使用 dplyr 在 function 中傳遞列名

Question

我知道在 function 中使用lazyeval以使用dplyr引用列名，但被卡住了。 In general, when creating a function that uses dplyr which also references column names from function arguments, what is the most idiomatic way to achieve that? 謝謝。

 library(lazyeval)

 ## Create data frame
 df0 <- data.frame(x=rnorm(100), y=runif(100))

 ##########################################
 ## Sample mean; this way works
 ##########################################
 df0 %>%
   filter(!is.na(x)) %>%
   summarize(mean=mean(x))

 ##########################################
 ## Sample mean via function; does not work
 ##########################################
 dfSummary2 <- function(df, var_y) { 
   p <- df %>%
        filter(!is.na(as.name(var_y))) %>%
        summarize(mean=mean(as.name(var_y)))
   return(p)
}

dfSummary(df0, "x")
#   mean
# 1   NA
# Warning message:
# In mean.default("x") : argument is not numeric or logical: returning NA

 ##########################################
 ## Sample mean via function; also does not work
 ##########################################
 dfSummary <- function(df, var_y) {
   p <- df %>%
        filter(!is.na(var_y)) %>%
        summarize(mean=mean(var_y))
  return(p)
}

 dfSummary(df0, "x")
 #   mean
 # 1   NA
 # Warning message:
 # In mean.default("x") : argument is not numeric or logical: returning NA

Answer 1

如果使用dplyr ，則使用summarize_和filter_的注釋是正確的方向，更多信息可通過vignette("nse")獲得。

盡管存在給定的問題，但這將提供一個 function 使用變量列名而不需要dplyr

dfSummary <- function(df, var_y) {
 mean(df[[var_y]], na.rm = TRUE) 
}

dfSummary(df0, "x")
[1] 0.105659

dfSummary(df0, "y")
[1] 0.4948618

Answer 2

filter_ summarize_已被棄用以獲取信息。 最好用

dfSummary <- function(df, var_y) {
   p <- df %>%
        filter(!is.na(var_y)) %>%
        summarize(mean=mean({{var_y}}))
  return(p)
}