在R中将一个变量拆分为多个变量答案

【问题标题】：Split one variable into multiple variables in R在R中将一个变量拆分为多个变量
【发布时间】：2018-05-13 17:24:05
【问题描述】：

我对 R 比较陌生。我的问题并不完全像标题那么简单。这是df 的样例：

id    amenities
1     wireless internet, air conditioning, pool, kitchen
2     pool, kitchen, washer, dryer
3     wireless internet, kitchen, dryer
4     
5     wireless internet

这就是我想要 df 的样子：

id    wireless internet   air conditioning   pool   kitchen   washer   dryer
1     1                   1                  1      1         0        0
2     0                   0                  1      1         1        1
3     1                   0                  0      1         0        1
4     0                   0                  0      0         0        0
5     1                   0                  0      0         0        0

重现数据的示例代码

df <- data.frame(id = c(1, 2, 3, 4, 5),
      amenities = c("wireless internet, air conditioning, pool, kitchen",  
                    "pool, kitchen, washer, dryer", 
                    "wireless internet, kitchen, dryer", 
                    "", 
                    "wireless internet"), 
      stringsAsFactors = FALSE)

【问题讨论】：

标签： r string variables dataframe split

【解决方案1】：

为了完整起见，这里也有一个简洁的data.table解决方案：

library(data.table)
setDT(df)[, strsplit(amenities, ", "), by = id][
  , dcast(.SD, id ~ V1, length)]

   id air conditioning dryer kitchen pool washer wireless internet
1:  1                1     0       1    1      0                 1
2:  2                0     1       1    1      1                 0
3:  3                0     1       1    0      0                 1
4:  5                0     0       0    0      0                 1

强制转换为 data.table 后，amenities 被 ", " 拆分为每个项目的单独行（长格式）。然后将其重新整形为宽格式，使用length() 函数进行聚合。

【讨论】：

【解决方案2】：

dummies 包在这里很有用。试试

library(dplyr); library(tidyr); library(dummies)
df2 <- df %>% separate_rows(amenities, sep = ",")
df2$amenities <- trimws(df2$amenities, "both") # remove spaces (left and right) - so that you will not have 2 "pool" columns in your final data frame
df2 <- dummy.data.frame(df2)[, -2]
colnames(df2) <- trimws(gsub("amenities", "", colnames(df2)), "both") # arrange colnames
df3 <- df2 %>% 
  group_by(id) %>%
  summarise_all(funs(sum)) ## aggregate by column and id
df3

# A tibble: 5 x 7
#id `air conditioning` dryer kitchen  pool washer `wireless internet`
#<dbl>              <int> <int>   <int> <int>  <int>               <int>
#     1                  1     0       1     1      0                   1
#     2                  0     1       1     1      1                   0
#     3                  0     1       1     0      0                   1
#     4                  0     0       0     0      0                   0
#     5                  0     0       0     0      0                   1

【讨论】：

【解决方案3】：

FWIW，这是一个基本的 R 方法（假设 df 包含您的数据，如问题所示）

dat <- with(df, strsplit(amenities, ', '))
df2 <- data.frame(id = factor(rep(df$id, times = lengths(dat)),
                              levels = df$id),
                  amenities = unlist(dat))
df3 <- as.data.frame(cbind(id = df$id,
                     table(df2$id, df2$amenities)))

这会导致

> df3
  id air conditioning dryer kitchen pool washer wireless internet
1  1                1     0       1    1      0                 1
2  2                0     1       1    1      1                 0
3  3                0     1       1    0      0                 1
4  4                0     0       0    0      0                 0
5  5                0     0       0    0      0                 1

分解正在发生的事情：

dat <- with(df, strsplit(amenities, ', ')) 在', ' 上拆分amenities 变量，导致

> dat
[[1]]
[1] "wireless internet" "air conditioning"  "pool"             
[4] "kitchen"          

[[2]]
[1] "pool"    "kitchen" "washer"  "dryer"  

[[3]]
[1] "wireless internet" "kitchen"           "dryer"            

[[4]]
character(0)

[[5]]
[1] "wireless internet"

第二行采用dat 并将其转换为向量，然后我们通过重复原始id 值与id 的便利设施数量一样多次来添加id 列。这导致

> df2
   id         amenities
1   1 wireless internet
2   1  air conditioning
3   1              pool
4   1           kitchen
5   2              pool
6   2           kitchen
7   2            washer
8   2             dryer
9   3 wireless internet
10  3           kitchen
11  3             dryer
12  5 wireless internet

使用table() 函数创建列联表，然后我们添加id 列。

【讨论】：

我有轻微的变化 - df2 <- stack(setNames(dat,seq_along(dat)))[2:1] 然后cbind(df["id"], unclass(table(df2)))

【解决方案4】：

使用dplyr 和tidyr 的解决方案。请注意，我将"" 替换为None，因为以后处理列名更容易。

library(dplyr)
library(tidyr)

df2 <- df %>%
  separate_rows(amenities, sep = ",") %>%
  mutate(amenities = ifelse(amenities %in% "", "None", amenities)) %>%
  mutate(value = 1) %>%
  spread(amenities, value , fill = 0) %>%
  select(-None)
df2
#   id  air conditioning  dryer  kitchen  pool  washer pool wireless internet
# 1  1                 1      0        1     1       0    0                 1
# 2  2                 0      1        1     0       1    1                 0
# 3  3                 0      1        1     0       0    0                 1
# 4  4                 0      0        0     0       0    0                 0
# 5  5                 0      0        0     0       0    0                 1

【讨论】：