如何将未加引号的列名列表馈入`lapply`（以便我可以将其与`dplyr`函数一起使用） [英] How to feed a list of unquoted column names into `lapply` (so that I can use it with a `dplyr` function)
问题描述
我试图在 tidyverse / dplyr 中编写一个函数，最终要与 lapply 一起使用（或地图）。 （我一直在将它处理到 ，但遇到一个有趣的结果/死角。请不要将其标记为重复-这个问题是您在此处看到的答案的延伸/偏离。）
I am trying to write a function in tidyverse/dplyr that I want to eventually use with lapply (or map). (I had been working on it to answer this question, but came upon an interesting result/dead-end. Please don't mark this as a duplicate - this question is an extension/departure from the answers that you see there.)
是否存在
1）一种获取带引号的变量列表以在dplyr函数中工作的方法
（并且不使用不推荐使用的 SE _ 函数）或在那里
2）通过 lapply 或 map
Is there
1) a way to get a list of quoted variables to work inside a dplyr function
(and not use the deprecated SE_ functions) or is there
2) some way to feed a list of unquoted strings through an lapply or map
我使用了 使用Dplyr 小插图进行编程以构造我认为最符合当前与NS配合使用的标准
的功能E。
I have used the Programming in Dplyr vignette to construct what I believe is a function most in line with the current standard
for working with the NSE.
sample_data <-
read.table(text = "REVENUEID AMOUNT YEAR REPORT_CODE PAYMENT_METHOD INBOUND_CHANNEL AMOUNT_CAT
1 rev-24985629 30 FY18 S Check Mail 25,50
2 rev-22812413 1 FY16 Q Other Canvassing 0.01,10
3 rev-23508794 100 FY17 Q Credit_card Web 100,250
4 rev-23506121 300 FY17 S Credit_card Mail 250,500
5 rev-23550444 100 FY17 S Credit_card Web 100,250
6 rev-21508672 25 FY14 J Check Mail 25,50
7 rev-24981769 500 FY18 S Credit_card Web 500,1e+03
8 rev-23503684 50 FY17 R Check Mail 50,75
9 rev-24982087 25 FY18 R Check Mail 25,50
10 rev-24979834 50 FY18 R Credit_card Web 50,75
", header = TRUE, stringsAsFactors = FALSE)
报告生成功能
report <- function(report_cat){
report_cat <- enquo(report_cat)
sample_data %>%
group_by(!!report_cat, YEAR) %>%
summarize(num=n(),total=sum(AMOUNT)) %>%
rename(REPORT_VALUE = !!report_cat) %>%
mutate(REPORT_CATEGORY := as.character(quote(!!report_cat))[2])
}
对于生成单个报告可以很好地工作：
Which works fine for generating a single report:
> report(REPORT_CODE)
# A tibble: 7 x 5
# Groups: REPORT_VALUE [4]
REPORT_VALUE YEAR num total REPORT_CATEGORY
<chr> <chr> <int> <int> <chr>
1 J FY14 1 25 REPORT_CODE
2 Q FY16 1 1 REPORT_CODE
3 Q FY17 1 100 REPORT_CODE
4 R FY17 1 50 REPORT_CODE
5 R FY18 2 75 REPORT_CODE
6 S FY17 2 400 REPORT_CODE
7 S FY18 2 530 REPORT_CODE
当我尝试建立所有要生成的所有4个报告的列表时，一切都崩溃了。 （尽管诚然，函数最后一行所需的代码（返回用来填充列的字符串）应该足够聪明，以至于我在错误的方向上徘徊了。）
It is when I try and set up a list of all 4 of the reports to generate, that everything breaks down. (Though admittedly the code required in that last line of the function - to return a string with which to then fill the column - should be clue enough that I have wandered off in the wrong direction.)
