如何从单个向量安排2x2 data.frame

时间:2019-01-28 10:14:31

标签: r regex

我有以下data.frame:

v1<-c("8/12/2018", "hello, how are you", "9/9/2016", "What is going on?","the number three", "9/18/2015", "hello", "9/8/1999","not going", "where to next?")
(df<-as.data.frame(v1, stringsAsFactors=FALSE))
                v1
1           8/12/2018
2  hello, how are you
3            9/9/2016
4   What is going on?
5    the number three
6           9/18/2015
7               hello
8            9/8/1999
9           not going
10      where to next?

我想生成一个函数,该函数将读取每一行,并将紧跟日期的行之后的每一行移动到新列,并删除不紧随日期的所有行。根据上面的示例,我想要的输出如下:

                  v1     value
1           8/12/2018    hello, how are you
2           9/9/2016     What is going on?
3           9/18/2015    hello
4           9/8/1999     not going

我的直觉是先复制v1,然后再复制lead,然后使用ifelse创建一个新列,如下所示,但是我没有运气,也不知道从何而来甚至在那里。

df$value<-ifelse(v1="^\d{1,2}\/\d{1,2}\/\d{4}$", lead(v1),"NA")

2 个答案:

答案 0 :(得分:4)

使用grep的基本R选项。首先找出严格遵循日期格式的索引,然后使用该索引的下一行创建带有新列的新数据框。

inds <- grep("^\\d{1,2}/\\d{1,2}/\\d{4}$", df$v1)
with(df, data.frame(v1 = v1[inds], value = v1[inds + 1]))

#         v1              value
#1 8/12/2018 hello, how are you
#2  9/9/2016  What is going on?
#3 9/18/2015              hello
#4  9/8/1999          not going

答案 1 :(得分:3)

一个选择是从“ v1”创建新列,作为列的leadfilter仅创建以数字或“ v1”格式为Date的元素

library(tidyverse)
df %>% 
   mutate(value = lead(v1)) %>%
   filter(grepl("^\\d+", v1))
   #or
   #filter(!is.na(mdy(v1)))
#         v1              value
#1 8/12/2018 hello, how are you
#2  9/9/2016  What is going on?
#3 9/18/2015              hello
#4  9/8/1999          not going