循环以替换匹配值

Question

我正在寻找一种简单而优雅的方法来实现这一目标。 所以，如果我有数据集x和关系是A -> B -> Z -> Y和D -> H -> G，我想创建数据集y。不幸的是，它们不一定是有序的：

> x <- data.frame(
+     from = as.character(c("A", "E", "B", "D", "H", "Z")), 
+     to = as.character(c("B", "E", "Z", "H", "G", "Y")))
> 
> y <- data.frame(
+     from = as.character(c("A", "E", "B", "D", "H", "Z")), 
+     to = as.character(c("Y", "E", "Y", "G", "G", "Y")))
> 
> x
  from to
1    A  B
2    E  E
3    B  Z
4    D  H
5    H  G
6    Z  Y
> y
  from to
1    A  Y
2    E  E
3    B  Y
4    D  G
5    H  G
6    Z  Y

我有一个相当大的数据集（目前有500k行;将来会增长）并且实际上关心性能;我不确定是否有任何其他方法可以在没有for循环或甚至向量化/并行化过程的情况下执行此操作。 我正在考虑拆分和删除所有行from == to或创建指针以跳过某些行，这样循环不必每次都通过整个数据集。 我还想知道如果我创建一个循环，断点应该是什么;我不确定如何定义循环何时停止。 任何建议，将不胜感激。谢谢！

Answer 1

我们可以使用dplyr通过比较'to'和'from'的相邻元素来创建分组变量，并将'to'中的值更改为'to'的last元素

library(dplyr)
x %>% 
    group_by(grp = cumsum(lag(lead(from, default = last(from)) != 
      as.character(to), default = TRUE))) %>% 
    mutate(to = last(to)) %>%
    ungroup %>%
    select(-grp)
# A tibble: 4 x 2
#  from   to    
# <fctr> <fctr>
#1 A      D     
#2 B      D     
#3 C      D     
#4 E      E    

Answer 2

使用来自lag的dplyr和来自fill的tidyr可以实现另一种解决方案：

library(tidyverse)

x %>% arrange(from) %>%
  mutate(samegroup = ifelse(from == lag(to), 1, 0)) %>%
  mutate(group = ifelse(samegroup == 0 | is.na(samegroup), row_number(), NA)) %>%
  fill(group) %>%
  group_by(group) %>%
  mutate(to = last(to)) %>%
  ungroup() %>%
  select(-samegroup, - group)

# A tibble: 6 x 2
#  from  to   
#  <chr> <chr>
#1 A     D    
#2 B     D    
#3 C     D    
#4 E     E    
#5 F     H 
#6 G     H 

Data used

x <- data.frame(from = as.character(c("A", "B", "F", "C", "G", "E")), 
   to = as.character(c("B", "C", "G", "D", "H", "E")), 
   stringsAsFactors = FALSE)