Question

我发现data.table和dplyr在尝试做同样的事情时会有不同的结果。我想使用dplyr语法，但让它以data.table的方式进行计算。用例是我想在表格中添加小计。为此，我需要对每个变量进行一些聚合，但是保留相同的变量名称（在转换后的版本中）。 Data.table允许我对变量执行一些聚合并保持相同的名称。然后用同一个变量做另一个聚合。它将继续使用未转换的版本。但是，Dplyr将使用转换后的版本。

在摘要文档中，它说：

# 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。我发现的一件事就是将变换后的变量命名为其他东西，然后在最后重命名它，但这对我来说并不是很好。如果有一个很好的方法来做小计，那也很好。我环顾了这个网站，没有看到这个确切的情况。任何帮助将不胜感激！

这里我做了一个简单的例子，一次使用data.table的结果，一次使用dplyr。我想采用这个简单的表并附加一个小计行，它是感兴趣的列的加权平均值（总计）。

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！

Answer 1

一种可能的解决方案是跳过mutate步骤并使用transmute作为第一个mutate / select步骤，直接从原始变量计算所需的变量为第二个mutate创建一个中间变量 - 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)))

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在dplyr中计算新变量的顺序，然后使用select将列顺序恢复到您最初想要的位置：

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))

如果您想要包含Count - 列，您也可以（根据我在下面的评论）：

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)))

Answer 2

另一种方法是使用${...}两次来计算mutate甚至Weighted_Total，并使用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

和

    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

使用dplyr汇总并保持相同的变量名称

2 个答案: