Borrowing from another question's answer :
library(dplyr)
df <- data.frame(
color = c("blue", "black", "blue", "blue", "black"),
value = 1:5)
The typically presented example takes the form of ...
# As of dplyr 0.7, new functions were introduced to simplify the situation
col_name <- quo(color)
df %>% filter((!!col_name) == val)
# Remember to use enquo within a function
filter_col <- function(df, col_name, val){
col_name <- enquo(col_name)
df %>% filter((!!col_name) == val)
}
filter_col(df, color, 'blue')
... but what if you wanted to have the name of the color column specified by a string?
Eg
column_name <- "color"
col_name <- quo(column_name) # <- something else goes here
df %>% filter((!!col_name) == "blue")
Or call a function with a signature like, filter_col(df, "color", "blue")
?
Following aosmith's link takes us to lukeA's answer... which modified for this use case is:
library(dplyr)
library(rlang)
df <- data.frame(
color = c("blue", "black", "blue", "blue", "black"),
value = 1:5)
# In interactive mode
column_name <- rlang::sym("color")
df %>% filter((!!column_name) == "blue")
# In a function
filter_col <- function(df, col_name_as_string, val){
col_name <- rlang::sym(col_name_as_string)
df %>% filter((!!col_name) == val)
}
filter_col(df, 'color', 'blue')
The key part is that rlang::sym()
produces an object that can be expanded by the unquo operator !!
. Although the answer to this question is a duplicate of another question I'm going to leave it for now because I believe this is a bit more on point in how the question is specified / there can't be too many right answers to this problem. :)
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