将行（不同名称）的值放入列中 - 复杂的扩散？

Question

我有这个数据框，每个问题有两个子部分（q1，q1_p2）。我想将第二个子部分的答案移到同一行。

question     answer
q1           bleh
q1_p2        bah
q2           meh
q2_p2        bleh

基本上是这样的。

question     answer     p2
q1           bleh       bah
q2           meh        bleh

我通常会使用像传播这样的东西，但我不知道如何结合这个事实，即每个问题的价值将不相同。

有什么想法吗？

Answer 1

如果您的完整数据集遵循示例的结构，那么这就足够了，

library(dplyr)
library(tidyr)

df %>% 
 group_by(question = sub('_.*', '', question)) %>% 
 mutate(new = seq(n())) %>% 
 spread(new, answer) %>% 
 rename(answer = `1`, p2 = `2`) %>% 
 ungroup()

# A tibble: 2 × 3
#  question answer     p2
#*    <chr> <fctr> <fctr>
#1       q1   bleh    bah
#2       q2    meh   bleh

Answer 2

好吧，这不像我希望的那样整洁，但它有效。

library(data.table)
dt = data.table("question" = c("q1", "q1_p2", "q2", "q2_p2"), "answer" = c("bleh","bah","meh","bleh"))
dt$q = sapply(dt$question ,function(x) strsplit(x, "_")[[1]][1])
dt[ , "Row" := 1:.N]

dt[ , "New" := ifelse(nchar(gsub("\\D","",question)) == 1, "answer", gsub("(.+(?=p\\d+))", "",question, perl = T)), by = .(Row)]

dt = dcast(dt, q ~ New, value.var = "answer")
> dt
    q answer   p2
1: q1   bleh  bah
2: q2    meh bleh

Answer 3

以下是tidyverse

的另一个选项

library(tidyverse)
separate(df1, question, into = c("question", "value")) %>% 
       mutate(value = replace(value, is.na(value), "answer")) %>% 
       spread(value, answer)
#  question answer   p2
#1       q1   bleh  bah
#2       q2    meh bleh