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在R中总结组意味时如何有条件地创建新组

tibble data.table dplyr r

TimTeaFan · 技术社区 · 7 年前

我有一些数据,我想对这些数据进行总结。然后我想重新分组一些较小的组(匹配某个n<x条件)分为一组,称为“其他”。我找到了一个办法。但感觉外面有更有效的解决方案。我想知道data.table方法如何解决这个问题。

# preps
library(tibble)
library(dplyr)
set.seed(7)

# generate 4 groups with more observations
tbl_1  <- tibble(group = rep(sample(letters[1:4], 150, TRUE), each = 4),
                 score = sample(0:10, size = 600, replace = TRUE))

# generate 3 groups with less observations
tbl_2 <- tibble(group = rep(sample(letters[5:7], 50, TRUE), each = 3),
                score = sample(0:10, size = 150, replace = TRUE)) 

# put them into one data frame
tbl <- rbind(tbl_1, tbl_2)

# aggregate the mean scores and count the observations for each group
tbl_agg1 <- tbl %>%
  group_by(group) %>%
  summarize(MeanScore = mean(score),
            n = n())

到目前为止很容易。

# First, calculate summary stats for groups less then n < 100
tbl_agg2 <- tbl_agg1 %>%
   filter(n<100) %>%
      summarize(MeanScore = weighted.mean(MeanScore, n),
                sumN = sum(n))

注:以上计算有误,现已更正(@Frank:谢谢你发现!)

# Second, delete groups less then n < 100 from the aggregate table and add a row containing the summary statistics calculated above instead
tbl_agg1 <- tbl_agg1 %>%
   filter(n>100) %>%
      add_row(group = "others", MeanScore = tbl_agg2[["MeanScore"]], n = tbl_agg2[["sumN"]])

tbl\u agg1基本上显示了我希望它显示的内容,但我想知道是否有一种更平滑、更有效的方法来实现这一点。同时,我想知道data.table方法如何处理手头的问题。

我欢迎任何建议。

1 回复 | 直到 7 年前

Frank 7 年前

你对“其他”组的计算是错误的,我猜。。。应该是。。。

tbl_agg1 %>% {bind_rows(
   filter(., n>100),
   filter(., n<100) %>%
   summarize(group = "other", MeanScore = weighted.mean(MeanScore, n), n = sum(n))
)}

但是,通过使用不同的分组变量,可以从一开始就让事情简单得多:

tbl %>% 
  group_by(group) %>% 
  group_by(g = replace(group, n() < 100, "other")) %>% 
  summarise(n = n(), m = mean(score))

# A tibble: 5 x 3
  g         n     m
  <chr> <int> <dbl>
1 a       136  4.79
2 b       188  4.49
3 c       160  5.32
4 d       116  4.78
5 other   150  5.42

或使用data.table

library(data.table)
DT = data.table(tbl)
DT[, n := .N, by=group]
DT[, .(.N, m = mean(score)), keyby=.(g = replace(group, n < 100, "other"))]

       g   N        m
1:     a 136 4.786765
2:     b 188 4.489362
3:     c 160 5.325000
4:     d 116 4.784483
5: other 150 5.420000