字符串拆分数字图案

我有一个如下所示的数据框:

V1                        V2
peanut butter sandwich    2 slices of bread 1 tablespoon peanut butter

我的目标是:

V1                        V2
peanut butter sandwich    2 slices of bread
peanut butter sandwich    1 tablespoon peanut butter

我试图分裂字符串使用strsplit(df$v2, " "),但我只能拆分" ".我不确定你是否只能在第一个数字处拆分字符串,然后取字符直到下一个数字.

1> Jota..：

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您可以按如下方式拆分字符串:

txt <- "2 slices of bread 1 tablespoon peanut butter"

strsplit(txt, " (?=\\d)", perl=TRUE)[[1]]
#[1] "2 slices of bread"          "1 tablespoon peanut butter"

这里使用的正则表达式是查找后跟数字的空格.它使用零宽度正向前瞻(?=)来表示如果空格后跟一个数字(\\d),那么它就是我们要分割的空间类型.为什么零宽度前瞻？这是因为我们不想将数字用作分裂字符,我们只想匹配任何后跟数字的空格.

要使用该想法并构建数据框,请参阅以下示例:

item <- c("peanut butter sandwich", "onion carrot mix", "hash browns")
txt <- c("2 slices of bread 1 tablespoon peanut butter", "1 onion 3 carrots", "potato")
df <- data.frame(item, txt, stringsAsFactors=FALSE)

# thanks to Ananda for recommending setNames
split.strings <- setNames(strsplit(df$txt, " (?=\\d)", perl=TRUE), df$item) 
# alternately: 
#split.strings <- strsplit(df$txt, " (?=\\d)", perl=TRUE)
#names(split.strings) <- df$item

stack(split.strings)
#                      values                    ind
#1          2 slices of bread peanut butter sandwich
#2 1 tablespoon peanut butter peanut butter sandwich
#3                    1 onion       onion carrot mix
#4                  3 carrots       onion carrot mix
#5                     potato            hash browns

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并且,对于管道痴迷:`library(dplyr); 库(tidyr); df%>%mutate(txt = strsplit(txt,"(？= \\ d)",perl = TRUE))%>%unnest(txt)`....

2> A5C1D2H2I1M1..：

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让我们想象你正在处理的事情如下:

mydf <- data.frame(
  V1 = c("peanut butter sandwich", "peanut butter and jam sandwich"), 
  V2 = c("2 slices of bread 1 tablespoon peanut butter", 
         "2 slices of bread 1 tablespoon peanut butter 1 tablespoon jam"))  

mydf
##                               V1
## 1         peanut butter sandwich
## 2 peanut butter and jam sandwich
##                                                              V2
## 1                  2 slices of bread 1 tablespoon peanut butter
## 2 2 slices of bread 1 tablespoon peanut butter 1 tablespoon jam

您可以先在"V2"中添加一个您不期望的分隔符,并使用cSplit我的"splitstackshape"获取"长"数据集格式.

library(splitstackshape)
mydf$V2 <- gsub(" (\\d+)", "|\\1", mydf$V2)
cSplit(mydf, "V2", "|", "long")
##                                V1                         V2
## 1:         peanut butter sandwich          2 slices of bread
## 2:         peanut butter sandwich 1 tablespoon peanut butter
## 3: peanut butter and jam sandwich          2 slices of bread
## 4: peanut butter and jam sandwich 1 tablespoon peanut butter
## 5: peanut butter and jam sandwich           1 tablespoon jam

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以下不足以自己发布作为答案,因为它们是@Jota方法的变体,但我在这里分享它们是为了完整性:

strsplit 在"data.table"中

拆分list自动展平为一列....

library(data.table)
as.data.table(mydf)[, list(
  V2 = unlist(strsplit(as.character(V2), '\\s(?=\\d)', perl=TRUE))), by = V1]

"dplyr"+"tidyr"

您可以使用unnest"tidyr"将列表列扩展为长格式....

library(dplyr)
library(tidyr)
mydf %>% 
  mutate(V2 = strsplit(as.character(V2), " (?=\\d)", perl=TRUE)) %>% 
  unnest(V2)