英文:
Order elements into different columns using separate()
问题
我有一些来自“选择所有适用”的调查数据，现在我有类似以下的数据：
ID <- c(1,2,3,4,5)answer <- c("apple, orange", "pinneaple, apple", "orange, pinneaple", "pinneaple, orange","apple")df <- data.frame(ID,answer)> dfID answer1 1 apple, orange2 2 pinneaple, apple3 3 orange, pinneaple4 4 pinneaple, orange5 5 apple
现在，我正在尝试将其分开并展开结果，为此我正在使用dplyr中的separate，但它只部分实现了我想要的效果：
df %>%separate(answer, into = c("ans1", "ans2","ans3"), sep = ", ")> dfID ans1 ans2 ans31 1 apple orange <NA>2 2 pinneaple apple <NA>3 3 orange pinneaple <NA>4 4 pinneaple orange <NA>5 5 apple <NA> <NA>
现在，我想要按顺序排列响应。在示例中，将ans1分配给apple等等，直到我们最终得到这样的结果：
> target_dfID ans1 ans2 ans31 1 apple orange <NA>2 2 apple <NA> pinneaple3 3 <NA> orange pinneaple4 4 <NA> orange pinneaple5 5 apple <NA> <NA>
有没有办法向separate函数添加此排序属性？
英文:
I have some data from a "select all that apply" survey data and I now have something like this:
ID <- c(1,2,3,4,5)answer <- c("apple, orange", "pinneaple, apple", "orange, pinneaple", "pinneaple, orange","apple")df <- data.frame(ID,answer)> dfID answer1 1 apple, orange2 2 pinneaple, apple3 3 orange, pinneaple4 4 pinneaple, orange5 5 apple
Now, I am trying to separate this and spread the results, for this I am using separate from dplyr and it does only partially what I am looking for:
df %>%separate(answer, into = c("ans1", "ans2","ans3"), sep = ", ")> dfID ans1 ans2 ans31 1 apple orange <NA>2 2 pinneaple apple <NA>3 3 orange pinneaple <NA>4 4 pinneaple orange <NA>5 5 apple <NA> <NA>
Now, I would like to have the responses ordered. In the example that would be assigning ans1 to apple and so on, until we would end up with something like this:
> target_dfID ans1 ans2 ans31 1 apple orange <NA>2 2 apple <NA> pinneaple3 3 <NA> orange pinneaple4 4 <NA> orange pinneaple5 5 apple <NA> <NA>
Is there a way to add this order attribute to the separate function?
答案1
得分: 1
一个可能的方法如下，尽管它不使用 separate：
library(dplyr)library(tidyr)library(stringr)ID <- c(1,2,3,4,5)answer <- c("apple, orange", "pinneaple, apple", "orange, pinneaple", "pinneaple, orange","apple")df1 <- data.frame(ID,answer)df1 |>separate_rows(answer) |>mutate(value = answer,value = str_replace_all(value, c("apple" = "ans1", "orange" = "ans2", "pinneaple" = "ans3"))) |>pivot_wider(names_from = value, values_from = answer)
创建于2023-07-04，使用 reprex v2.0.2
英文:
One approach could be as follows, although it does not use separate:
library(dplyr)library(tidyr)library(stringr)ID <- c(1,2,3,4,5)answer <- c("apple, orange", "pinneaple, apple", "orange, pinneaple", "pinneaple, orange","apple")df1 <- data.frame(ID,answer)df1 |>separate_rows(answer) |>mutate(value = answer,value = str_replace_all(value, c("apple" = "ans1", "orange" = "ans2", "pinneaple" = "ans3"))) |>pivot_wider(names_from = value, values_from = answer)#> # A tibble: 5 × 4#> ID ans1 ans2 ans3#> <dbl> <chr> <chr> <chr>#> 1 1 apple orange <NA>#> 2 2 apple <NA> pinneaple#> 3 3 <NA> orange pinneaple#> 4 4 <NA> orange pinneaple#> 5 5 apple <NA> <NA>
<sup>Created on 2023-07-04 with reprex v2.0.2</sup>
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