r - 搜索带条件的行并更改它们

Question

我有以下数据框

head(ca.spe)

 Dataset recording_date grid_cell
1       2 September 1987   7914/2 
2       1 September 1937   8025/3 
3       1 September 1933   8114/1 
4       1 September 1926   8021/4 
5       1 September 1926   8021/4 
6       1 September 1865   7522/21

在这个数据框中有一些行，我想改变。

网格单元格的行在/之后有两个或三个数字需要更改。

例如第 6 行有 7522/21

我想将此行转换为 7522/1 和 7522/2 两行，该行的所有其他信息都应保持不变。

运行结果应该是这样的:

head(ca.spe)

 Dataset recording_date grid_cell
1       2 September 1987   7914/2 
2       1 September 1937   8025/3 
3       1 September 1933   8114/1 
4       1 September 1926   8021/4 
5       1 September 1926   8021/4 
6       1 September 1865   7522/1
7       1 September 1865   7522/2

有人知道怎么做吗？

干杯迈克

Answer 1

这是 tidyverse 的一个选项 - separate 通过在 / 处拆分列 'grid_cell' 一分为二，然后通过拆分创建行在“grid_cell2”列中的每个字符(或数字)之间，最后联合“grid_cell”列以创建“grid_cell”列

library(dplyr)
library(tidyr)
ca.spe %>%
   separate(grid_cell, into = c("grid_cell1", "grid_cell2"), sep = "/") %>% 
   separate_rows(grid_cell2, sep = "(?<=.)(?=.)") %>% 
   unite(grid_cell, starts_with('grid_cell'), sep = "/")

-输出

# A tibble: 7 × 3
  Dataset recording_date grid_cell
    <int> <chr>          <chr>    
1       2 September 1987 7914/2   
2       1 September 1937 8025/3   
3       1 September 1933 8114/1   
4       1 September 1926 8021/4   
5       1 September 1926 8021/4   
6       1 September 1865 7522/2   
7       1 September 1865 7522/1  

数据

ca.spe <- structure(list(Dataset = c(2L, 1L, 1L, 1L, 1L, 1L), 
recording_date = c("September 1987", 
"September 1937", "September 1933", "September 1926", "September 1926", 
"September 1865"), grid_cell = c("7914/2", "8025/3", "8114/1", 
"8021/4", "8021/4", "7522/21")), class = "data.frame", row.names = c("1", 
"2", "3", "4", "5", "6"))