![](/img/trans.png)
[英]Using `:=` from rlang to assign column names using lapply function inputs
[英]rlang: Get names from … with colon shortcut in NSE function
我正在編寫一套用於制作人口統計數據表的函數。 我有一個函數,縮寫如下,我需要在幾列( ...
)中gather
數據框。 訣竅是我想保留這些列的名稱,因為我需要在收集后按順序放置一列。 在這種情況下,這些列是estimate
, moe
, share
, sharemoe
。
library(tidyverse)
library(rlang)
race <- structure(list(region = c("New Haven", "New Haven", "New Haven", "New Haven", "Outer Ring", "Outer Ring", "Outer Ring", "Outer Ring"),
variable = c("white", "black", "asian", "latino", "white", "black", "asian", "latino"),
estimate = c(40164, 42970, 6042, 37231, 164150, 3471, 9565, 8518),
moe = c(1395, 1383, 697, 1688, 1603, 677, 896, 1052),
share = c(0.308, 0.33, 0.046, 0.286, 0.87, 0.018, 0.051, 0.045),
sharemoe = c(0.011, 0.011, 0.005, 0.013, 0.008, 0.004, 0.005, 0.006)),
class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, -8L))
race
#> # A tibble: 8 x 6
#> region variable estimate moe share sharemoe
#> <chr> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 New Haven white 40164 1395 0.308 0.011
#> 2 New Haven black 42970 1383 0.33 0.011
#> 3 New Haven asian 6042 697 0.046 0.005
#> 4 New Haven latino 37231 1688 0.286 0.013
#> 5 Outer Ring white 164150 1603 0.87 0.008
#> 6 Outer Ring black 3471 677 0.018 0.004
#> 7 Outer Ring asian 9565 896 0.051 0.005
#> 8 Outer Ring latino 8518 1052 0.045 0.006
在函數gather_arrange
,我通過映射rlang::exprs(...)
並轉換為字符來獲取...
列的名稱。 將這些列的名稱提取為字符串是很困難的,所以這可能是改進或重寫的地方。 但這符合我的要求,將列type
作為具有級別estimate
, moe
, share
, sharemoe
的因子。
gather_arrange <- function(df, ..., group = variable) {
gather_cols <- rlang::quos(...)
grp_var <- rlang::enquo(group)
gather_names <- purrr::map_chr(rlang::exprs(...), as.character)
df %>%
tidyr::gather(key = type, value = value, !!!gather_cols) %>%
dplyr::mutate(!!rlang::quo_name(grp_var) := !!grp_var %>%
forcats::fct_inorder() %>% forcats::fct_rev()) %>%
dplyr::mutate(type = as.factor(type) %>% forcats::fct_relevel(gather_names)) %>%
arrange(type)
}
race %>% gather_arrange(estimate, moe, share, sharemoe)
#> # A tibble: 32 x 4
#> region variable type value
#> <chr> <fct> <fct> <dbl>
#> 1 New Haven white estimate 40164
#> 2 New Haven black estimate 42970
#> 3 New Haven asian estimate 6042
#> 4 New Haven latino estimate 37231
#> 5 Outer Ring white estimate 164150
#> 6 Outer Ring black estimate 3471
#> 7 Outer Ring asian estimate 9565
#> 8 Outer Ring latino estimate 8518
#> 9 New Haven white moe 1395
#> 10 New Haven black moe 1383
#> # ... with 22 more rows
但我想選擇使用冒號表示法來選擇列,即estimate:sharemoe
相當於輸入所有這些列名。
race %>% gather_arrange(estimate:sharemoe)
#> Error: Result 1 is not a length 1 atomic vector
這失敗了,因為它無法從rlang::exprs(...)
提取列名。 如何使用此表示法獲取列名? 提前致謝!
我認為您正在尋找的函數是tidyselect::vars_select()
,它由select和rename在內部使用來完成此任務。 它返回變量名稱的字符向量。 例如:
> tidyselect::vars_select(letters, g:j)
g h i j
"g" "h" "i" "j"
這允許您使用對dplyr::select
有效的所有相同語法。
我們可以使用以下方法為這些情況創建if
條件:
從select
獲取列名('gather_names')以在fct_relevel
gather_arrange <- function(df, group = variable, ...) {
gather_cols <- quos(...)
grp_var <- enquo(group)
if(length(gather_cols)==1 && grepl(":", quo_name(gather_cols[[1]]))) {
gather_cols <- parse_expr(quo_name(gather_cols[[1]]))
}
gather_names <- df %>%
select(!!! gather_cols) %>%
names
df %>%
gather(key = type, value = value, !!!gather_cols) %>%
mutate(!!rlang::quo_name(grp_var) := !!grp_var %>%
fct_inorder() %>%
fct_rev()) %>%
mutate(type = as.factor(type) %>%
fct_relevel(gather_names)) %>%
arrange(type)
}
-檢查
out1 <- gather_arrange(df = race, group = variable,
estimate, moe, share, sharemoe)
out1
# A tibble: 32 x 4
# region variable type value
# <chr> <fct> <fct> <dbl>
# 1 New Haven white estimate 40164
# 2 New Haven black estimate 42970
# 3 New Haven asian estimate 6042
# 4 New Haven latino estimate 37231
# 5 Outer Ring white estimate 164150
# 6 Outer Ring black estimate 3471
# 7 Outer Ring asian estimate 9565
# 8 Outer Ring latino estimate 8518
# 9 New Haven white moe 1395
#10 New Haven black moe 1383
# ... with 22 more rows
out2 <- gather_arrange(df = race, group = variable, estimate:sharemoe)
identical(out1, out2)
#[1] TRUE
如果我們在...
