如何将一列中的值复制到R中不同列中具有相同值的行？ [复制]答案

【问题标题】：How can copy a value in one column to rows with the same value in a different column in R? [duplicate]如何将一列中的值复制到R中不同列中具有相同值的行？ [复制]
【发布时间】：2021-03-01 23:15:33
【问题描述】：

如果我有这样的事情：

id date        type     location
1  1/2/2021    student   Georgia
1                        Maine
1                        New York
2  2/20/2021   teacher   Florida 
2                        Louisiana 
3  2/15/2021   student   New York
3                        Maine
3                        California
3                        Florida

我想根据“id”的值复制“日期”和“类型”列中的值：

id date        type     location
1  1/2/2021    student   Georgia
1  1/2/2001    student   Maine
1  1/2/2001    student   New York
2  2/20/2021   teacher   Florida 
2  2/20/2021   teacher   Louisiana 
3  2/15/2021   student   New York
3  2/15/2021   student   Maine
3  2/15/2021   student   California
3  2/15/2021   student   Florida

我知道这可能已经问过了，但我不知道如何问这个问题。感谢您的帮助！

【问题讨论】：

标签： r

【解决方案1】：

我们可以将空白"" 转换为NA 并使用fill

library(dplyr)
library(tidyr)
df1 %>%
    mutate(across(c(date, type), na_if, "")) %>%
    group_by(id) %>%
    fill(date, type)

或者另一种选择是通过'id'将其更改为first元素

df %>%
   group_by(id) %>%
   mutate(across(c(date, type), first))

【讨论】：

【解决方案2】：

这是一个data.table 选项

> setDT(df)[, lapply(.SD, function(x) replace(x, is.na(x), na.omit(x))), id]
   id      date    type   location
1:  1  1/2/2021 student    Georgia
2:  1  1/2/2021 student      Maine
3:  1  1/2/2021 student   New York
4:  2 2/20/2021 teacher    Florida
5:  2 2/20/2021 teacher  Louisiana
6:  3 2/15/2021 student   New York
7:  3 2/15/2021 student      Maine
8:  3 2/15/2021 student California
9:  3 2/15/2021 student    Florida

数据

> dput(df)
structure(list(id = c(1L, 1L, 1L, 2L, 2L, 3L, 3L, 3L, 3L), date = c("1/2/2021",
NA, NA, "2/20/2021", NA, "2/15/2021", NA, NA, NA), type = c("student",
NA, NA, "teacher", NA, "student", NA, NA, NA), location = c("Georgia",
"Maine", "New York", "Florida", "Louisiana", "New York", "Maine",
"California", "Florida")), class = "data.frame", row.names = c(NA,
-9L))

【讨论】：