Male
#  <chr> <chr>       <int> <int>
#1 Cali  05/06/2005      3     2
#2 Cali  10/06/2005      4     3
#3 NY    11/06/2005     10     5
#4 NY    12/06/2005     11     6
#5 Texas 01/01/2004      5     3
#6 Texas 02/01/2004      5     4

或者使用来自base R的aggregate

aggregate(.~ State + Date, df, sum, na.rm = TRUE)

数据

df <-  structure(list(State = c("Texas", "Texas", "Texas", "Cali", "Cali", 
"Cali", "Cali", "NY", "NY"), Female = c(2L, 3L, 5L, 1L, 2L, 3L, 
1L, 10L, 11L), Male = c(2L, 1L, 4L, 1L, 1L, 1L, 2L, 5L, 6L), 
    Date = c("01/01/2004", "01/01/2004", "02/01/2004", "05/06/2005", 
    "05/06/2005", "10/06/2005", "10/06/2005", "11/06/2005", "12/06/2005"
    )), class = "data.frame", row.names = c(NA, -9L))

【讨论】：

【解决方案2】：

试试这个。您可以使用summarise_all() 将多个变量聚合到所需的函数。代码如下：

library(dplyr)
#Code
df %>% group_by(State,Date) %>%
  summarise_all(.funs = sum,na.rm=T)

输出：

# A tibble: 6 x 4
# Groups:   State [3]
  State Date       Female  Male
  <chr> <chr>       <int> <int>
1 Cali  05/06/2005      3     2
2 Cali  10/06/2005      4     3
3 NY    11/06/2005     10     5
4 NY    12/06/2005     11     6
5 Texas 01/01/2004      5     3
6 Texas 02/01/2004      5     4

使用的一些数据：

#Data
df <- structure(list(State = c("Texas", "Texas", "Texas", "Cali", "Cali", 
"Cali", "Cali", "NY", "NY"), Female = c(2L, 3L, 5L, 1L, 2L, 3L, 
1L, 10L, 11L), Male = c(2L, 1L, 4L, 1L, 1L, 1L, 2L, 5L, 6L), 
    Date = c("01/01/2004", "01/01/2004", "02/01/2004", "05/06/2005", 
    "05/06/2005", "10/06/2005", "10/06/2005", "11/06/2005", "12/06/2005"
    )), class = "data.frame", row.names = c(NA, -9L))

【讨论】：