dplyr

Question

我不理解的是，使用下面的例子，我认为：

df2 <- df1 %>% left_join(code_label, by = c("team"="code"))

并且，在函数中，

df2 <- df1 %>% left_join(code_label, by = c(x='code'))

会被相同地评估，但它们不是，有没有办法包装 'x' 所以它们是？

df1_id <- c(1, 2, 3, 4, 5, 6)
team <- c(1, 2, 1, 6, 4, 1)
year <- c(2014, 2014, 2009, 2020, 2015, 2017)

df1 <- data.frame(df1_id, team, year)

code_id <- c(1,2,3,4,5,6)
code <- c(1,2,3,4,5,6)
label <- c("team_A", "team_B","team_C","team_D","team_E","team_F")

code_label <- data.frame(code_id, code,label)

df2 <- df1 %>% left_join(code_label, by = c("team"="code"))

给出一个 result.But 我想将 df1 的列名“team”作为未命名对象（我认为是正确的描述？）传递给如下函数：

f_show_labels_with_codes(x = "team")  



f_show_labels_with_codes <- function(x) 
  
{

print(x)

    df2 <- df1 %>% left_join(code_label, by = c(x='code')) 

}

但它产生错误：

Error: Join columns must be present in data.
x Problem with `x`

我查看了有关 enquo 等的文档，这对我来说是新的。但是函数中的 print(x) 已经 returns [1] “team”。但和我想的不太一样。

Answer 1

我们可以使用具有 setNames

的命名向量

f_show_labels_with_codes <- function(x) {

   print(x)

    df1 %>%
     left_join(code_label, by = setNames('code', x)) 
    

}

-测试

> f_show_labels_with_codes(x = "team")  
[1] "team"
  df1_id team year code_id  label
1      1    1 2014       1 team_A
2      2    2 2014       2 team_B
3      3    1 2009       1 team_A
4      4    6 2020       6 team_F
5      5    4 2015       4 team_D
6      6    1 2017       1 team_A

dplyr - 在函数 left_join(y, by = c(x="value")) 中评估字符串

dplyr - evaluating a string in a function left_join(y, by = c(x="value"))

r

left-join