在回答
this question,时,我写了以下代码：
df <- data.frame(Call_Num = c("HV5822.H4 C47 Circulating Collection,3rd Floor","QE511.4 .G53 1982 Circulating Collection,"TL515 .M63 Circulating Collection,"D753 .F4 Circulating Collection,"DB89.F7 D4 Circulating Collection,3rd Floor"))
require(stringr)
matches = str_match(df$Call_Num,"([A-Z]+)(\\d+)\\s*\\.")
df2 <- data.frame(df,letter=matches[,2],number=matches[,3])
现在我的问题是：有没有一种简单的方法将最后两行合并为一个dplyr调用,大概是使用mutate()？或者,我也对do()的解决方案感兴趣.对于mutate()方法,由于我们正在提取2个组,因此我将采用一个解决方案,使用不同的正则表达式调用str_match()两次,每个所需的组一个.
编辑：为了澄清,我在这里看到的主要挑战是str_match返回一个矩阵,我想知道如何在mutate()或do()中处理它.我对使用其他提取信息的方法解决原始问题不感兴趣.已经有很多这样的解决方案已经给出了here.
解决方法
你可以尝试做
df %>%
do(data.frame(.,str_match(.$Call_Num,"([A-Z]+)(\\d+)\\s*\\.")[,-1],stringsAsFactors=FALSE)) %>%
rename_(.dots=setNames(names(.)[-1],c('letter','number')))
# Call_Num letter number
#1 HV5822.H4 C47 Circulating Collection,3rd Floor HV 5822
#2 QE511.4 .G53 1982 Circulating Collection,3rd Floor QE 511
#3 TL515 .M63 Circulating Collection,3rd Floor TL 515
#4 D753 .F4 Circulating Collection,3rd Floor D 753
#5 DB89.F7 D4 Circulating Collection,3rd Floor DB 89
--- %>% setNames(.,c(names(.)[1],"letter","number"))
