在R数据框中的指定列范围内将大于阈值的值替换为0

Question

我有一个数据集 df1，其中的列从 S_2018_p 到 S_2021_p 我想将 >= 10 的值替换为 0。我希望有一个像 df2 这样的数据集。

library(data.table)
df1 = data.table(
  ID = c("a1", "a2", "a3", "a4", "a5", "a6", "a7"),
  "string1" = c("x2", "g3", "n2", "m3", "2w", "ps2", "kg2"),
  "S_2018_p" = c(3,5,11,3,9,22,6),
  "S_2019_p" = c(3,5,6,21,1,4,0),
  "S_2020_p" = c(0,4,13,9,16,7,9),
  "S_2021_p" = c(4,0,3,8,5,4,6),
  "string2" = c("si", "q2", "oq", "mx", "ix", "p2", "2q"))

  ID string1 S_2018_p S_2019_p S_2020_p S_2021_p string2
1: a1      x2        3        3        0        4      si
2: a2      g3        5        5        4        0      q2
3: a3      n2       11        6       13        3      oq
4: a4      m3        3       21        9        8      mx
5: a5      2w        9        1       16        5      ix
6: a6     ps2       22        4        7        4      p2
7: a7     kg2        6        0        9        6      2q

df2 = data.table(
  ID = c("a1", "a2", "a3", "a4", "a5", "a6", "a7"),
  "string1" = c("x2", "g3", "n2", "m3", "2w", "ps2", "kg2"),
  "S_2018_p" = c(3,5,0,3,9,0,6),
  "S_2019_p" = c(3,5,6,0,1,4,0),
  "S_2020_p" = c(0,4,0,9,0,7,9),
  "S_2021_p" = c(4,0,3,8,5,4,6),
  "string2" = c("si", "q2", "oq", "mx", "ix", "p2", "2q"))

   ID string1 S_2018_p S_2019_p S_2020_p S_2021_p string2
1: a1      x2        3        3        0        4      si
2: a2      g3        5        5        4        0      q2
3: a3      n2        0        6        0        3      oq
4: a4      m3        3        0        9        8      mx
5: a5      2w        9        1        0        5      ix
6: a6     ps2        0        4        7        4      p2
7: a7     kg2        6        0        9        6      2q

我尝试使用 mutete_if 执行此操作，但没有得到所需的输出。

df1 %>% 
  mutate_if(is.numeric, ~1 * (. >= 10))

Answer 1

Dplyr（使用最新版本）有一个很好的“across()”函数可以与 mutate 一起使用。请务必更新您的 dplyr 软件包，因为它是最新的

library(dplyr)

df1 %>% mutate(across(where(is.numeric), function(x) ifelse(x >= 10, 0, x)))

   ID string1 S_2018_p S_2019_p S_2020_p S_2021_p string2
1: a1      x2        3        3        0        4      si
2: a2      g3        5        5        4        0      q2
3: a3      n2        0        6        0        3      oq
4: a4      m3        3        0        9        8      mx
5: a5      2w        9        1        0        5      ix
6: a6     ps2        0        4        7        4      p2
7: a7     kg2        6        0        9        6      2q

Answer 2

另一种解决方案，基于dplyr：

library(tidyverse)
library(data.table)

df1 = data.table(
  ID = c("a1", "a2", "a3", "a4", "a5", "a6", "a7"),
  "string1" = c("x2", "g3", "n2", "m3", "2w", "ps2", "kg2"),
  "S_2018_p" = c(3,5,11,3,9,22,6),
  "S_2019_p" = c(3,5,6,21,1,4,0),
  "S_2020_p" = c(0,4,13,9,16,7,9),
  "S_2021_p" = c(4,0,3,8,5,4,6),
  "string2" = c("si", "q2", "oq", "mx", "ix", "p2", "2q"))

df1 %>% 
  mutate(
    across(
      where(is.numeric),
      ~ if_else(get(cur_column()) > 10, 0, get(cur_column()))))

#>    ID string1 S_2018_p S_2019_p S_2020_p S_2021_p string2
#> 1: a1      x2        3        3        0        4      si
#> 2: a2      g3        5        5        4        0      q2
#> 3: a3      n2        0        6        0        3      oq
#> 4: a4      m3        3        0        9        8      mx
#> 5: a5      2w        9        1        0        5      ix
#> 6: a6     ps2        0        4        7        4      p2
#> 7: a7     kg2        6        0        9        6      2q

---

还有另一个解决方案，基于purrr::map_if：

library(tidyverse)
library(data.table)

df1 = data.table(
  ID = c("a1", "a2", "a3", "a4", "a5", "a6", "a7"),
  "string1" = c("x2", "g3", "n2", "m3", "2w", "ps2", "kg2"),
  "S_2018_p" = c(3,5,11,3,9,22,6),
  "S_2019_p" = c(3,5,6,21,1,4,0),
  "S_2020_p" = c(0,4,13,9,16,7,9),
  "S_2021_p" = c(4,0,3,8,5,4,6),
  "string2" = c("si", "q2", "oq", "mx", "ix", "p2", "2q"))

df1 %>% map_if(is.numeric, ~ ifelse(.x > 10 , 0, .x)) %>% as.data.table

#>    ID string1 S_2018_p S_2019_p S_2020_p S_2021_p string2
#> 1: a1      x2        3        3        0        4      si
#> 2: a2      g3        5        5        4        0      q2
#> 3: a3      n2        0        6        0        3      oq
#> 4: a4      m3        3        0        9        8      mx
#> 5: a5      2w        9        1        0        5      ix
#> 6: a6     ps2        0        4        7        4      p2
#> 7: a7     kg2        6        0        9        6      2q

Answer 3

您可以在感兴趣的特定列上使用apply 和ifelse 函数。例如：

apply(df1[,c(3,4,5,6)], MARGIN = c(1,2), FUN = function(x) ifelse(x >= 10, 0, x))

apply 函数将作用于数据框的选定行 (df1[,c(3,4,5,6)]) 并将应用函数 FUN = ifelse(x >= 10, 0, x)（如果 x 大于或等于 10，则将其替换为 0，否则，将其替换用它自己（不要替换它））在数据框的每个单元格上（MARGIN = c(1,2)）。

你当然可以用更正的部分替换数据框部分：

df1[,c(3,4,5,6)] <- apply(df1[,c(3,4,5,6)], MARGIN = c(1,2), FUN = function(x) ifelse(x >= 10, 0, x))