Question

我有以下数据框作为示例

   df <- data.frame(score=letters[1:15], total1=1:15, total2=16:30)
> df
   score total1 total2
1      a      1     16
2      b      2     17
3      c      3     18
4      d      4     19
5      e      5     20
6      f      6     21
7      g      7     22
8      h      8     23
9      i      9     24
10     j     10     25
11     k     11     26
12     l     12     27
13     m     13     28
14     n     14     29
15     o     15     30

我想通过对具有不同名称的行进行分组来总结aggregate我的数据框，即

  groups  sum1 sum2
 'a-b-c'  6     51
 'c-d-e'  21    60
etc

这类问题的所有给定答案都假设字符串在行中重复。

我用来获取摘要的常用aggregate函数会产生不同的结果：

aggregate(df$total1, by=list(sum1=df$score %in% c('a','b','c'), sum2=df$score %in% c('d','e','f')), FUN=sum)
   sum1  sum2  x
1 FALSE FALSE 99
2  TRUE FALSE  6
3 FALSE  TRUE 15

Answer 1

如果你想要一个tidyverse解决方案，这里有一种可能性：

df <- data.frame(score=letters[1:15], total1=1:15, total2=16:30)

df %>%
  mutate(groups = case_when(
    score %in% c("a","b","c") ~ "a-b-c",
    score %in% c("d","e","f") ~ "d-e-f"
  )) %>%
  group_by(groups) %>%
  summarise_if(is.numeric, sum)

返回

# A tibble: 3 x 3
  groups total1 total2
  <chr>   <int>  <int>
1 a-b-c       6     51
2 d-e-f      15     60
3 <NA>       99    234

Answer 2

添加＆＃34;群组＆＃34;具有类别值的列。

df$groups = NA

然后像这样定义每个组：

df$groups[df$score=="a" | df$score=="b" | df$score=="c" ] = "a-b-c"

最后按该列汇总。

Answer 3

这是适用于任何规模数据框架的解决方案。

df <- data.frame(score=letters[1:15], total1=1:15, total2=16:30)

# I'm adding a row to demonstrate that the grouping pattern works when the 
# number of rows is not equally divisible by 3.
df <- rbind(df, data.frame(score = letters[16], total1 = 16, total2 = 31))

# A vector that represents the correct groupings for the data frame.
groups <- c(rep(1:floor(nrow(df) / 3), each = 3), 
            rep(floor(nrow(df) / 3) + 1, nrow(df) - length(1:(nrow(df) / 3)) * 3))

# Your method of aggregation by `groups`. I'm going to use `data.table`.
require(data.table)
dt <- as.data.table(df)
dt[, group := groups]

aggDT <- dt[, list(score = paste0(score, collapse = "-"), 
          total1 = sum(total1), total2 = sum(total2)), by = group][
            , group := NULL]
aggDT

   score total1 total2
1: a-b-c      6     51
2: d-e-f     15     60
3: g-h-i     24     69
4: j-k-l     33     78
5: m-n-o     42     87
6:     p     16     31

在3行的滚动块中聚合数据帧

3 个答案: