我正在阅读dplyr的插图，试图弄清楚如何dplyr在我的功能代码中使用它。在其中途中讨论了如何使用enquoson...以便将多个参数传递给group_by。
它如何工作的简短示例
grp <- rlang::enquos(...)
df %>%
group_by(!!!grp)
我不知道是否有一种方法可以分配多个表达式值而不进行保留，...而不进行一些可疑的编码。
要了解调用的外观，请使用以下示例：
#reproducable data
df <- datasets::USJudgeRatings
df$name <- rownames(df)
df <- tidyr::gather(df, key = "key", value = "value", -name)
df$dummy <- c("1","2")
test_summarize <- function(df, sum.col, grp = NULL, filter = NULL) {
filter <- rlang::enquo(filter)
sum.col <- rlang::enquo(sum.col)
if(!is.null(rlang::get_expr(filter))){
df <- dplyr::filter(df, !!filter)
}
#how grp is turned into a character vector to be passed to .dots in group_by
grp <- substitute(grp)
if(!is.null(grp)){
grp <- deparse(grp)
grp <- strsplit(gsub(pattern = "list\\(|c\\(|\\)|", replacement = "", x = grp), split =",")[[1]]
grp <- gsub(pattern = "^ | $", replacement = "", x = grp)
df %>%
dplyr::group_by(.dots=grp) %>%
dplyr::summarise(mean = mean(!!sum.col), sum = sum(!!sum.col), n = n())
} else{
df %>%
dplyr::summarise(mean = mean(!!sum.col), sum = sum(!!sum.col), n = n())
}
}
test_summarize(df, sum.col=value, grp = c(name, dummy))
# A tibble: 86 x 5
# Groups: name [?]
name dummy mean sum n
<chr> <fct> <dbl> <dbl> <int>
1 AARONSON,L.H. 1 7.17 43 6
2 AARONSON,L.H. 2 7.42 44.5 6
3 ALEXANDER,J.M. 1 8.35 50.1 6
4 ALEXANDER,J.M. 2 7.95 47.7 6
5 ARMENTANO,A.J. 1 7.53 45.2 6
6 ARMENTANO,A.J. 2 7.7 46.2 6
7 BERDON,R.I. 1 8.67 52 6
8 BERDON,R.I. 2 8.25 49.5 6
9 BRACKEN,J.J. 1 5.65 33.9 6
10 BRACKEN,J.J. 2 5.82 34.9 6
# ... with 76 more rows
这对我想做的事情有效，但我想知道是否有更好的方法来接受参数并处理它们。我所做的每一次尝试都将原始grp呼叫转换为类似于enquos(...)失败的内容，因此我进行了一次装帧并将其转换为字符向量，说实话，我应该只希望用户传递字符？
我选择不使用字符向量作为预期输入，因为考虑到该函数的sum.col和filter参数期望使用NSE表达式，因此我试图保持一致。也许rlang包中有一些东西可以将原始表达式的每个元素转换成一个quosures列表？
编辑：修复了可重现的示例并提供了预期的输出
如果使用group_by_at，则可能不需要if/else参数
test_summarize <- function(df, sum.col, grp = NULL, filter = NULL) {
df %>%
group_by_at(grp) %>%
summarise(mean = mean({{sum.col}}),
sum = sum({{sum.col}}), n = n())
}
test_summarize(df, sum.col=value, grp = c("name", "dummy"))
# A tibble: 86 x 5
# Groups: name [43]
# name dummy mean sum n
# <chr> <chr> <dbl> <dbl> <int>
# 1 AARONSON,L.H. 1 7.17 43 6
# 2 AARONSON,L.H. 2 7.42 44.5 6
# 3 ALEXANDER,J.M. 1 8.35 50.1 6
# 4 ALEXANDER,J.M. 2 7.95 47.7 6
# 5 ARMENTANO,A.J. 1 7.53 45.2 6
# 6 ARMENTANO,A.J. 2 7.7 46.2 6
# 7 BERDON,R.I. 1 8.67 52 6
# 8 BERDON,R.I. 2 8.25 49.5 6
# 9 BRACKEN,J.J. 1 5.65 33.9 6
#10 BRACKEN,J.J. 2 5.82 34.9 6
# … with 76 more rows
test_summarize(df, sum.col=value)
# A tibble: 1 x 3
# mean sum n
# <dbl> <dbl> <int>
#1 7.57 3908. 516
这与
df %>%
summarise(mean = mean(value), sum = sum(value), n = n())
# mean sum n
#1 7.57345 3907.9 516
如果使用filter，则一个选项是，...并通过尽可能多的过滤条件
test_summarize <- function(df, sum.col, grp = NULL, ...) {
df %>%
filter(!!! rlang::enexprs(...)) %>%
group_by_at(grp) %>%
summarise(mean = mean({{sum.col}}), sum = sum({{sum.col}}), n = n())
}
test_summarize(df, sum.col=value, grp = c("name", "dummy"),
key %in% c("CONT", "INTG"), value > 6.5)
# A tibble: 77 x 5
# Groups: name [43]
# name dummy mean sum n
# <chr> <chr> <dbl> <dbl> <int>
# 1 AARONSON,L.H. 2 7.9 7.9 1
# 2 ALEXANDER,J.M. 1 8.9 8.9 1
# 3 ALEXANDER,J.M. 2 6.8 6.8 1
# 4 ARMENTANO,A.J. 1 7.2 7.2 1
# 5 ARMENTANO,A.J. 2 8.1 8.1 1
# 6 BERDON,R.I. 1 8.8 8.8 1
# 7 BERDON,R.I. 2 6.8 6.8 1
# 8 BRACKEN,J.J. 1 7.3 7.3 1
# 9 BURNS,E.B. 1 8.8 8.8 1
#10 CALLAHAN,R.J. 1 10.6 10.6 1
# … with 67 more rows
并且还将在没有过滤器参数的情况下进行评估
test_summarize(df, sum.col=value, grp = c("name", "dummy"))
# A tibble: 86 x 5
# Groups: name [43]
# name dummy mean sum n
# <chr> <chr> <dbl> <dbl> <int>
# 1 AARONSON,L.H. 1 7.17 43 6
# 2 AARONSON,L.H. 2 7.42 44.5 6
# 3 ALEXANDER,J.M. 1 8.35 50.1 6
# 4 ALEXANDER,J.M. 2 7.95 47.7 6
# 5 ARMENTANO,A.J. 1 7.53 45.2 6
# 6 ARMENTANO,A.J. 2 7.7 46.2 6
# 7 BERDON,R.I. 1 8.67 52 6
# 8 BERDON,R.I. 2 8.25 49.5 6
# 9 BRACKEN,J.J. 1 5.65 33.9 6
#10 BRACKEN,J.J. 2 5.82 34.9 6
# … with 76 more rows
与您的第一个输出相同
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