按子组汇总 dplyr答案

【问题标题】：summarise by subgroups dplyr按子组汇总 dplyr
【发布时间】：2020-04-13 11:09:15
【问题描述】：

我对 dyplr 还很陌生，我正在努力使用 summarise 函数来获得像比例一样简单的东西。

我的数据集的一部分

city = rep(c("A","B"), each = 5)
month = c("Jan","Jan","Jan","Feb","Feb","Jan","Jan","Feb","Feb","Feb")
species = c("aegypti","gambiae","quinque","aegypti","quinque",
            "aegypti","gambiae","quinque","aegypti","quinque")
total = c(20, 5, 25, 12, 23, 10, 10, 0, 20, 0)
df = data.frame(city, month, species, total)

然后我要计算每个城市，每个月每个物种的比例。我可以计算每个城市和每个月的总数

df %>% group_by(city, month) %>% summarise(total = sum(total))

# A tibble: 4 x 3
# Groups:   city [2]
#  city  month total
#  <fct> <fct> <dbl>
#1 A     Feb      35
#2 A     Jan      50
#3 B     Feb      20
#4 B     Jan      20

那我可以按物种分别计算

df %>% group_by(city, month, species) %>% summarise(total = sum(total)

# A tibble: 9 x 4
# Groups:   city, month [4]
#  city  month species total
#  <fct> <fct> <fct>   <dbl>
#1 A     Feb   aegypti    12
#2 A     Feb   quinque    23
#3 A     Jan   aegypti    20
#4 A     Jan   gambiae     5
#5 A     Jan   quinque    25
#6 B     Feb   aegypti    20
#7 B     Feb   quinque     0
#8 B     Jan   aegypti    10
#9 B     Jan   gambiae    10

但我需要这样的东西来计算比例。我希望表格看起来与此类似。

# A tibble: 9 x 4
# Groups:   city, month [4]
#  city  month species total total1 prop
#  <fct> <fct> <fct>   <dbl> <dbl>  <dbl>
#1 A     Feb   aegypti    12   35    34.29
#2 A     Feb   quinque    23   35    65.71
#3 A     Jan   aegypti    20   50    40.00
#4 A     Jan   gambiae     5   50    10.00
#5 A     Jan   quinque    25   50    50.00
#6 B     Feb   aegypti    20   20   100.00
#7 B     Feb   quinque     0   20     0
#8 B     Jan   aegypti    10   20    50.00
#9 B     Jan   gambiae    10   20    50.00

我一直在玩，但我找不到这样做的方法.. 谢谢！

【问题讨论】：

标签： r group-by dplyr frequency

【解决方案1】：

group_bycity 和month 并在组中将total 除以sum 以获得比例。

library(dplyr)
df %>% group_by(city, month) %>% mutate(total = total/sum(total) * 100)

#  city  month species total
#   <fct> <fct> <fct>   <dbl>
# 1 A     Jan   aegypti  40  
# 2 A     Jan   gambiae  10  
# 3 A     Jan   quinque  50  
# 4 A     Feb   aegypti  34.3
# 5 A     Feb   quinque  65.7
# 6 B     Jan   aegypti  50  
# 7 B     Jan   gambiae  50  
# 8 B     Feb   quinque   0  
# 9 B     Feb   aegypti 100  
#10 B     Feb   quinque   0

或者在基础 R 中：

df$total <- with(df, total/ave(total, city, month, FUN = sum))

【讨论】：

谢谢，然后我按物种分组并汇总(sum(total))，这样我就可以将这些月份与重复的物种行合并。

【解决方案2】：

我们可以使用data.table

library(data.table)
setDT(df)[, total := total/sum(total) * 100, .(city, month)]

【讨论】：