中傳遞多組列
gather_arrange2 <- function(df, group = variable, ...) {
gather_cols <- quos(...)
grp_var <- enquo(group)
gather_names <- df %>%
select(!!! gather_cols) %>%
names
gather_colsN <- lapply(gather_cols, function(x) parse_expr(quo_name(x)))
df %>%
gather(key = type, value = value, !!!gather_colsN) %>%
mutate(!!rlang::quo_name(grp_var) := !!grp_var %>%
fct_inorder() %>%
fct_rev()) %>%
mutate(type = as.factor(type) %>%
fct_relevel(gather_names)) %>%
arrange(type)
}
-檢查
out1 <- gather_arrange2(df = race, group = variable,
estimate, moe, share, sharemoe, region)
out2 <- gather_arrange2(df = race, group = variable, estimate:sharemoe, region)
identical(out1, out2)
#[1] TRUE
或者只檢查一組列
out1 <- gather_arrange2(df = race, group = variable,
estimate, moe, share, sharemoe)
out2 <- gather_arrange2(df = race, group = variable, estimate:sharemoe)
identical(out1, out2)
#[1] TRUE
fun <- function(df, ...){
as.character(substitute(list(...)))[-1] %>%
lapply(function(x)
if(!grepl(':', x)) x
else strsplit(x, ':')[[1]] %>%
lapply(match, names(df)) %>%
{names(df)[do.call(seq, .)]})%>%
unlist
}
names(race)
# [1] "region" "variable" "estimate" "moe" "share" "sharemoe"
fun(race, estimate:sharemoe, region)
# [1] "estimate" "moe" "share" "sharemoe" "region"
fun(race, estimate, moe, share, sharemoe, region)
# [1] "estimate" "moe" "share" "sharemoe" "region"
fun(race, moe, region:variable)
# [1] "moe" "region" "variable"
這涉及兩個:
符號表達式和其他列名稱作為參數,例如fun(race, estimate:sharemoe, region)
。
有趣的是,這種hacky解決方案似乎比tidyselect
選擇更快(並不是變量選擇可能是整體速度的痛點)
fun <- function(y, ...){
as.character(substitute(list(...)))[-1] %>%
lapply(function(x)
if(!grepl(':', x)) x
else strsplit(x, ':')[[1]] %>%
lapply(match, y) %>%
{y[do.call(seq, .)]})%>%
unlist
}
library(microbenchmark)
microbenchmark(
tidy = tidyselect::vars_select(letters, b, g:j, a),
fun = fun(letters, b, g:j, a),
unit = 'relative')
# Unit: relative
# expr min lq mean median uq max neval
# tidy 19.90837 18.10964 15.32737 14.28823 13.86212 14.44013 100
# fun 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 100
原始功能
gather_arrange <- function(df, ..., group = variable) {
gather_cols <- rlang::quos(...)
grp_var <- rlang::enquo(group)
gather_names <- purrr::map_chr(rlang::exprs(...), as.character)
df %>%
tidyr::gather(key = type, value = value, !!!gather_cols) %>%
dplyr::mutate(!!rlang::quo_name(grp_var) := !!grp_var %>%
forcats::fct_inorder() %>% forcats::fct_rev()) %>%
dplyr::mutate(type = as.factor(type) %>% forcats::fct_relevel(gather_names)) %>%
arrange(type)
}
功能使用上面定義的fun
:
my_gather_arrange <- function(df, ..., group = variable) {
gather_cols <- gather_names <-
as.character(substitute(list(...)))[-1] %>%
lapply(function(x){
if(grepl(':', x)){
strsplit(x, ':')[[1]] %>%
lapply(match, names(df)) %>%
{names(df)[do.call(seq, .)]}}
else x}) %>%
unlist
grp_var <- rlang::enquo(group)
df %>%
tidyr::gather(key = type, value = value, !!!gather_cols) %>%
dplyr::mutate(!!rlang::quo_name(grp_var) := !!grp_var %>%
forcats::fct_inorder() %>% forcats::fct_rev()) %>%
dplyr::mutate(type = as.factor(type) %>% forcats::fct_relevel(gather_names)) %>%
arrange(type)
}
out1 <- gather_arrange(race, estimate, moe, share, sharemoe, region)
out2 <- my_gather_arrange(race, estimate:sharemoe, region)
#
identical(out1, out2)
# [1] TRUE
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.