#the other reports
cat.list <- c("REPORT_CODE","PAYMENT_METHOD","INBOUND_CHANNEL","AMOUNT_CAT")
# Applying and Mapping attempts
lapply(cat.list, report)
map_df(cat.list, report)
这将导致：
> lapply(cat.list, report)
Error in (function (x, strict = TRUE) :
the argument has already been evaluated
> map_df(cat.list, report)
Error in (function (x, strict = TRUE) :
the argument has already been evaluated
我还尝试过将字符串列表转换为名称，然后再将其交给 apply 和 map ：
I have also tried to convert the list of strings to names before handing it over to apply and map:
library(rlang)
cat.names <- lapply(cat.list, sym)
lapply(cat.names, report)
map_df(cat.names, report)
> lapply(cat.names, report)
Error in (function (x, strict = TRUE) :
the argument has already been evaluated
> map_df(cat.names, report)
Error in (function (x, strict = TRUE) :
the argument has already been evaluated
无论如何，我问这个问题的原因是我认为我已经按照当前记录的标准编写了该函数，但是最终我看不到任何利用 apply 甚至是成员的方法。 purrr :: map 系列具有这样的功能。只需像 useR 那样使用名称来重写函数，就可以在 https://stackoverflow.com/a/47316151/5088194 是否可以使此功能与 apply 或<$ c $一起使用c> map ？
In any case, the reason I am asking this question is that I think that I have written the function to the currently documented standards, but ultimately I can then see no way to utilize a member of the apply or even of the purrr::map family with such a function. Short of rewriting the function to use names like useR has done here https://stackoverflow.com/a/47316151/5088194 is there a way to get this function to work with apply or map?
我希望看到这样的结果：
I am hoping to see this as a result:
# A tibble: 27 x 5
# Groups: REPORT_VALUE [16]
REPORT_VALUE YEAR num total REPORT_CATEGORY
<chr> <chr> <int> <int> <chr>
1 J FY14 1 25 REPORT_CODE
2 Q FY16 1 1 REPORT_CODE
3 Q FY17 1 100 REPORT_CODE
4 R FY17 1 50 REPORT_CODE
5 R FY18 2 75 REPORT_CODE
6 S FY17 2 400 REPORT_CODE
7 S FY18 2 530 REPORT_CODE
8 Check FY14 1 25 PAYMENT_METHOD
9 Check FY17 1 50 PAYMENT_METHOD
10 Check FY18 2 55 PAYMENT_METHOD
# ... with 17 more rows
推荐答案
as.name 会将字符串转换为名称，并且可以将其传递给 report ：
as.name will convert a string to a name and that can be passed to report:
lapply(cat.list, function(x) do.call("report", list(as.name(x))))
字符参数的另一种方法是重写 report ，以便它接受一个字符字符串参数：
character argument An alternative is to rewrite report so that it accepts a character string argument:
report_ch <- function(colname) {
report_cat <- rlang::sym(colname) # as.name(colname) would also work here
sample_data %>%
group_by(!!report_cat, YEAR) %>%
summarize(num = n(), total = sum(AMOUNT)) %>%
rename(REPORT_VALUE = !!report_cat) %>%
mutate(REPORT_CATEGORY = colname)
}
lapply(cat.list, report_ch)
wrapr 使用包装程序包的 report 可以替代rlang / tidyeval：
wrapr An alternate approach is to rewrite report using the wrapr package which is an alternative to rlang/tidyeval:
library(dplyr)
library(wrapr)
report_wrapr <- function(colname)
let(c(COLNAME = colname),
sample_data %>%
group_by(COLNAME, YEAR) %>%
summarize(num = n(), total = sum(AMOUNT)) %>%
rename(REPORT_VALUE = COLNAME) %>%
mutate(REPORT_CATEGORY = colname)
)
lapply(cat.list, report_wrapr)
当然，如果您使用其他框架，例如
Of course, this whole problem would go away if you used a different framework, e.g.
plyr
library(plyr)
report_plyr <- function(colname)
ddply(sample_data, c(REPORT_VALUE = colname, "YEAR"), function(x)
data.frame(num = nrow(x), total = sum(x$AMOUNT), REPORT_CATEOGRY = colname))
lapply(cat.list, report_plyr)
sqldf
library(sqldf)
report_sql <- function(colname, envir = parent.frame(), ...)
fn$sqldf("select [$colname] REPORT_VALUE,
YEAR,
count(*) num,
sum(AMOUNT) total,
'$colname' REPORT_CATEGORY
from sample_data
group by [$colname], YEAR", envir = envir, ...)
lapply(cat.list, report_sql)
基础-
report_base_by <- function(colname)
do.call("rbind",
by(sample_data, sample_data[c(colname, "YEAR")], function(x)
data.frame(REPORT_VALUE = x[1, colname],
YEAR = x$YEAR[1],
num = nrow(x),
total = sum(x$AMOUNT),
REPORT_CATEGORY = colname)
)
)
lapply(cat.list, report_base_by)
data.table data.table包提供了另一种选择，但已经被另一个答案所涵盖。
data.table The data.table package provides another alternative but that has already been covered by another answer.
更新：：添加了其他替代方法。
Update: Added additional alternatives.
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