collapse：逐行修改列以及来自多列的组合值

Question

我想使用 tidytable into collapse 翻译以下 R 代码：高级和快速数据转换。

tidytable代码

library(tidytable)
library(collapse)
Out1 <- 
  wlddev %>% 
  mutate_rowwise.(New1 = sum(c_across.(PCGDP:GINI), na.rm = TRUE))
Out1 %>% 
  select.(New1)
# A tidytable: 13,176 x 1
    New1
   <dbl>
 1  32.4
 2  33.0
 3  33.5
 4  34.0
 5  34.5
 6  34.9
 7  35.4
 8  35.9
 9  36.4
10  36.9
# ... with 13,166 more rows

收起代码

library(collapse)
Out2 <- 
  wlddev %>% 
  ftransform(New1 = fsum(across(PCGDP:GINI), na.rm = TRUE))

  Error in `context_peek()`:
  ! `across()` must only be used inside dplyr verbs.
  Run `rlang::last_error()` to see where the error occurred.

任何提示请。

Answer 1

collapse 中的 ?fsum 按列求和

fsum is a generic function that computes the (column-wise) sum of all values in x, (optionally) grouped by g and/or weighted by w (e.g. to calculate survey totals).

基于tidytable代码，它是rowwise，所以一种选择是select（slt）感兴趣的列，t ranspose，转换为 tibble/data.frame 并使用 fsum 并创建一个新列

library(collapse)
Out2 <- wlddev %>%
    slt(PCGDP:GINI) %>%
    t %>%
    as_tibble %>%
    fsum(.) %>% 
    ftransform(wlddev, New1 = .)

sum returns 0 当所有元素都是 NA 而 fsum 默认使用 na.rm = TRUE 并且它 returns NA 如果所有元素是 NA

> fsum(c(NA, NA))
[1] NA
> sum(c(NA, NA), na.rm = TRUE)
[1] 0

因此，如果我们将第二个数据中的NA改为0，输出结果将与OP的'Out1'

相同

> Out2$New1[is.na(Out2$New1)] <- 0
> all.equal(Out1, Out2, check.attributes = FALSE)
[1] TRUE

Answer 2

根据@akrun 的回答，我想出了一个更快的解决方案。

Out3 <- 
  wlddev %>%
  slt(PCGDP:GINI) %>%
  qDT() %>% 
  t %>%
  fsum(.) %>% 
  ftransform(.data = wlddev, New1 = .) %>%
  qDT() %>% 
  replace_NA(X = ., value = 0, cols = "New1")

速度比较

library(microbenchmark)

microbenchmark(
  Out1 = 
    wlddev %>% 
    mutate_rowwise.(New1 = sum(c_across.(PCGDP:GINI), na.rm = TRUE))
, Out2 =
    wlddev %>%
    slt(PCGDP:GINI) %>%
    t %>%
    as_tibble %>%
    fsum(.) %>% 
    ftransform(wlddev, New1 = .)
, Out3 = 
    wlddev %>%
    slt(PCGDP:GINI) %>%
    qDT() %>% 
    t %>%
    fsum(.) %>% 
    ftransform(.data = wlddev, New1 = .) %>%
    qDT() %>% 
    replace_NA(X = ., value = 0, cols = "New1")
)

Unit: microseconds
 expr     min       lq      mean   median       uq      max neval
 Out1 72618.0 78268.75 81296.992 79888.50 81671.10 162397.8   100
 Out2 33549.7 35520.75 37763.537 37728.25 39021.90  55001.3   100
 Out3   241.2   310.85   360.225   357.40   387.35    780.1   100

Answer 3

我想知道你为什么需要想出这么复杂的东西。您在 base R 中有类似 rowSums 的函数，并且在 kit:

中有并行统计函数

library(collapse)
library(magrittr)
library(kit, include.only = "psum")  
library(microbenchmark)
  
microbenchmark(
A = wlddev %>%
  ftransform(New1 = rowSums(qM(slt(., PCGDP:GINI)), na.rm = TRUE)),
B = wlddev %>%
  ftransform(New1 = psum(slt(., PCGDP:GINI), na.rm = TRUE)), 
C = wlddev %>%
  ftransform(New1 = psum(PCGDP, LIFEEX, GINI, na.rm = TRUE))
)

#> Unit: microseconds
#>  expr   min      lq      mean   median       uq      max neval
#>     A 68.88 97.8875 194.24037 102.2335 113.8775 4646.366   100
#>     B 25.83 30.1350  35.43548  34.9115  38.6630   56.416   100
#>     C 22.55 25.8095  29.99396  30.5860  32.9025   53.792   100

^{由 reprex package (v2.0.1)}

于 2022-02-05 创建

collapse：逐行修改列以及来自多列的组合值

collapse: Modifying Columns by row along with combine values from multiple columns

r

data.table

tidyverse

collapse

tidytable