dplyr :: mutate中的非标准eval [英] Non-standard eval in dplyr::mutate
问题描述
从理论上讲，这应该可行，因为我已经阅读了NSE上的tidyverse指南，但是如本例底部所示，它给我抛出了一个错误。为什么是这样？我了解如何对对象进行简单的准报价，但不了解如何计算两个准报价对象的一部分。有人可以帮忙吗？
In theory this should work, as I've read the tidyverse guide on NSE, but it throws me an error as seen in the bottom of this example. Why is this? I understand how to do a simple quasiquotation of an object, but I do not understand how to evaluate a fraction of two quasiquoted objects. Can anyone help with this?
tmp <- structure(list(qa11a = structure(c(1616, 7293, 1528, 1219, 2049, 286),
label = "Total voters removed from Nov. 2008 to Nov. 2010",
class = c("labelled","numeric")),
state_abbv = c("AL", "AL", "AL", "AL", "AL", "AL"),
fipscode = c("0100100000", "0100300000", "0100500000",
"0100700000", "0100900000", "0101100000"),
qa1a = structure(c(34727, 114952, 16450, 12239, 31874, 7650),
label = "Total registered & eligible to vote November 2010",
class = c("labelled", "numeric")),
reg.pct = structure(c(0.0465343968669911, 0.0634438722249287,
0.092887537993921, 0.0995996404935044,
0.0642843697057163, 0.0373856209150327),
label = "Total voters removed from Nov. 2008 to Nov. 2010",
class = c("labelled", "numeric")),
precleared = c(1, 1, 1, 1, 1, 1),
year = c(2010, 2010, 2010, 2010, 2010, 2010),
shelby = c(0, 0, 0, 0, 0, 0),
pres = c(FALSE, FALSE, FALSE, FALSE, FALSE, FALSE)),
.Names = c("qa11a", "state_abbv", "fipscode", "qa1a",
"reg.pct", "precleared", "year", "shelby", "pres"),
row.names = c(NA, -6L),
class = c("tbl_df", "tbl", "data.frame"))
require(tidyverse)
require(rlang)
upper="qa11a"
lower="qa1a"
pct.name="rej.reg"
pct.reg <- quo_name(enquo(pct.name))
upper <- enquo(upper)
lower <- enquo(lower)
h <- tmp %>%
mutate(!! pct.reg := (!!upper)/(!!lower))
#> Error in mutate_impl(.data, dots): Evaluation error: non-numeric argument to binary operator.
推荐答案
我们可以将字符串转换为符号，然后求值
We can convert the string to symbol and then evaluate
tmp %>%
mutate(!! pct.name := (!! sym(upper)/(!! sym(lower))))
# A tibble: 6 x 10
# qa11a state_abbv fipscode qa1a reg.pct precleared year shelby pres rej.reg
# <dbl> <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <lgl> <dbl>
#1 1616 AL 0100100000 34727 0.0465 1.00 2010 0 F 0.0465
#2 7293 AL 0100300000 114952 0.0634 1.00 2010 0 F 0.0634
#3 1528 AL 0100500000 16450 0.0929 1.00 2010 0 F 0.0929
#4 1219 AL 0100700000 12239 0.0996 1.00 2010 0 F 0.0996
#5 2049 AL 0100900000 31874 0.0643 1.00 2010 0 F 0.0643
#6 286 AL 0101100000 7650 0.0374 1.00 2010 0 F 0.0374
当我们应用 enquo 在字符串上，它转换为带引号的单价货币
when we apply enquo on a string, it is converting to a quosure with quotes
enquo(upper)
# <quosure>
# expr: ^"qa11a"
# env: empty
从字符串转换，这样做会更容易
Instead of converting from a string, it could be easier to do
upper <- quo(qalla)
lower <- quo(qala)
在OP的代码中，调用 enquo ，即在字符串对象上转换为quasure会导致字符串quosure，这不是故意的。
In the OP's code, calling enquo i.e. converting to quosure on a string object will result in string quosure and that is not intended
upper <- "qa11a"
lower <- "qa1a"
enquo(upper)
#<quosure>
# expr: ^"qa11a"
# env: empty
将其与
upper <- quo(qa11a)
lower <- quo(qa1a)
upper
# <quosure>
# expr: ^qalla
# env: global
并执行
tmp %>%
mutate(!! pct.name := (!! upper)/ (!! lower))
# A tibble: 6 x 10
# qa11a state_abbv fipscode qa1a reg.pct precleared year shelby pres rej.reg
# <dbl> <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <lgl> <dbl>
#1 1616 AL 0100100000 34727 0.0465 1.00 2010 0 F 0.0465
#2 7293 AL 0100300000 114952 0.0634 1.00 2010 0 F 0.0634
#3 1528 AL 0100500000 16450 0.0929 1.00 2010 0 F 0.0929
#4 1219 AL 0100700000 12239 0.0996 1.00 2010 0 F 0.0996
#5 2049 AL 0100900000 31874 0.0643 1.00 2010 0 F 0.0643
#6 286 AL 0101100000 7650 0.0374 1.00 2010 0 F 0.0374
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