使用变量名向量

时间:2019-02-24 11:36:56

标签: r dplyr tidyeval

在连接并删除重复项后合并重复列的非NA值时,我经常遇到问题。它类似于所描述的in this questionthis one。我想在coalesce(并可能包括left_join)周围创建一个小函数,以便每当遇到它时都可以一行处理(当然,该函数本身可以根据需要的长度而定)。

在这样做时,我遇到了quo_names所描述的here等价的quos

对于reprex,请使用包含标识信息的数据框,将其与包含正确值但通常拼写错误的ID的其他信息合并。

library(dplyr)
library(rlang)

iris_identifiers <- iris %>% 
  select(contains("Petal"), Species)

iris_alt_name1 <- iris %>% 
  mutate(Species = recode(Species, "setosa" = "stosa")) 

iris_alt_name2 <- iris %>%
  mutate(Species = recode(Species, "versicolor" = "verscolor"))

此更简单的功能有效:

replace_xy <- function(df, var) {

  x_var <- paste0(var, ".x")
  y_var <- paste0(var, ".y")

  df %>% 
    mutate(!! quo_name(var) := coalesce(!! sym(x_var), !! sym(y_var))) %>% 
    select(-(!! sym(x_var)), -(!! sym(y_var)))

}


iris_full <- iris_identifiers %>% 
  left_join(iris_alt_name1, by = c("Species", "Petal.Length", "Petal.Width")) %>% 
  left_join(iris_alt_name2, by = c("Species", "Petal.Length", "Petal.Width")) %>% 
  replace_xy("Sepal.Length") %>% 
  replace_xy("Sepal.Width")


head(iris_full)
#>   Petal.Length Petal.Width Species Sepal.Length Sepal.Width
#> 1          1.4         0.2  setosa          5.1         3.5
#> 2          1.4         0.2  setosa          4.9         3.0
#> 3          1.4         0.2  setosa          5.0         3.6
#> 4          1.4         0.2  setosa          4.4         2.9
#> 5          1.4         0.2  setosa          5.2         3.4
#> 6          1.4         0.2  setosa          5.5         4.2

但是我对如何实现几个变量的概括有些迷惑,我认为这将是更容易的部分。下面的代码段只是一次绝望的尝试-在尝试了多种变体之后-大致捕获了我要实现的目标。

replace_many_xy <- function(df, vars) {

  x_var <- paste0(vars, ".x")
  y_var <- paste0(vars, ".y")

  df %>% 
    mutate_at(vars(vars), funs(replace_xy(.data, .))) %>% 
    select(-(!!! syms(x_var)), -(!!! syms(y_var)))

}

new_cols <- colnames(iris_alt_name1)
diff_cols <- new_cols [!(new_cols %in% colnames(iris_identifiers))]

iris_full <- iris_identifiers %>% 
  left_join(iris_alt_name1, by = c("Species", "Petal.Length", "Petal.Width")) %>% 
  left_join(iris_alt_name2, by = c("Species", "Petal.Length", "Petal.Width")) %>% 
  replace_many_xy(diff_cols)
#> Warning: Column `Species` joining factors with different levels, coercing
#> to character vector

#> Warning: Column `Species` joining character vector and factor, coercing
#> into character vector
#> Error: Unknown columns `Sepal.Length` and `Sepal.Width`

任何帮助将不胜感激!

1 个答案:

答案 0 :(得分:2)

我写了a package that does just that,现在应该稳定了。

# devtools::install_github("moodymudskipper/safejoin")
library(safejoin)
iris_full <- iris_identifiers %>%
  left_join(iris_alt_name1, by = c("Species", "Petal.Length", "Petal.Width")) %>%
  safe_left_join(iris_alt_name2, by = c("Species", "Petal.Length", "Petal.Width"), conflict  = coalesce) %>%
  head

iris_full
#   Petal.Length Petal.Width Species Sepal.Length Sepal.Width
# 1          1.4         0.2  setosa          5.1         3.5
# 2          1.4         0.2  setosa          4.9         3.0
# 3          1.4         0.2  setosa          5.0         3.6
# 4          1.4         0.2  setosa          4.4         2.9
# 5          1.4         0.2  setosa          5.2         3.4
# 6          1.4         0.2  setosa          5.5         4.2

safe_left_join是经过改进的left_join,允许通过 check参数以及通过conflict参数处理列冲突的一些方法,就像我们在这里所做的那样。

conflict参数是一个函数,该函数一个接一个地处理一对冲突的列,以从您需要的conflict = ~coalesce(.y, .x)右边合并。


这是使函数正常工作的一种方法:

replace_many_xy <- function(tbl, vars){
  for(var in vars){
    x <- paste0(var,".x")
    y <-  paste0(var,".y")
    tbl <- mutate(tbl, !!sym(var) := coalesce(!!sym(x) , !!sym(y) )) %>%
     select(-one_of(x,y))
  }
  tbl
}
iris_full <- iris_identifiers %>%
  left_join(iris_alt_name1, by = c("Species", "Petal.Length", "Petal.Width")) %>%
  left_join(iris_alt_name2, by = c("Species", "Petal.Length", "Petal.Width")) %>%
  replace_many_xy(diff_cols) %>% as_tibble()
# # A tibble: 372 x 5
#    Petal.Length Petal.Width Species Sepal.Length Sepal.Width
#           <dbl>       <dbl> <chr>          <dbl>       <dbl>
#  1          1.4         0.2 setosa           5.1         3.5
#  2          1.4         0.2 setosa           4.9         3  
#  3          1.4         0.2 setosa           5           3.6
#  4          1.4         0.2 setosa           4.4         2.9
#  5          1.4         0.2 setosa           5.2         3.4
#  6          1.4         0.2 setosa           5.5         4.2
#  7          1.4         0.2 setosa           4.6         3.2
#  8          1.4         0.2 setosa           5           3.3
#  9          1.4         0.2 setosa           5.1         3.5
# 10          1.4         0.2 setosa           4.9         3  
# # ... with 362 more rows