如何取消引用str_detect()中的quosure来过滤r中函数的数据？

Question

我想要创建一个函数，这将节省我大量的时间打字或复制和粘贴。我正在创建这个函数，以便以后可以将其用于其他分析。

函数

extractItemsForScales <- function(df=NULL, col_name_to_detect_scales=NULL, col_name_to_extract=NULL, scales=NULL, name_suffix=""){



  if(is.null(df))
    stop("Dataframe must not be null")

  if(is.null(scales))
    stop("Scales must not be null")

  if(is.null(col_name_to_detect_scales))
    stop("You must provide the column name to detect scales")

  if(is.null(col_name_to_extract))
    stop("You must provide the column name to extract the items from")

  col_name_to_detect_scales <- enquo(col_name_to_detect_scales)
  col_name_to_extract <- enquo(col_name_to_extract)

  cat("Extracting items")

  vars <- scales %>%
    map(function(x){
      df %>% filter(str_detect(!!col_name_to_detect_scales, x)) %>% pull(!!col_name_to_extract)
    }) %>%
    setNames(paste0(tolower(scales),name_suffix))

  return(vars)
}

数据

df <- structure(list(Scale = c("S01", "S05", "S05", "S01", "S01", "S11", 
"S15", "S15", "S16", "S16", "S05", "S04", "S10", "S13", "S13", 
"S05", "S10", "S09", "S07", "S07", "S06", "S06", "S06", "S07", 
"S11", "S09", "S11", "S04", "S09", "S09", "S07", "S06", "S05", 
"S05", "S06", "S05", "S01", "S01", "S01", "S12", "S01", "S02", 
"S08", "S12", "S08", "S08", "S05", "S04"), SectionNo = c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), 
    Item = c("S1_23", "S1_29", "S1_36", "S1_41", "S1_46", "S1_6", 
    "S1_81", "S1_89", "S1_40", "S1_51", "S1_12", "S1_15", "S1_34", 
    "S1_44", "S1_50", "S1_78", "S1_73", "S1_77", "S1_31", "S1_59", 
    "S1_67", "S1_83", "S1_86", "S1_90", "S1_10", "S1_11", "S1_19", 
    "S1_26", "S1_30", "S1_45", "S1_49", "S1_61", "S1_62", "S1_91", 
    "S1_20", "S1_8", "S1_9", "S1_14", "S1_22", "S1_25", "S1_33", 
    "S1_53", "S1_54", "S1_92", "S1_63", "S1_64", "S1_71", "S1_47"
    )), class = "data.frame", row.names = c(NA, -48L))

期望输出

    list(S01_48 = c("S1_23", "S1_41", "S1_46", 
"S1_9", "S1_14", "S1_22", "S1_33"), S05_48 = c("S1_29", 
"S1_36", "S1_12", "S1_78", "S1_62", "S1_91", 
"S1_8", "S1_71"), S11_48 = c("S1_6", "S1_10", 
"S1_19"), S15_48 = c("S1_81", "S1_89"), S16_48 = c("S1_40", 
"S1_51"), S04_48 = c("S1_15", "S1_26", "S1_47"
), S10_48 = c("S1_34", "S1_73"), S13_48 = c("S1_44", 
"S1_50"), S09_48 = c("S1_77", "S1_11", "S1_30", 
"S1_45"), S07_48 = c("S1_31", "S1_59", 
"S1_90", "S1_49"), S06_48 = c("S1_67", "S1_83", 
"S1_86", "S1_61", "S1_20"), S12_48 = c("S1_25", 
"S1_92"), S02_48 = "S1_53", S08_48 = c("S1_54", 
"S1_63", "S1_64"))

函数调用

我会按如下方式调用该函数：

scales <- structure(list(Scale = c("S01", "S05", "S11", "S15", "S16", "S04", 
    "S10", "S13", "S09", "S07", "S06", "S12", "S02", "S08")), class = "data.frame", row.names = c(NA, 
    -14L))

df %>% extractItemsForScales(col_name_to_detect_scales = Scale, col_name_to_extract = Item, scales = scales, name_suffix = "_48")

但是我不断地发现这个错误：

Error in extractItemsForScales(., col_name_to_detect_scales = Scale, col_name_to_extract = Item,  : 
  object 'Item' not found

我为什么要犯这个错误？我怎么才能解决这个问题？

更新

调试时，似乎在is.null()检查col_name_to_extract期间抛出了错误。但是，我不知道为什么在此检查中出现错误，而不是上面对col_name_to_detect_scales的错误。

Answer 1

我认为你的不引用是可以的。如果将if(is.null更改为if(missing，此函数将按预期工作。您还为scale传递了错误的值，因为您传递的对象似乎是一个数据框架，所以不是按Map中的每个字符串映射，而是按列进行映射。

所以如果你有

extractItemsForScales <- function(df, 
                                  col_name_to_detect_scales, 
                                  col_name_to_extract, 
                                  scales, 
                                  name_suffix=""){

  if(missing(df)) 
    stop("Dataframe must not be null")

  if(missing(scales)) 
    stop("Scales must not be null")

  if(missing(col_name_to_detect_scales))
    stop("You must provide the column name to detect scales")

  if(missing(col_name_to_extract))
    stop("You must provide the column name to extract the items from")

  col_name_to_detect_scales <- enquo(col_name_to_detect_scales)
  col_name_to_extract <- enquo(col_name_to_extract)

  cat("Extracting items\n\n")

  vars <- scales %>%
    map(function(x){
      df %>% 
        filter(str_detect(!!col_name_to_detect_scales, x)) %>% 
        pull(!!col_name_to_extract)
    }) %>%
    setNames(paste0(tolower(scales),name_suffix))

  return(vars)
}

而你却有

df %>% extractItemsForScales(col_name_to_detect_scales = Scale, 
                             col_name_to_extract = Item, 
                             scales = scales$Scale, 
                             name_suffix = "_48")

你会得到

#> Extracting items
#> 
#> $s01_48
#> [1] "S1_23" "S1_41" "S1_46" "S1_9"  "S1_14" "S1_22" "S1_33"
#> 
#> $s05_48
#> [1] "S1_29" "S1_36" "S1_12" "S1_78" "S1_62" "S1_91" "S1_8"  "S1_71"
#> 
#> $s11_48
#> [1] "S1_6"  "S1_10" "S1_19"
#> 
#> $s15_48
#> [1] "S1_81" "S1_89"
#> 
#> $s16_48
#> [1] "S1_40" "S1_51"
#> 
#> $s04_48
#> [1] "S1_15" "S1_26" "S1_47"
#> 
#> $s10_48
#> [1] "S1_34" "S1_73"
#> 
#> $s13_48
#> [1] "S1_44" "S1_50"
#> 
#> $s09_48
#> [1] "S1_77" "S1_11" "S1_30" "S1_45"
#> 
#> $s07_48
#> [1] "S1_31" "S1_59" "S1_90" "S1_49"
#> 
#> $s06_48
#> [1] "S1_67" "S1_83" "S1_86" "S1_61" "S1_20"
#> 
#> $s12_48
#> [1] "S1_25" "S1_92"
#> 
#> $s02_48
#> [1] "S1_53"
#> 
#> $s08_48
#> [1] "S1_54" "S1_63" "S1_64"