在R中按模式合并行

Question

我正在尝试按模式合并行。

数据框只有一列（字符串），通常应遵循日期、公司名称和薪水的模式。但是，有些情况就是没有薪水。

有没有一种方法可以按日期模式合并行？通过这样做，我可以稍后将它们拆分为列。我不想早点做 pivot_wider 的原因是公司名称和薪水之间可能不匹配 - 不平衡的行。所以我认为最好按日期模式合并行，因为日期永远不会丢失并遵循模式。

数据集：

# A tibble: 10 x 1
   detail                                        
   <chr>                                         
 1 26 January 2021                               
 2 NatWest Group - Bristol, BS2 0PT              
 3 26 January 2021                               
 4 NatWest Group - Manchester, M3 3AQ            
 5 15 February 2021                              
 6 Brook Street - Liverpool, Merseyside, L21AB   
 7 £13.84 per hour                               
 8 16 February 2021                              
 9 Anglo Technical Recruitment - London, WC2N 5DU
10 £400.00 per day   

数据集的输入：

structure(list(detail = c("26 January 2021", "NatWest Group - Bristol, BS2 0PT", 
"26 January 2021", "NatWest Group - Manchester, M3 3AQ", "15 February 2021", 
"Brook Street - Liverpool, Merseyside, L21AB", "£13.84 per hour", 
"16 February 2021", "Anglo Technical Recruitment - London, WC2N 5DU", 
"£400.00 per day")), row.names = c(NA, -10L), class = c("tbl_df", 
"tbl", "data.frame"))

预期结果：

 detail                                                                         
 <chr>                                                                          
1 26 January 2021 NatWest Group - Bristol, BS2 0PT                               
2 26 January 2021 NatWest Group - Manchester, M3 3AQ                             
3 15 February 2021 Brook Street - Liverpool, Merseyside, L21AB £13.84 per hour   
4 16 February 2021 Anglo Technical Recruitment - London, WC2N 5DU £400.00 per day

预期结果的输入：

df <- structure(list(detail = c("26 January 2021 NatWest Group - Bristol, BS2 0PT", 
                          "26 January 2021 NatWest Group - Manchester, M3 3AQ", "15 February 2021 Brook Street - Liverpool, Merseyside, L21AB £13.84 per hour", 
                          "16 February 2021 Anglo Technical Recruitment - London, WC2N 5DU £400.00 per day")), row.names = c(NA, -4L), class = c("tbl_df", 
                                                                                  "tbl", "data.frame"))

Answer 1

在每一行前加上一个标签，然后使用read.dcf 创建一个3 列字符矩阵mat。最后，我们将其转换为每个逻辑记录一个元素的字符向量，但您可能只想使用mat，因为这似乎是一种更有用的格式。

我们假设日期具有 %d %B %Y 格式（有关百分比代码，请参见 ?strptime），工资行以 £ 开头，其他行是地址行。

library(dplyr)

mat <- dat %>%
  mutate(detail = case_when(
    !is.na(as.Date(detail, "%d %B %Y")) ~ paste("\nDate:", detail), 
    grepl("^£", detail) ~ paste("Salary:", detail),
    TRUE ~ paste("Address:", detail))) %>%
  { read.dcf(textConnection(.$detail)) }

mat %>%
  apply(1, toString) %>%
  sub(", NA$", "", .)

更新

简化的假设和代码。

Answer 2

另一个解决方案假设仅第一行包含日期。无论两个日期之间的行数如何，它都会起作用..

library(tidyverse)
df %>% group_by(d = cumsum(str_detect(detail, "^(^\\d\\d? \\w+ \\d{4})$"))) %>%
       mutate(c = paste0("Col", as.character(row_number()))) %>%
  pivot_wider(id_cols = d, values_from = detail, names_from = c)

# A tibble: 4 x 4
# Groups:   d [4]
      d Col1             Col2                                           Col3           
  <int> <chr>            <chr>                                          <chr>          
1     1 26 January 2021  NatWest Group - Bristol, BS2 0PT               NA             
2     2 26 January 2021  NatWest Group - Manchester, M3 3AQ             NA             
3     3 15 February 2021 Brook Street - Liverpool, Merseyside, L21AB    £13.84 per hour
4     4 16 February 2021 Anglo Technical Recruitment - London, WC2N 5DU £400.00 per day

Answer 3

这是一个纯粹的data.table 方法

library( data.table )
#make it a data.table
setDT( df )
#first, summarise by block separated by days, collapse the text, using @@ as separator
ans <- df[, .( paste0( detail, collapse = "@@") ), 
          by = .(d = cumsum( ( grepl( "[0-9]{2} [a-zA-Z]+ [0-9]{4}", detail) ) ) ) ]
#split text again to cols, based on te @@ introduced in the collapse/ Number of cols is dynamic!
ans[, paste0( "Col", 1:length( tstrsplit(ans$V1, "@@" ))) := tstrsplit( V1, "@@" )][, V1 := NULL ][]

#    d             Col1                                           Col2            Col3
# 1: 1  26 January 2021               NatWest Group - Bristol, BS2 0PT            <NA>
# 2: 2  26 January 2021             NatWest Group - Manchester, M3 3AQ            <NA>
# 3: 3 15 February 2021    Brook Street - Liverpool, Merseyside, L21AB £13.84 per hour
# 4: 4 16 February 2021 Anglo Technical Recruitment - London, WC2N 5DU £400.00 per day

Answer 4

这是一种 data.table 方法，它使用dcast() 和rowid() 重塑为宽格式。它返回一个包含四列的 data.table：记录号、日期、 公司名称和薪水。

library(data.table)
setDT(df1)[, rn := cumsum(!is.na(lubridate::dmy(detail)))]
dcast(df1, rn ~ rowid(rn, prefix = "Col"), value.var = "detail")

   rn             Col1                                           Col2            Col3
1:  1  26 January 2021               NatWest Group - Bristol, BS2 0PT            <NA>
2:  2  26 January 2021             NatWest Group - Manchester, M3 3AQ            <NA>
3:  3 15 February 2021    Brook Street - Liverpool, Merseyside, L21AB £13.84 per hour
4:  4 16 February 2021 Anglo Technical Recruitment - London, WC2N 5DU £400.00 per day

为了检测开始新记录的行，即带有日期的行，这种方法借鉴了Anil's answer 以及G.Grothendieck's。

dcast() 允许将所有内容打包成一个“单行”（如果不计算 library() 调用）：

library(data.table)
library(lubridate)
dcast(setDT(df1), cumsum(!is.na(dmy(detail))) ~ rowid(cumsum(!is.na(dmy(detail))), prefix = "Col"), 
      value.var = "detail")