如何在R中将新行添加到分组数据框中？

xvw2m8pv 于 2023-10-13 发布在其他

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我想知道是否有一种方法可以基于两个变量对数据框进行分组，然后在分组数据中不存在某个字符串（如名称）时添加新行。
我有一个虚拟名称列表（name_list），我希望确保将其包含在分组数据中，而不管它们是否首先出现在那里（基于name列）。例如，我想按date和drill分组，但您会注意到2023-01-01/Activity 1中缺少Person A和Person B。我想确保date/drill的每个组合都包含了name_list中的每个名称，如果添加了名称，则将它们分配为0的duration，以基本上表示它们不存在于date/drill组合中。希望你说得通。谢谢.

library(tidyverse)
# List of names to be joined within grouping variables.
name_list <- c(paste("Person", LETTERS[1:8]))
set.seed(10)
name <- c(name_list[3:8], name_list[1:6], name_list[2:7], name_list[3:8])
date <- rep(seq(as.Date('2023/01/01'), as.Date('2023/01/02'), by = "day"),
            each = 12)
drill <- rep(paste("Activity", 1:2), each = 6, times = 2)
duration <- rep(c(5, 8), each = 6, times = 2)
df <- data.frame(name, date, drill, duration)
       name       date      drill duration
1  Person C 2023-01-01 Activity 1        5
2  Person D 2023-01-01 Activity 1        5
3  Person E 2023-01-01 Activity 1        5
4  Person F 2023-01-01 Activity 1        5
5  Person G 2023-01-01 Activity 1        5
6  Person H 2023-01-01 Activity 1        5
7  Person A 2023-01-01 Activity 2        8
8  Person B 2023-01-01 Activity 2        8
9  Person C 2023-01-01 Activity 2        8
10 Person D 2023-01-01 Activity 2        8
11 Person E 2023-01-01 Activity 2        8
12 Person F 2023-01-01 Activity 2        8
13 Person B 2023-01-02 Activity 1        5
14 Person C 2023-01-02 Activity 1        5
15 Person D 2023-01-02 Activity 1        5
16 Person E 2023-01-02 Activity 1        5
17 Person F 2023-01-02 Activity 1        5
18 Person G 2023-01-02 Activity 1        5
19 Person C 2023-01-02 Activity 2        8
20 Person D 2023-01-02 Activity 2        8
21 Person E 2023-01-02 Activity 2        8
22 Person F 2023-01-02 Activity 2        8
23 Person G 2023-01-02 Activity 2        8
24 Person H 2023-01-02 Activity 2        8

r

来源：https://stackoverflow.com/questions/77218473/how-can-i-add-new-rows-into-a-grouped-data-frame-in-r

1条答案

按热度按时间

xwbd5t1u1#

我想你正在寻找complete()功能。

编辑：

正如@LMc所提到的，只有当name的所有值都已经存在于数据中时，使用complete才有效。将name列设置为包含所有可能名称name_list的因子可以解决此问题。

df %>% 
        mutate(name = factor(name, levels = name_list)) %>% 
        complete(name, date, drill, fill = list(duration = 0))
# A tibble: 32 × 4
   name     date       drill      duration
   <fct>    <date>     <chr>         <dbl>
 1 Person A 2023-01-01 Activity 1        0
 2 Person A 2023-01-01 Activity 2        8
 3 Person A 2023-01-02 Activity 1        0
 4 Person A 2023-01-02 Activity 2        0
 5 Person B 2023-01-01 Activity 1        0
 6 Person B 2023-01-01 Activity 2        8
 7 Person B 2023-01-02 Activity 1        5
 8 Person B 2023-01-02 Activity 2        0
 9 Person C 2023-01-01 Activity 1        5
10 Person C 2023-01-01 Activity 2        8
11 Person C 2023-01-02 Activity 1        5
12 Person C 2023-01-02 Activity 2        8
13 Person D 2023-01-01 Activity 1        5
14 Person D 2023-01-01 Activity 2        8
15 Person D 2023-01-02 Activity 1        5
16 Person D 2023-01-02 Activity 2        8
17 Person E 2023-01-01 Activity 1        5
18 Person E 2023-01-01 Activity 2        8
19 Person E 2023-01-02 Activity 1        5
20 Person E 2023-01-02 Activity 2        8
21 Person F 2023-01-01 Activity 1        5
22 Person F 2023-01-01 Activity 2        8
23 Person F 2023-01-02 Activity 1        5
24 Person F 2023-01-02 Activity 2        8
25 Person G 2023-01-01 Activity 1        5
26 Person G 2023-01-01 Activity 2        0
27 Person G 2023-01-02 Activity 1        5
28 Person G 2023-01-02 Activity 2        8
29 Person H 2023-01-01 Activity 1        5
30 Person H 2023-01-01 Activity 2        0
31 Person H 2023-01-02 Activity 1        0
32 Person H 2023-01-02 Activity 2        8

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如何在R中将新行添加到分组数据框中？

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