根據因子和數值數據創建新列

Question

我有一個數據框，我想使用相鄰列中的數據將一列中的值轉換為新列。 df$species中的每個因素都將成為一個新列，新列的數據將是df$fish_num中的相應數據但是，我真的很困惑 go 如何做到這一點並且不知道在哪里開始！

這是我目前的 df：

     site treatment section species fish_num
 1 Site 1   Control       A    parr        7
 2 Site 1   Control       A  salmon        6
 3 Site 1   Control       B   trout        4
 4 Site 1   Control       B  salmon       12
 5 Site 1 Treatment       A    parr        8
 6 Site 1 Treatment       A  salmon        5
 7 Site 1 Treatment       B   trout       15
 8 Site 1 Treatment       B  salmon        9

df <- structure(list(site = c("Site 1", "Site 1", "Site 1", "Site 1", 
"Site 1", "Site 1", "Site 1", "Site 1"), treatment = c("Control", 
"Control", "Control", "Control", "Treatment", "Treatment", "Treatment",
"Treatment"), section = c("A", "A", "B", "B", "A", "A", "B",
"B"), species = c("parr", "salmon", "trout", "salmon", "parr",
"salmon", "trout", "salmon"), fish_num = c(7L, 6L, 4L, 12L, 8L,
5L, 15L, 9L)), class = "data.frame", row.names = c("1", "2",
"3", "4", "5", "6", "7", "8"))

我希望能夠制作以下內容：

   site treatment section fish_num parr salmon trout
1 Site 1   Control       A        7    7      0     0
2 Site 1   Control       A        6    0      6     0
3 Site 1   Control       B        4    0      0     4
4 Site 1   Control       B       12    0     12     0
5 Site 1 Treatment       A        8    8      0     0
6 Site 1 Treatment       A        5    0      5     0
7 Site 1 Treatment       B       15    0      0    15
8 Site 1 Treatment       B        9    0      9     0

我不確定最好的方法！

Answer 1

如果您想留在tidyr中，一種方法是使用 tidyr 的pivot_wider() 。 我還添加了對mutate_at()和mutate()的調用，以用零替換缺失值並計算列fish_num 。

library(tidyverse)

df %>%
   pivot_wider(names_from = species,
             values_from = fish_num) %>%
   mutate_at(c("parr", "salmon", "trout"), ~replace(., is.na(.), 0)) %>%
   mutate(fish_num = parr+salmon+trout)

# A tibble: 4 x 7
site   treatment section  parr salmon trout fish_num
<chr>  <chr>     <chr>   <dbl>  <dbl> <dbl>    <dbl>
1 Site 1 Control   A           7      6     0       13
2 Site 1 Control   B           0     12     4       16
3 Site 1 Treatment A           8      5     0       13
4 Site 1 Treatment B           0      9    15       24

Answer 2

使用tidyverse的pivot_wider

df %>% 
 mutate(id=1:nrow(.)) %>%
 pivot_wider(names_from = species, values_from=fish_num, id_cols=everything())


#>   site   treatment section    id  parr salmon trout
#>   <fct>  <fct>     <fct>   <int> <int>  <int> <int>
#> 1 Site 1 Control   A           1     7     NA    NA
#> 2 Site 1 Control   A           2    NA      6    NA
#> 3 Site 1 Control   B           3    NA     NA     4
#> 4 Site 1 Control   B           4    NA     12    NA
#> 5 Site 1 Treatment A           5     8     NA    NA
#> 6 Site 1 Treatment A           6    NA      5    NA
#> 7 Site 1 Treatment B           7    NA     NA    15
#> 8 Site 1 Treatment B           8    NA      9    NA

Answer 3

您可以使用 R 中的 TRUE 和 FALSE 在數字上表示零和一的技巧：

df$salmon <- df$fish_num * (df$species == "salmon")
df$trout <- df$fish_num * (df$species == "trout")

要對所有出現的物種執行此操作，請將其置於levels(df$species)的循環中。

Answer 4

一個簡單的基礎 R 選項正在使用xtabs

cbind(df,unclass(t(xtabs(fish_num ~species + q,cbind(df,q = 1:nrow(df))))))

這使

    site treatment section species fish_num parr salmon trout
1 Site 1   Control       A    parr        7    7      0     0
2 Site 1   Control       A  salmon        6    0      6     0
3 Site 1   Control       B   trout        4    0      0     4
4 Site 1   Control       B  salmon       12    0     12     0
5 Site 1 Treatment       A    parr        8    8      0     0
6 Site 1 Treatment       A  salmon        5    0      5     0
7 Site 1 Treatment       B   trout       15    0      0    15
8 Site 1 Treatment       B  salmon        9    0      9     0

數據

> dput(df)
structure(list(site = c("Site 1", "Site 1", "Site 1", "Site 1", 
"Site 1", "Site 1", "Site 1", "Site 1"), treatment = c("Control", 
"Control", "Control", "Control", "Treatment", "Treatment", "Treatment",
"Treatment"), section = c("A", "A", "B", "B", "A", "A", "B",
"B"), species = c("parr", "salmon", "trout", "salmon", "parr",
"salmon", "trout", "salmon"), fish_num = c(7L, 6L, 4L, 12L, 8L,
5L, 15L, 9L)), class = "data.frame", row.names = c("1", "2",
"3", "4", "5", "6", "7", "8"))