bind_rows() 在 R 中绑定每个数据帧时创建它们的副本

Question

假设这是我的数据框：

df <- data.frame(grp = c("ab -10", "ab 0", "ab 8", "ab -1",
                         "ab 6", "ab 6", "ab -10", "ab 1",
                         "ab -10", "ab 0", "ab 8", "ab -1",
                         "ab 6", "ab 6", "ab -10", "ab 1",
                         "d", "e", "e", "e"),
                 freq = c(1,0,0,1,0,1,2,0,1,0,2,2,1,1,0,1,0,2,2,1))
df
      grp freq
1  ab -10    1
2    ab 0    0
3    ab 8    0
4   ab -1    1
5    ab 6    0
6    ab 6    1
7  ab -10    2
8    ab 1    0
9  ab -10    1
10   ab 0    0
11   ab 8    2
12  ab -1    2
13   ab 6    1
14   ab 6    1
15 ab -10    0
16   ab 1    1
17      d    0
18      e    2
19      e    2
20      e    1

我想要：

> finaldf
     grp freq
1 ab < 0    7
2 ab 0-5    1
3  ab 5+    5
4      d    0
5      e    5

这是我试过的：

df %>%
  bind_rows(df %>%
              filter(!grepl("ab", grp)),
            
            df %>%
              filter(grepl("ab", grp)) %>%
              mutate(grp = parse_number(grp)) %>%
              mutate(grp = cut(as.numeric(grp),
                                          breaks = c(-999, 0, 6, 999),
                                          labels = c("ab < 0", "ab 0-5", "ab 5+"),
                                          right = F))) %>%
              group_by(grp) %>%
              summarise(N =n())

但似乎 bind_rows 正在复制数据帧。

      grp freq
1  ab -10    1
2    ab 0    0
3    ab 8    0
4   ab -1    1
5    ab 6    0
6    ab 6    1
7  ab -10    2
8    ab 1    0
9  ab -10    1
10   ab 0    0
11   ab 8    2
12  ab -1    2
13   ab 6    1
14   ab 6    1
15 ab -10    0
16   ab 1    1
17      d    0
18      e    2
19      e    2
20      e    1
21      d    0
22      e    2
23      e    2
24      e    1
25 ab < 0    1
26 ab 0-5    0
27  ab 5+    0
28 ab < 0    1
29  ab 5+    0
30  ab 5+    1
31 ab < 0    2
32 ab 0-5    0
33 ab < 0    1
34 ab 0-5    0
35  ab 5+    2
36 ab < 0    2
37  ab 5+    1
38  ab 5+    1
39 ab < 0    0
40 ab 0-5    1

我可以 slice() 一半的行，但我更想知道我做错了什么？

任何其他简洁漂亮的方法也非常感谢！

Answer 1

这是一种方法，将列拆分为 'two' 和 separate，重新编码数值，unite，然后按 sum[= 进行分组21=]

library(dplyr)
library(tidyr)
df %>% 
  separate(grp, into = c('grp1', 'value'), sep = "(?<=ab)\s+",
   fill = "right", convert = TRUE) %>% 
  mutate(value = case_when(value <0 ~ '< 0', 
     between(value, 0, 5) ~ '0-5', value > 5 ~ '5+')) %>%
  unite(grp, grp1, value, na.rm = TRUE, sep=" ") %>% 
  group_by(grp) %>%
  summarise(freq = sum(freq), .groups = 'drop')

-输出

# A tibble: 5 × 2
  grp     freq
  <chr>  <dbl>
1 ab < 0     7
2 ab 0-5     1
3 ab 5+      5
4 d          0
5 e          5

在 OP 的代码中，这是开头 df %>% 需要删除，因为我们要在 bind_rows 中传递两个 filtered 数据集。当我们添加 df %>% 时，它将作为第一个参数传递给 bind_rows，从而复制行

library(readr)
bind_rows(df %>%
              filter(!grepl("ab", grp)),
            df %>%
              filter(grepl("ab", grp)) %>%
              mutate(grp = parse_number(grp)) %>%
              mutate(grp = cut(as.numeric(grp),
                                          breaks = c(-999, 0, 6, 999),
                                          labels = c("ab < 0", "ab 0-5", "ab 5+"),
                                          right = FALSE))) %>% 
    group_by(grp) %>%
    summarise(N =sum(freq))
# A tibble: 5 × 2
  grp        N
  <chr>  <dbl>
1 ab < 0     7
2 ab 0-5     1
3 ab 5+      5
4 d          0
5 e          5

Answer 2

tidyverse 解决方案 count() + str_replace():

library(dplyr)
library(stringr)

df %>%
  count(grp = str_replace(grp, "-?\d+", function(x) {
    x <- as.numeric(x)
    case_when(x < 0 ~ '< 0', between(x, 0, 5) ~ '0-5', x > 5 ~ '5+')
  }), wt = freq)

#      grp n
# 1 ab < 0 7
# 2 ab 0-5 1
# 3  ab 5+ 5
# 4      d 0
# 5      e 5

诀窍是将一个函数传递给 str_replace()：它会为每个匹配项调用一次，它的 return 值将用于替换匹配项。另外，df %>% count(a, wt = b)是df %>% group_by(a) %>% summarise(n = sum(b))的快捷方式。

bind_rows() 在 R 中绑定每个数据帧时创建它们的副本

bind_rows() creates duplicate of each dataframe when binding them in R

r

dataframe

dplyr

rows