将数据帧列表取消列出到一个具有更改结构的数据帧

Question

我有一个包含大量数据框的列表（每个数据框代表一年）。每个数据框由两个变量组成：(1) ID 变量，和 (2) 表示当年是否发生事件的伯努利变量。我想取消列出该列表，而是创建一个数据框，其中每年的结果变量是它自己的列。

您可以在下面找到生成示例数据的代码，这些示例数据与我的数据具有相同的特征。

set.seed(123)
df <- list()
year <- (1950:1960)

# Simulating data. 
for (i in 1:length(year)) {
  id <- sample(1:20, 10)
  outcome <- sample(0:1, 10, replace = T)
  
  df[[i]] <- cbind(id, outcome)
}

所以实际上，我想取消列出 df 并创建一个看起来像这样的数据框：

      id outcome1950 outcome1951 outcome1952 outcome1953 outcome1954 outcome1955 [...]
 [1,]  1               
 [2,]  2       
 [3,]  3       
 [4,]  4       
 [5,]  5       
 [6,]  6       
 [7,]  7       
 [8,]  8       
 [9,]  9       
[10,] 10       
[11,] 11       
[12,] 12       
[13,] 13       
[14,] 14       
[15,] 15       
[16,] 16       
[17,] 17       
[18,] 18       
[19,] 19       
[20,] 20       

当然，结果应该具有每年的每个值。

注意：并非所有年份都存在所有 ID，在这种情况下，我想添加一个 NA。

Answer 1

应该这样做。

set.seed(123)
df <- list()
year <- (1950:1960)

# Simulating data. 
for (i in 1:length(year)) {
  id <- sample(1:20, 10)
  outcome <- sample(0:1, 10, replace = T)
  
  df[[i]] <- cbind(id, outcome)
}

## add names to df
names(df) <- year

## turn each element in df into a tibble (or data frame) and make a 
## variable called "year" that has is "outcomeYYYY"
for(i in 1:length(df)){
  df[[i]] <- as_tibble(df[[i]]) %>% mutate(year = paste0("outcome", names(df)[i]))
}

## turn the elements of the list into a single data tibble
df <- bind_rows(df)

## pivot to wider and arrange by id
dfwide <- df %>% 
  pivot_wider(names_from="year", values_from="outcome") %>% 
  arrange(id)

head(dfwide)
# A tibble: 6 x 12
#      id outcome1950 outcome1951 outcome1952 outcome1953 outcome1954
#   <int>       <int>       <int>       <int>       <int>       <int>
# 1     1          NA           0          NA           0          NA
# 2     2           1          NA           0           1          NA
# 3     3           1           0          NA          NA           1
# 4     4           0           0          NA          NA          NA
# 5     5           0          NA           0          NA          NA
# 6     6           0          NA           0           1          NA
# … with 6 more variables: outcome1955 <int>, outcome1956 <int>,
#   outcome1957 <int>, outcome1958 <int>, outcome1959 <int>,
#   outcome1960 <int>

Answer 2

如果我正确理解了您的问题，可能的解决方案是：

library(dplyr)
library(tidyr)
library(purrr)
library(plyr)

purrr::map2(df, year, ~ cbind(.y, .x))  %>% 
  plyr::ldply(data.frame) %>% 
  dplyr::rename(YEAR = 1) %>% 
  tidyr::pivot_wider(names_from = "YEAR", names_prefix = "outcome", values_from = "outcome") %>% 
  dplyr::arrange(id)

Answer 3

library(tidyverse)

names(df) <- paste0('outcome', year)

df %>% 
  purrr::map_df(as.data.frame, .id = 'name') %>% 
  tidyr::pivot_wider(names_from = name, values_from = outcome) %>% 
  dplyr::arrange(id)

# A tibble: 20 x 12
      id outcome1950 outcome1951 outcome1952 outcome1953 outcome1954 outcome1955 outcome1956 outcome1957 outcome1958 outcome1959 outcome1960
   <int>       <int>       <int>       <int>       <int>       <int>       <int>       <int>       <int>       <int>       <int>       <int>
 1     1          NA           0          NA           0          NA           0          NA          NA           0          NA          NA
 2     2           1          NA           0           1          NA           1          NA           1          NA           0          NA
 3     3           1           0          NA          NA           1          NA          NA           0           0          NA          NA
 4     4           0           0          NA          NA          NA           0          NA           1           0           1          NA
 5     5           0          NA           0          NA          NA          NA           1          NA          NA          NA          NA
 6     6           0          NA           0           1          NA           1           1          NA           0           1          NA
 7     7          NA           1           0           0           1          NA           1           0          NA          NA           1
 8     8          NA           0           0           0           1          NA          NA           1           1           0           0
 9     9          NA           0           0          NA           1          NA          NA           1          NA           0           0
10    10           0           0           0          NA          NA           0           1          NA          NA           1           1
11    11           0          NA          NA           1          NA          NA           1          NA           1           0          NA
12    12          NA          NA          NA          NA           0           1          NA          NA           1           0           0
13    13          NA          NA           0          NA           1          NA          NA           1           0           1           0
14    14           0           1          NA          NA           0          NA           0          NA          NA           0           0
15    15           1          NA           1           0          NA           0           1          NA           1          NA          NA
16    16          NA          NA          NA           0           0           0          NA          NA           0          NA           0
17    17          NA          NA          NA          NA          NA          NA          NA           1          NA          NA          NA
18    18          NA          NA          NA          NA          NA          NA           1          NA          NA          NA           1
19    19           1           0          NA           0           0           1           0           1          NA          NA          NA
20    20          NA           1           1           1           0           0           0           1          NA          NA           0