R - dplyr/purrr语言 - 从现有列对的函数创建新列

Question

我遇到了几次 dplyr：：mutate 的绊脚石，因为我无法弄清楚如何基于函数创建新列(例如，求和或其他任何东西(，该函数将基于两组输入列的所有对创建新列。部分演示如下：

#Input data
set.seed(100)
in_dat <- tibble(x1 = sample(x = c(1:10, NA_real_), size = 1000, replace = TRUE),
x2 = sample(x = c(1:10, NA_real_), size = 1000, replace = TRUE),
x3 = sample(x = c(1:10, NA_real_), size = 1000, replace = TRUE),
x4 = sample(x = c(1:10, NA_real_), size = 1000, replace = TRUE),
y1 = sample(x = c(1, 0, NA_real_), size = 1000, replace = TRUE),
y2 = sample(x = c(1, 0, NA_real_), size = 1000, replace = TRUE),
y3 = sample(x = c(1, 0, NA_real_), size = 1000, replace = TRUE),
y4 = sample(x = c(1, 0, NA_real_), size = 1000, replace = TRUE),
y5 = sample(x = c(1, 0, NA_real_), size = 1000, replace = TRUE),
y6 = sample(x = c(1, 0, NA_real_), size = 1000, replace = TRUE))
#Output data with 1 column pair; all pairs between x and y should be computed
out_dat_1col <- in_dat %>% 
mutate(miss_x1y1 = if_else(is.na(x1) & is.na(y1), TRUE, FALSE))

这将检查 x 和 y 列对在新列中都有缺失值并标记为 TRUE。不过这只是一对，我想要一种方法来对 x 和 y 列之间的所有对执行此操作，而不是在它们自己的变异行中手动编码它们中的每一个。我认为咕噜声应该能够做到这一点，但我还没有弄清楚地图变体的正确语法，或者可能也减少。我目前收到来自map2_dfc(将新列附加到带有bind_cols的现有列(和reduce2.x(x 变量(和.y(y 变量(长度不一致的错误，我不确定如何规避这一点。任何想法都非常感谢。

#Produces error
out_dat <- in_dat %>% 
bind_cols(map2_dfc(
.x = in_dat %>% select(starts_with('x')),
.y = in_dat %>% select(starts_with('y')),
.f = ~if_else(is.na(.x) & is.na(.y), TRUE, FALSE)
))
Error: Mapped vectors must have consistent lengths:
* `.x` has length 4
* `.y` has length 6

Answer 1

以下是使用lapply、sapply和mapply创建数据框的简短基本 R 方法：

all_cols <- lapply(in_dat, function(y) sapply(in_dat, function(x) is.na(y) & is.na(x)))
all_cols <- mapply(function(x, y) {colnames(x) <- paste(y, colnames(x), sep = "_"); x}, 
all_cols, names(all_cols), SIMPLIFY = FALSE)
df <- as_tibble(cbind(in_dat, do.call(cbind, all_cols)))
df
#> # A tibble: 1,000 x 110
#>       x1    x2    x3    x4    y1    y2    y3    y4    y5    y6 x1_x1 x1_x2 x1_x3 x1_x4
#>    <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <lgl> <lgl> <lgl> <lgl>
#>  1     3     7     2     5     1     1     0     1     0    NA FALSE FALSE FALSE FALSE 
#>  2     7     5    10     3    NA     0    NA    NA     0    NA FALSE FALSE FALSE FALSE
#>  3     3     3     3     7     1     1    NA     1     1     1 FALSE FALSE FALSE FALSE
#>  4     7     3     1     8     1    NA     1     0    NA     1 FALSE FALSE FALSE FALSE 
#>  5     5     2    10     7     0    NA    NA     0    NA     1 FALSE FALSE FALSE FALSE 
#>  6     7     8    10     8    NA     1     1     1     1     1 FALSE FALSE FALSE FALSE 
#>  7    10     8     3     5     0     1    NA     1     1     1 FALSE FALSE FALSE FALSE 
#>  8     1    10     5    10     1    NA    NA     0     1     1 FALSE FALSE FALSE FALSE
#>  9     7     2     5     9    NA     0     0    NA     1     1 FALSE FALSE FALSE FALSE
#> 10     8     9     1     4     1    NA    NA     1    NA     0 FALSE FALSE FALSE FALSE
#> # ... with 990 more rows, and 96 more variables

唯一的问题是您还针对自身检查了每一行，因此要删除它们，您可以执行以下操作：

df <- df[sapply(strsplit(names(df), "_"), function(x) {!any(duplicated(x))})]