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在dplyr::group_by中,获取多个分组变量之一的观测次数

group-by dplyr r

0

Eliot Dixon tmfmnk · 技术社区 · 3 年前

这很可能以前有人问过,但我很难阐明我的问题。

在我的数据中,我有3个变量,LOCATION、TOPIC和RESPONSE。我想按位置计算TOPIC和RESPONSE的每个组合的分布。

创建玩具数据并执行初始数据准备

responses <- data.frame(LOCATION = c("LOC_A", "LOC_A", "LOC_A", "LOC_A", "LOC_A", 
                                     "LOC_A", "LOC_A", "LOC_A", 
                                     "LOC_B", "LOC_B", "LOC_B", "LOC_B", "LOC_B", 
                                     "LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C", 
                                     "LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C", 
                                     "LOC_C", "LOC_C", "LOC_C", "LOC_C", "LOC_C"),
                        TOPIC = c("Dogs", "Dogs", "Dogs", "Dogs", "Dogs", "Dogs", 
                                  "Lizards", "Lizards", "Lizards",
                                  "Lizards", "Lizards", "Lizards", "Lizards", "Lizards", 
                                   "Lizards", "Lizards", "Snakes", "Snakes", "Snakes", "Snakes", "Snakes", 
                                  "Snakes", "Dogs", "Snakes", "Dogs", "Snakes", "Dogs", 
                                  "Snakes", "Dogs", "Snakes"),
                        RESP = c("Agree", "Disagree", "Agree", "Disagree", "Agree", 
                                 "Disagree", "Agree", "Disagree", 
                                 "Agree", "Disagree", "Agree", "Disagree", "Neither", "Agree",
                                 "Neither", "Agree", "Neither", "Agree", "Neither", 
                                 "Agree", "Neither", "Agree", "Agree", "Neither", 
                                 "Agree", "Neither", "Agree", "Disagree", "Disagree",
                                 "Neither"))

# Obtain counts for each combination of levels
distribution <- responses %>% 
  table() %>% 
  as.data.frame() %>% 
  # Make it more readable
  dplyr::arrange(LOCATION, TOPIC, RESP)

下面是一个使用循环创建所需输出的示例解决方案:

# ugly loop solution :(
# Initialize output container
out <- list()
# Iterate over each location
for(loc in unique(distribution$LOCATION)){
  # Subset distribution for this location
  thisDist <- dplyr::filter(distribution, LOCATION == loc)
  
  # Calculate percent of each response for this location
  thisDist$percent <- thisDist$Freq/sum(thisDist$Freq)
  
  # Store distribution df with percent column
  out[[loc]] <- thisDist

}

# combine output into single df
out <- do.call("rbind", out)

我想要的是一个简洁明了的解决方案。下面是一些伪代码,它描述了我想象中的解决方案。

# Imaginary tidyverse solution :)
out <- distribution %>% 
  group_by(LOCATION, TOPIC, RESP) %>% 
  summarise(#percent = Freq/(sum(<all-Freq-values-for-this-group's-LOCATION-value>))
            )

我在这里要做的是获得当前组的LOCATION值的所有Freq值的总和。有没有一种很好的方法可以在一个小组中做到这一点?

谢谢你的阅读,我希望这不是完全无法理解的。

2 回复 | 直到 3 年前

1

r2evans 3 年前

这就是你要找的吗?

distribution %>%
  mutate(percent = Freq/sum(Freq), .by = LOCATION)
#    LOCATION   TOPIC     RESP Freq    percent
# 1     LOC_A    Dogs    Agree    3 0.37500000
# 2     LOC_A    Dogs Disagree    3 0.37500000
# 3     LOC_A    Dogs  Neither    0 0.00000000
# 4     LOC_A Lizards    Agree    1 0.12500000
# 5     LOC_A Lizards Disagree    1 0.12500000
# 6     LOC_A Lizards  Neither    0 0.00000000
# 7     LOC_A  Snakes    Agree    0 0.00000000
# 8     LOC_A  Snakes Disagree    0 0.00000000
# 9     LOC_A  Snakes  Neither    0 0.00000000
# 10    LOC_B    Dogs    Agree    0 0.00000000
# 11    LOC_B    Dogs Disagree    0 0.00000000
# 12    LOC_B    Dogs  Neither    0 0.00000000
# 13    LOC_B Lizards    Agree    2 0.40000000
# 14    LOC_B Lizards Disagree    2 0.40000000
# 15    LOC_B Lizards  Neither    1 0.20000000
# 16    LOC_B  Snakes    Agree    0 0.00000000
# 17    LOC_B  Snakes Disagree    0 0.00000000
# 18    LOC_B  Snakes  Neither    0 0.00000000
# 19    LOC_C    Dogs    Agree    3 0.17647059
# 20    LOC_C    Dogs Disagree    1 0.05882353
# 21    LOC_C    Dogs  Neither    0 0.00000000
# 22    LOC_C Lizards    Agree    2 0.11764706
# 23    LOC_C Lizards Disagree    0 0.00000000
# 24    LOC_C Lizards  Neither    1 0.05882353
# 25    LOC_C  Snakes    Agree    3 0.17647059
# 26    LOC_C  Snakes Disagree    1 0.05882353
# 27    LOC_C  Snakes  Neither    6 0.35294118

如果你有 dplyr 1.1以上,然后使用

distribution %>%
  group_by(LOCATION) %>%
  mutate(percent = Freq/sum(Freq))

2

1

William Wong 3 年前

钥匙不用 summarise 但是 mutate 。

out <- distribution %>% 
  ungroup() %>% 
  group_by(LOCATION) %>% 
  mutate(percent = Freq/ sum(Freq))