dplyr`left_join()`不适用于字符对象作为LHS变量 [英] dplyr `left_join()` does not work with a character objects as the LHS variable

问题描述

我可以使用dplyr::left_join(..., by = c("name1" = "name2")联接包含两个具有不同名称的变量的两个数据集.

I can join two datasets that contain two variables with different names using dplyr::left_join(..., by = c("name1" = "name2").

我想使用角色对象left_join(..., by = c(nameOb1 = nameOb2)加入.奇怪的是:这适用于by = c("name1", nameOb2)，，但不适用于 by = c(nameOb1, "name2").

I want to join using character objects, left_join(..., by = c(nameOb1 = nameOb2). Oddly: this works for by = c("name1", nameOb2), but not for by = c(nameOb1, "name2").

这是为什么?

在下面复制我的问题.非常感谢.

Replication of my issue below. Many thanks.

    orig <- tibble(name1 = c("a", "b", "c"),
                   n     = c(10, 20, 30))  

    tojoin <- tibble(name2 = c("a", "b", "c"),
                     pc    = c(.4, .1, .2))    

Works:对by参数使用字符串

Works: using character strings for the by arguments

    left_join(orig, tojoin, by = c("name1" = "name2"))

    # A tibble: 3 x 3
      name1     n    pc
      <chr> <dbl> <dbl>
    1 a        10   0.4
    2 b        20   0.1
    3 c        30   0.2

不起作用:使用object作为 first by参数的字符串

Does not work: using object as the character string for the first by argument

    firstname <- "name1"

    left_join(orig, tojoin, by = c(firstname = "name2"))

    # Error: `by` can't contain join column `firstname` which is missing from LHS
    # Call `rlang::last_error()` to see a backtrace

Works:使用对象作为 second by参数的字符串

Works: using object as the character string for the second by argument

    secondname <- "name2"

    left_join(orig, tojoin, by = c("name1" = secondname))

    # A tibble: 3 x 3
      name1     n    pc
      <chr> <dbl> <dbl>
    1 a        10   0.4
    2 b        20   0.1
    3 c        30   0.2

包装:

dplyr 0.8.0.1

dplyr 0.8.0.1

推荐答案

'left_join'函数需要在by参数中使用命名的字符向量.第二次尝试:

Hy, the 'left_join' function needs a named character vector in the by argument. In your second try:

firstname <- "name1"
left_join(orig, tojoin, by = c(firstname = "name2"))

您将字符向量的名称设置为firstname，该名称不适用于联接. 为了解决这个问题，您可以先生成一个命名的字符向量，然后将其传递给join函数的by参数

You set the name of the character vector to firstname which does not work for the join. For solving this you can first generate a named character vector and pass it then to the by argument of the join function

firstname <- "name1"
join_cols = c("name2")
names(join_cols) <- firstname

dplyr::left_join(orig, tojoin, by = join_cols)

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