在 R 中扩展电子邮件数据集的标题
Expand tibble of email dataset in R
我有大量电子邮件数据,如下所示:
library(dplyr)
emails <- tibble(
from = c('employee.1@xtra.co','employee.5@xtra.co','employee.1@xtra.co',
'employee.3@xtra.co','employee.1@xtra.co'),
to = list(
c('employee.5@xtra.co', 'employee.3xtra.co'),
c('employee.3@xtra.co', 'employee.1@xtra.co'),
c('employee.2@xtra.co'),
c('employee.1@xtra.co'),
c('employee.3@xtra.co','employee.5@xtra.co','employee.6@xtra.co')),
cc = list(
c('employee.2xtra.co', 'employee.4xtra.co', 'employee.6xtra.co'),
c('employee.1xtra.co', 'employee.8xtra.co', 'employee.6xtra.co'),
NA,
c('employee.2xtra.co', 'employee.4xtra.co'),
c('employee.2xtra.co', 'employee.6xtra.co'))
)
emails
# A tibble: 5 x 3
from to cc
<chr> <list> <list>
1 employee.1@xtra.co <chr [2]> <chr [3]>
2 employee.5@xtra.co <chr [2]> <chr [3]>
3 employee.1@xtra.co <chr [1]> <lgl [1]>
4 employee.3@xtra.co <chr [1]> <chr [2]>
5 employee.1@xtra.co <chr [3]> <chr [2]>
我需要你的帮助才能为每个组合扩展每条记录。例如,我想为第 1 行实现的是:
from to cc
employee.1@xtra.co employee.5@xtra.co employee.2xtra.co
employee.1@xtra.co employee.5@xtra.co employee.4xtra.co
employee.1@xtra.co employee.5@xtra.co employee.6xtra.co
employee.1@xtra.co employee.3xtra.co employee.2xtra.co
employee.1@xtra.co employee.3xtra.co employee.4xtra.co
employee.1@xtra.co employee.3xtra.co employee.6xtra.co
非常感谢您的宝贵时间。
我们可以申请unnest
两次。
library(dplyr)
library(tidyr)
emails2 <- emails %>%
unnest(cols = "to") %>%
unnest(cols = "cc")
head(emails2)
# # A tibble: 6 x 3
# from to cc
# <chr> <chr> <chr>
# 1 employee.1@xtra.co employee.5@xtra.co employee.2xtra.co
# 2 employee.1@xtra.co employee.5@xtra.co employee.4xtra.co
# 3 employee.1@xtra.co employee.5@xtra.co employee.6xtra.co
# 4 employee.1@xtra.co employee.3xtra.co employee.2xtra.co
# 5 employee.1@xtra.co employee.3xtra.co employee.4xtra.co
# 6 employee.1@xtra.co employee.3xtra.co employee.6xtra.co
如果要展开的列超过两列,下面是一种方法。首先确定列出的列。将列名存储在 names_target
中,然后使用 for 循环重复应用 unnest
函数。
names_target <- emails %>%
select(where(is.list)) %>%
names()
temp <- emails
for (i in names_target){
temp <- temp %>% unnest(cols = all_of(i))
}
identical(temp, emails2)
# [1] TRUE
我有大量电子邮件数据,如下所示:
library(dplyr)
emails <- tibble(
from = c('employee.1@xtra.co','employee.5@xtra.co','employee.1@xtra.co',
'employee.3@xtra.co','employee.1@xtra.co'),
to = list(
c('employee.5@xtra.co', 'employee.3xtra.co'),
c('employee.3@xtra.co', 'employee.1@xtra.co'),
c('employee.2@xtra.co'),
c('employee.1@xtra.co'),
c('employee.3@xtra.co','employee.5@xtra.co','employee.6@xtra.co')),
cc = list(
c('employee.2xtra.co', 'employee.4xtra.co', 'employee.6xtra.co'),
c('employee.1xtra.co', 'employee.8xtra.co', 'employee.6xtra.co'),
NA,
c('employee.2xtra.co', 'employee.4xtra.co'),
c('employee.2xtra.co', 'employee.6xtra.co'))
)
emails
# A tibble: 5 x 3
from to cc
<chr> <list> <list>
1 employee.1@xtra.co <chr [2]> <chr [3]>
2 employee.5@xtra.co <chr [2]> <chr [3]>
3 employee.1@xtra.co <chr [1]> <lgl [1]>
4 employee.3@xtra.co <chr [1]> <chr [2]>
5 employee.1@xtra.co <chr [3]> <chr [2]>
我需要你的帮助才能为每个组合扩展每条记录。例如,我想为第 1 行实现的是:
from to cc
employee.1@xtra.co employee.5@xtra.co employee.2xtra.co
employee.1@xtra.co employee.5@xtra.co employee.4xtra.co
employee.1@xtra.co employee.5@xtra.co employee.6xtra.co
employee.1@xtra.co employee.3xtra.co employee.2xtra.co
employee.1@xtra.co employee.3xtra.co employee.4xtra.co
employee.1@xtra.co employee.3xtra.co employee.6xtra.co
非常感谢您的宝贵时间。
我们可以申请unnest
两次。
library(dplyr)
library(tidyr)
emails2 <- emails %>%
unnest(cols = "to") %>%
unnest(cols = "cc")
head(emails2)
# # A tibble: 6 x 3
# from to cc
# <chr> <chr> <chr>
# 1 employee.1@xtra.co employee.5@xtra.co employee.2xtra.co
# 2 employee.1@xtra.co employee.5@xtra.co employee.4xtra.co
# 3 employee.1@xtra.co employee.5@xtra.co employee.6xtra.co
# 4 employee.1@xtra.co employee.3xtra.co employee.2xtra.co
# 5 employee.1@xtra.co employee.3xtra.co employee.4xtra.co
# 6 employee.1@xtra.co employee.3xtra.co employee.6xtra.co
如果要展开的列超过两列,下面是一种方法。首先确定列出的列。将列名存储在 names_target
中,然后使用 for 循环重复应用 unnest
函数。
names_target <- emails %>%
select(where(is.list)) %>%
names()
temp <- emails
for (i in names_target){
temp <- temp %>% unnest(cols = all_of(i))
}
identical(temp, emails2)
# [1] TRUE