r语言 - 如何从单个向量排列 2x2 数据帧

Question

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，该函数将通读每一行并将带有日期的行后面的每一行移动到新列，并删除所有不遵循日期的行。基于上述示例，我所需的输出如下：

                  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

我的直觉是复制v1然后lead它并使用ifelse创建一个新列，如下所示，但我没有运气，甚至不知道从那里开始。

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

Answer 1

使用 grep 的基本 R 选项。首先找出严格遵循日期模式的索引，然后使用该索引的下一行创建一个具有新列的新数据帧。

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

Answer 2

一种选择是从"v1"创建一个新列作为列的lead，并且仅filter以数字开头的元素或"v1"中的Date格式

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