如何在 R 中创建这些列？

Question

这是示例数据

 stfips <- c("39","39","39")
 year <- c("2023", "2023","2023")
 industry_code <- c(112, 113, 114)
 first_quarter_establishments <- c(987,654,321)
 county <- data.frame(stfips, year, industry_code, first_quarter_establishments)

当前的任务是创建一个名为 period 的新列，其值为 01。01 的原因是它代表第一季度。如果第四列的名称中包含“Second”一词，则句点将为“02”，依此类推。以下是我从 ChatGPT 获得的信息。错误如下。知道如何根据专栏的措辞创建这个时期专栏吗？

first_columns <- grepl("first", names(county), ignore.case = TRUE)
county$period <- ifelse(first_columns, "01", "") 

  Error in `$<-.data.frame`(`*tmp*`, period, value = c("", "", "", "01")) : 
  replacement has 4 rows, data has 3

想要的最终结果

 stfips     year    industry_code    first_quarter_establishments   period
  39        2023         112                 987                    01
  39        2023         113                 654                    01
  39        2023         114                 321                    01

Answer 1

我确信有更优雅的解决方案，但在基础 R 中，您可以将

match

与

gsub

结合使用，从数据框的

names

中识别季度：

quarters <- c("first" = 1, "second" = 2, 
              "third" = 3, "fourth" = 4)

county$quarter <- quarters[match(gsub("(.+?)(\\_.*)", "\\1", names(county[4])), 
                                 names(quarters))]

输出：

#   stfips year industry_code first_quarter_establishments quarter
# 1     39 2023           112                          987       1
# 2     39 2023           113                          654       1
# 3     39 2023           114                          321       1

如果将其更改为第二个：

second_quarter_establishments <- c(987,654,321)
county <- data.frame(stfips, year, industry_code, second_quarter_establishments)

county$quarter <- quarters[match(gsub("(.+?)(\\_.*)", "\\1", names(county[4])), 
                                 names(quarters))]

#   stfips year industry_code second_quarter_establishments quarter
# 1     39 2023           112                           987       2
# 2     39 2023           113                           654       2
# 3     39 2023           114                           321       2