在 R 中使用正则表达式拆分列

Question

我有一个类型的字符数据

a[1:5]  
"Tampa Bay Buccaneers / 1st / 1st pick / 2015" 
"Tennessee Titans / 1st / 2nd pick / 2015"      
"Oakland Raiders / 1st / 4th pick / 2015"      
"Washington Redskins / 1st / 5th pick / 2015"  
"New York Jets / 1st / 6th pick / 2015"

第一季度。我想将该列分成四个单独的列并存储 每个都在您的数据框中。例如：

Tampa Bay Buccaneers, Tennessee Titans ... in data frame1 , 
1st, 1st, ... in data frame2
1st pick, 2nd pick, ... in data frame 3
2015, 2015, ... in data frame 4

第二季度。我们需要另一个数据框，其中包含数字和变量“a”中删除的所有文本。 例如：112015、122015、132015 等。

我被正则表达式困住了。请帮忙。

Answer 1

我们可以使用" \\/ ") 将“a”向量拆分为list、rbind 并转换为data.frame。

as.data.frame(do.call(rbind, strsplit(a, " \\/ ")), stringsAsFactors=FALSE)

如果我们需要多个data.frames，transposelist输出（'lst'），将list中的每个vector转换为data.frame，更改list名称并使用list2env 在全局环境中创建多个 'data.frame' 对象（虽然不推荐）

library(data.table)
lst <- transpose(strsplit(a, " \\/ "))
list2env(setNames(Map(function(x,y) setNames(data.frame(x),y) , lst, 
        "Col1"), paste0("df", seq_along(lst))), envir = .GlobalEnv)
 df1
#               Col1
#1 Tampa Bay Buccaneers
#2     Tennessee Titans
#3      Oakland Raiders
#4  Washington Redskins
#5        New York Jets

数据

a <- c("Tampa Bay Buccaneers / 1st / 1st pick / 2015", "Tennessee Titans / 1st / 2nd pick / 2015", 
"Oakland Raiders / 1st / 4th pick / 2015", "Washington Redskins / 1st / 5th pick / 2015", 
"New York Jets / 1st / 6th pick / 2015")

Answer 2

a<-c("Tampa Bay Buccaneers / 1st / 1st pick / 2015", "Tennessee Titans / 1st / 2nd pick / 2015",
"Oakland Raiders / 1st / 4th pick / 2015" , "Washington Redskins / 1st / 5th pick / 2015" , "New York Jets / 1st / 6th pick / 2015")

library(stringr)
library(plyr)
options(stringsAsFactors = FALSE)

Q1<-do.call(rbind,str_split(a, "/"))
Q1b<-apply(Q1,2,data.frame)
Q2<-as.data.frame(gsub("[A-Za-z///]+[[:blank:]]","",a))

Answer 3

stringr 包有几个有用的函数：str_split 和 str_split_fixed。

library(stringr)
data.frame(str_split_fixed(a, "/", n=4))