使用 pmap 和 c(...) 第 2 部分 [英] Using pmap with c(...) part 2
问题描述
我最近一直在探索使用 pmap 函数及其变体的各种应用，我对使用 c(...) 传递所有参数特别感兴趣进入.以下数据集属于我们今天早些时候讨论的另一个问题 有许多知识渊博的用户.我们应该根据 Days 列中的值沿着它们各自的行重复 weight 列中的值，以获得以下输出:
I have been exploring the various application of using pmap function and its variations recently and I am particularly interested in using c(...) to pass all the arguments into. The following data set belongs to another question that we discussed earlier today with a number of very knowledgeable users.
We were supposed to repeat the values in weight column based on values in Days column along their respective rows to get the following output:
df <- tribble(
~Name, ~School, ~Weight, ~Days,
"Antoine", "Bach", 0.03, 5,
"Antoine", "Ken", 0.02, 7,
"Barbara", "Franklin", 0.04, 3
)
输出:
df %>%
mutate(map2_dfr(Weight, Days, ~ set_names(rep(.x, .y), 1:.y))) %>%
select(-c(Weight, Days))
# A tibble: 3 x 9
Name School `1` `2` `3` `4` `5` `6` `7`
<chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Antoine Bach 0.03 0.03 0.03 0.03 0.03 NA NA
2 Antoine Ken 0.02 0.02 0.02 0.02 0.02 0.02 0.02
3 Barbara Franklin 0.04 0.04 0.04 NA NA NA NA
我的问题是可以通过各种解决方案实现此输出，但其中一位贡献者提出的以下解决方案引起了我的注意.我想知道如何通过 c(...)
My question is this output is achievable through various solutions but the following one proposed by one of the contributors caught my attention. I would like to know how I could rewrite it by means of c(...)
# This is not my code and it works:
pmap_dfr(df, function(Weight, Days, ...) c(..., setNames(rep(Weight, Days), 1:Days)))
# And I can also rewrite it in the following way which also works:
df %>%
mutate(data = pmap(list(Weight, Days), ~ setNames(rep(.x, .y), 1:.y))) %>%
unnest_wider(data)
但我想知道为什么这些都不起作用:
But I would like to know why any of these doesn't work:
df %>%
mutate(pmap_dfr(., ~ c(..., setNames(rep(Weight, Days), 1:Days))))
df %>%
pmap_dfr(., ~ c(..., setNames(rep(Weight, Days), 1:Days)))
在此先非常感谢您，对冗长的描述深表歉意.
Thank you very much in advance and so sorry for the long description.
推荐答案
问题似乎是混合自定义匿名/lambda 函数 (function(Weight, Days, ...) - 其中参数的命名与列名相同)使用默认的 lambda 函数(~ - 其中参数是 .x, .y 如果只有两个元素，或者如果超过两个 - ..1、..2、..3 等).在 OP 的代码中
The issue seems to be mixing the custom anonymous/lambda function (function(Weight, Days, ...) - where the arguments are named as the same as the column name) with the default lambda function (~ - where the arguments are .x, .y if only two elements or if more than two - ..1, ..2, ..3 etc). In the OP's code
library(dplyr)
library(purrr)
df %>%
mutate(pmap_dfr(., ~ c(..., setNames(rep(Weight, Days), 1:Days))))
'Weight'、'Days' 返回来自原始数据集而非行的完整列值.如果我们还想使用上面的命令，我们需要将每行捕获的数据转换成tibble并使用with
The 'Weight', 'Days' returns the full column values from original dataset and not from rows. If we want to still make use of the above command, we need to convert the data captured in each row to a tibble and use with
df %>%
pmap_dfr(., ~ with(as_tibble(list(...)),
setNames(rep(Weight, Days), seq_len(Days))))
# A tibble: 3 x 7
# `1` `2` `3` `4` `5` `6` `7`
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 0.03 0.03 0.03 0.03 0.03 NA NA
#2 0.02 0.02 0.02 0.02 0.02 0.02 0.02
#3 0.04 0.04 0.04 NA NA NA NA
如果我们想要其他列，
If we want the other columns,
df %>%
pmap_dfr(., ~ c(list(...)[-(3:4)], with(as_tibble(list(...)),
setNames(rep(Weight, Days), seq_len(Days)))))
# A tibble: 3 x 9
# Name School `1` `2` `3` `4` `5` `6` `7`
# <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 Antoine Bach 0.03 0.03 0.03 0.03 0.03 NA NA
#2 Antoine Ken 0.02 0.02 0.02 0.02 0.02 0.02 0.02
#3 Barbara Franklin 0.04 0.04 0.04 NA NA NA NA
或者使用rowwise
library(tidyr)
df %>%
rowwise %>%
mutate(out = list(setNames(rep(Weight, Days), seq_len(Days)))) %>%
ungroup %>%
unnest_wider(c(out)) %>%
select(-Weight, -Days)
# A tibble: 3 x 9
# Name School `1` `2` `3` `4` `5` `6` `7`
# <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 Antoine Bach 0.03 0.03 0.03 0.03 0.03 NA NA
#2 Antoine Ken 0.02 0.02 0.02 0.02 0.02 0.02 0.02
#3 Barbara Franklin 0.04 0.04 0.04 NA NA NA NA
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