如何使用R清洁城市和州（包括完整和缩写）

Question

我有一份未清洁的城市和州的名单来自＆＃34;位置＆＃34;在推特中，例如：

location <- c("the Great Lake State", "PA", "Harrisburg, Pennsylvania",
              "Pennsylvania", "MI", "Detroit,MI")

如何清理数据以生成包含城市和州的两列的清洁列表？

Answer 1

你可以这样做：

splitted_list <- strsplit(location,",")
wide_matrix   <- sapply(splitted_list,function(x) c(rep(NA,length(x)==1),x))
res <- setNames(data.frame(t(wide_matrix),stringsAsFactors = FALSE),c("city","state"))
res
#         city                state
# 1       <NA> the Great Lake State
# 2       <NA>                   PA
# 3 Harrisburg         Pennsylvania
# 4       <NA>         Pennsylvania
# 5       <NA>                   MI
# 6    Detroit                   MI

Answer 2

假设您的数据（location）已经是要清理的data.frame的一部分，那么tidyr::separate可能是合适的选项。

location <- c("the Great Lake State", "PA", "Harrisburg, Pennsylvania",
              "Pennsylvania", "MI", "Detroit,MI")


library(tidyverse)

as.data.frame(location) %>% # I created a data.frame, which is not needed in actual data 
  tidyr::separate(location, c("City", "State"), sep=",", fill="left")

#         City                State
# 1       <NA> the Great Lake State
# 2       <NA>                   PA
# 3 Harrisburg         Pennsylvania
# 4       <NA>         Pennsylvania
# 5       <NA>                   MI
# 6    Detroit                   MI