如何将具有多个分类值的单列的长格式转换为宽格式

Question

我的数据包含一个名为“states”的列，其中包含多个逗号分隔值，如下所示

test <- structure(list(states = c("WA", "SC", "IN", "IN", "WI", "NY",
"CA, CO, CT, DE, FL, GA, IA, ID, IL, IN, LA, MD, MI, MT, NJ, NV, OH, PA, SC, TX, UT, VA, WA", 
"CA, CO, DE, GA, IL, LA, MA, MD, MI, MO, NJ, NV, NY, PA, VA, TX, WA", 
"LA, MS", "DC, MD, VA", "AL, GA, NC", "MN WI", "MN WI", "KS, OK, TX", 
"KS, MO, OK, TX", "IN, MI, NY, OH, PA", "CO, NE", "CO", "CO, NE", 
"AZ, CA, CO, NV, TX, WA", "AZ, CA, NV, TX, UT,WA", "AZ, CA, NV, TX, UT, WA", 
"CA, CT, IL, WA", "AL, AZ, CA, IL, MI, MO, MT, NJ, NM, OH, OK, PA, TX, VA, WI", 
"AL, NC, TX, VA", "IL, MO, NJ, OH", "AZ, CA, CO, MN", "CO, IA, KY, TX", 
"CO, IA, KY, MI, NC, NE, OH, PA, TX", "AR, GA, NC, NM, OK", "AL & WV", 
"KY, MN, ND, OH,OR,PA", "KS", "AL, AR, AZ, CA, CT, DE, FL, GA, HI, IA, IL, IN, KS, KY, LA, MA, MD, MI, MN, MO, MS, NC, NE, NJ, NM, NY, OH, OK, OR, PA, RI, SC, TN, TX, UT, VA, WI", 
"AR, CO, GA, IL, LA, MI, MN, MS, MT, NC, ND, NE, OH, PA, RI, SC, TX, WI", 
"AL, AR, AZ, CA, CT, DE, FL, GA, HI, IA, IL, IN, KS, KY, LA, MA, MD, MI, MN, MO, MS, NC, NE, NJ, NM, NY, OH, OK, OR, PA, RI, SC, TN, TX, UT, VA, WI", 
"AL, AR, AZ, CA, CT, DE, FL, GA, HI, IA, IL, IN, KS, KY, LA, MA, MD, MI, MN, MO, MS, NC, NE, NJ, NM, NY, OH, OK, OR, PA, RI, SC, TN, TX, UT, VA, WI", 
"AL, AZ, FL, KS, MI, MN, MO, NC, OK, WI", "GA, SC", "CA, CO, FL, IL, KY, NJ, OH, TX, VA", 
"AL, AZ, CA, FL, GA, NJ, NM, NV, OH, PA, TX, VA", "ALL 50 STATES", 
"ALL 50 STATES", "ALL 50 STATES", "AL, AZ, FL, GA, MI, NJ, NY, OH, OR, PA, TX, UT"
)), .Names = "states", row.names = c(NA, -45L), class = c("tbl_df", 
                                                            "tbl", "data.frame"))
test

我想将其转换为一种格式，其中每个“状态”作为一列，1 表示状态存在，否则为零。

谢谢

Answer 1

这可能是你想要的。由于您没有提供预期的输出，这是我根据您的描述做出的解释。想法是用rowid_to_column添加索引，用“ALL”替换“ALL 50 STATES”，用separate_rows分隔基于符号和空格的状态，然后spread数据框。

library(tidyverse)

test2 <- test %>%
  # Create index
  rowid_to_column() %>%
  # Replace ALL 50 STATES with ALL
  mutate(states = replace(states, states %in% "ALL 50 STATES", "ALL")) %>%
  # Separate states with punct and space
  separate_rows(states, sep = "[[:punct:][:space:]]+") %>%
  group_by(rowid) %>%
  mutate(Group_ID = row_number(), Present = 1L) %>%
  spread(states, Present, fill = 0L) %>%
  select(-Group_ID)

Answer 2

首先，我加载库。

# Load libraries
library(dplyr)
library(magrittr)
library(datasets)

接下来，我将您的数据集中的ALL 50 STATES 替换为所有 50 个州的缩写。 （state.abb 来自 datasets 包。）

# Change "ALL 50 STATES" to state abbreviations
test %<>%
  mutate(states = ifelse(states == "ALL 50 STATES", paste0(state.abb, collapse = ","), states))

最后，我遍历每个元素并使用strsplit 解析出状态，使用table 计算每个状态，使用bind_rows 将结果绑定到一个数据帧中，并将NAs 替换为零replace_na 和 mutate_all。

# Count assuming state only can appear once per row
do.call(bind_rows, lapply(test$states, function(x)table(strsplit(x, "[[:punct:][:space:]]+")))) %>% 
  mutate_all(replace_na, replace = 0)

[注意您的数据集有点乱：大多数状态用逗号分隔，但有些只有空格或 & 符号。我曾使用[[:punct:][:space:]]+ 来解释所有这些可能性。]

这是前 10 行和前 10 个状态的示例：

#      WA    SC    IN    WI    NY    CA    CO    CT    DE    FL
#   <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1     1     0     0     0     0     0     0     0     0     0
# 2     0     1     0     0     0     0     0     0     0     0
# 3     0     0     1     0     0     0     0     0     0     0
# 4     0     0     1     0     0     0     0     0     0     0
# 5     0     0     0     1     0     0     0     0     0     0
# 6     0     0     0     0     1     0     0     0     0     0
# 7     1     1     1     0     0     1     1     1     1     1
# 8     1     0     0     0     1     1     1     0     1     0
# 9     0     0     0     0     0     0     0     0     0     0
# 10    0     0     0     0     0     0     0     0     0     0