代码之家 › 专栏 › 技术社区 › Hutch3232

使用dplyr汇总并保持相同的变量名

summarize data.table dplyr variables r

5

Hutch3232 · 技术社区 · 6 年前

我已经找到了这些数据。table和dplyr在尝试执行相同的操作时会产生不同的结果。我想使用dplyr语法,但让它以数据的方式计算。表有。用例是我想将小计添加到表中。要做到这一点,我需要对每个变量进行聚合,但要保持相同的变量名(在转换后的版本中)。数据表允许我对变量执行一些聚合并保持相同的名称。然后使用相同的变量进行另一次聚合。它将继续使用未翻译的版本。然而,Dplyr将使用转换后的版本。

在总结文件说明:

# Note that with data frames, newly created summaries immediately
# overwrite existing variables
mtcars %>%
  group_by(cyl) %>%
  summarise(disp = mean(disp), sd = sd(disp))

这基本上就是我遇到的问题,但我想知道是否有一个很好的解决方法。我发现的一件事是将转换后的变量命名为其他名称 重命名 它在结尾,但这对我来说不是很好。如果有一个很好的方法做小计,那也很高兴知道。我环顾了一下这个网站,没有看到讨论过的确切情况。任何帮助都将不胜感激!

这里我举了一个简单的例子,有一次是关于数据的。我想用这个简单的表并附加一个小计行,它是感兴趣的列(Total)的加权平均数。

library(data.table)
library(dplyr)

dt <- data.table(Group = LETTERS[1:5],
                 Count = c(1000, 1500, 1200, 2000, 5000),
                 Total = c(50, 300, 600, 400, 1000))
dt[, Count_Dist := Count/sum(Count)]
dt[, .(Count_Dist = sum(Count_Dist), Weighted_Total = sum(Count_Dist*Total))]

dt <- rbind(dt[, .(Group, Count_Dist, Total)],
      dt[, .(Group = "All", Count_Dist = sum(Count_Dist), Total = sum(Count_Dist*Total))])
setnames(dt, "Total", "Weighted_Avg_Total")

dt

df <- data.frame(Group = LETTERS[1:5],
                 Count = c(1000, 1500, 1200, 2000, 5000),
                 Total = c(50, 300, 600, 400, 1000))

df %>%
  mutate(Count_Dist = Count/sum(Count)) %>%
  summarize(Count_Dist = sum(Count_Dist),
            Weighted_Total = sum(Count_Dist*Total))

df %>% 
  mutate(Count_Dist = Count/sum(Count)) %>%
  select(Group, Count_Dist, Total) %>% 
  rbind(df %>%
          mutate(Count_Dist = Count/sum(Count)) %>%
          summarize(Group = "All",
                    Count_Dist = sum(Count_Dist),
                    Total = sum(Count_Dist*Total))) %>% 
  rename(Weighted_Avg_Total = Total)

再次感谢您的帮助!

2 回复 | 直到 6 年前

1

3

Jaap 6 年前

一种可能的解决方案是跳过 mutate 步骤和使用 transmute 第一次 变异 / select -步骤并直接从原始变量计算所需变量,而无需为第二个变量创建中间变量 变异 -步骤:

df %>% 
  transmute(Group, Count_Dist = Count/sum(Count), Weighted_Avg_Total = Total) %>% 
  bind_rows(df %>%
              summarize(Group = "All",
                        Count_Dist = sum(Count/sum(Count)),
                        Weighted_Avg_Total = sum((Count/sum(Count))*Total)))

其中给出:

  Group Count_Dist Weighted_Avg_Total
1     A 0.09345794            50.0000
2     B 0.14018692           300.0000
3     C 0.11214953           600.0000
4     D 0.18691589           400.0000
5     E 0.46728972          1000.0000
6   All 1.00000000           656.0748

另一种可能的解决方案是改变计算新变量的顺序 dplyr 然后使用 选择 要将列顺序恢复为您最初想要的顺序,请执行以下操作:

df %>% 
  mutate(Count_Dist = Count/sum(Count)) %>%
  select(Group, Count_Dist, Weighted_Avg_Total = Total) %>% 
  bind_rows(df %>%
              mutate(Count_Dist = Count/sum(Count)) %>%
              summarize(Group = "All",
                        Weighted_Avg_Total = sum(Count_Dist*Total),
                        Count_Dist = sum(Count_Dist)) %>% 
              select(Group, Count_Dist, Weighted_Avg_Total))

如果要包括 Count -专栏,您也可以这样做(根据我下面的评论):

df %>% 
  transmute(Group = Group, Count_Dist = Count/sum(Count), Weighted_Avg_Total = Total, Count) %>% 
  bind_rows(df %>%
              summarize(Group = "All",
                        Count_Dist = sum(Count/sum(Count)),
                        Weighted_Avg_Total = sum((Count/sum(Count))*Total),
                        Count = sum(Count)))

2

1

MKR 6 年前

一种替代方法是 mutate 两次计算偶数 Weighted_Total 和使用 sum 中该列的 summarize .

df %>%
  mutate(Count_Dist = Count/sum(Count)) %>%
  mutate(Weighted_Total = Count_Dist*Total) %>%
  summarize(Count_Dist = sum(Count_Dist),
            Weighted_Total = sum(Weighted_Total))
Result:
  Count_Dist Weighted_Total
1          1     656.074766

以及:

    df %>% 
      mutate(Count_Dist = Count/sum(Count)) %>%
      select(Group, Count_Dist, Total) %>% 
      rbind(df %>%
              mutate(Count_Dist = Count/sum(Count)) %>%
              mutate(Weighted_Total = Count_Dist*Total) %>%
              summarize(Group = "All",
                        Count_Dist = sum(Count_Dist),
                        Total = sum(Weighted_Total))) %>% 
      rename(Weighted_Avg_Total = Total)

Result:

      Group   Count_Dist Weighted_Avg_Total
    1     A 0.0934579439          50.000000
    2     B 0.1401869159         300.000000
    3     C 0.1121495327         600.000000
    4     D 0.1869158879         400.000000
    5     E 0.4672897196        1000.000000
    6   All 1.0000000000         656.074766