r语言 - 基于不一致的列名合并两个数据集

Question

我想知道基于不一致的列合并数据集的最简洁的方法。

> head(team_measures)
# A tibble: 6 x 7
team_id geo_entropy job_entropy
<chr>         <dbl>       <dbl>
1 10012         1.79        1.79 
2 10027         0           1.25 
3 10044         1.79        0.650
4 10049         1.00        1.46 
5 10053         0.811       2.00 

> head(p_calc)
# A tibble: 6 x 2
team.id              p_average
<int>                <dbl>
1   10000                 4.75
2   10001                 4.98
3   10002                 4.17
4   10003                 4.32
5   10004                 4.22
6   10005                 4.44

我目前正在做这样一个简单的程序感觉非常笨拙的事情：

team_measures <- p_calc %>%
rename(team_id = team.id) %>%
select(team_id, p_average) %>%
left_join(team_measures, by = c('team_id')) %>%
na.omit()

它实际上比这更糟糕，因为我收到错误：

Error in left_join_impl(x, y, by_x, by_y, aux_x, aux_y, na_matches) : 
Can't join on 'team_id' x 'team_id' because of incompatible types (character / integer)

所以我必须将它们重新转换为相同的类型。

有没有更简单的方法来实现这一点？

Answer 1

这是你想要实现的目标吗？

DPLYR 联接有一个by=参数，其语法不是很直观，如by = c("xxx" = "xxxx")

full_join(team_measures, p_cal, by = c("team_id" = "team.id") )
team_id geo_entropy job_entropy p_average
1    10012       1.790        1.79        NA
2    10027       0.000        1.25        NA
3    10044       1.790        0.65        NA
4    10049       1.000        1.46        NA
5    10053       0.811        2.00        NA
6    10000          NA          NA      4.75
7    10001          NA          NA      4.98
8    10002          NA          NA      4.17
9    10003          NA          NA      4.32
10   10004          NA          NA      4.22
11   10005          NA          NA      4.44

Answer 2

只需将数字更改为字符即可

library(tidyverse)
data.frame(team_id = c("10012", '10027', '10044', '10049','10053'),
geo_entropy = c(1.79,0,1.79,1.00,0.811),
job_entropy = c(1.79,1.25,0.650,1.46,2.00)) -> team_measures
data.frame(team.id = 10000:10005,
p_average = c(4.75,4.98,4.17,4.32,4.22,4.44)) -> p_calc

p_calc %>% 
mutate(team.id = as.character(team.id)) %>% 
rename(team_id = team.id) %>% 
left_join(team_measures)

尽管这会为您提供geo_entropy和job_entropyNA，因为您的team_id值都不匹配。