在 sapply 中使用 rep 根据另一个向量拉伸一个向量

Question

我想生成一个data.frame 的边。当许多边在一个节点上结束时，就会出现问题。边在向量from 和to 中定义。

# Data
vertices <- data.frame(id = 1:3, label = c("a", "b", "c"), stringsAsFactors = FALSE)
to <- c("a", "b", "c")
from1 <- c("c", "a", "b")
from2 <- c("c", "a", "a,b,c")

我尝试了什么：

# Attempt 1
create_edges_1 <- function(from, to) {
  to <- sapply(to, function(x){vertices$id[vertices$label == x]})
  from <- sapply(from, function(x){vertices$id[vertices$label == x]})
  data.frame(from = from, to = to, stringsAsFactors = FALSE)
}

这适用于例如create_edges_1(from1, to)，输出为：

  from to
c    3  1
a    1  2
b    2  3

但是，例如from2，此尝试失败。

所以我尝试了以下方法：

# Attempt 2
create_edges_2 <- function(from, to) {
  to <- sapply(unlist(sapply(strsplit(to, ","), function(x){vertices$id[vertices$label == x]})), function(x){rep(x, sapply(strsplit(from2, ","), length))})
  from <- unlist(sapply(strsplit(from2, ","), function(x){vertices$id[vertices$label == x]}))
  data.frame(from = from, to = to, stringsAsFactors = FALSE)
}

这个想法是为不止一条边结束的每个节点“拉伸”to。但是create_edges_2(from1, to) 和create_edges_2(from2, to) 都会抛出错误

rep(x, sapply(strsplit(from2, ","), length)) 中的错误： 'times' 参数无效

我在sapply 语句中做错了什么？

create_edges_2(from2, to) 的预期输出是：

  from to
     3  1
     1  2
     1  3
     2  3
     3  3

Answer 1

您可以为此使用连接或match

f2 <- strsplit(from2, ',')

df <- data.frame(from = unlist(f2)
                 , to = rep(to, lengths(f2))
                 , stringsAsFactors = FALSE)

与match

library(tidyverse)

map_dfc(df, ~ with(vertices, id[match(.x, label)]))

# # A tibble: 5 x 2
#    from    to
#   <int> <int>
# 1     3     1
# 2     1     2
# 3     1     3
# 4     2     3
# 5     3     3

有连接

library(dplyr)

df %>% 
  inner_join(vertices, by = c(from = 'label')) %>% 
  inner_join(vertices, by = c(to = 'label')) %>% 
  select_at(vars(matches('.x|.y')))

#   id.x id.y
# 1    3    1
# 2    1    2
# 3    1    3
# 4    2    3
# 5    3    3

Answer 2

这是一种方法：

# Attempt 3
library(dplyr)
to <- sapply(to, function(x){vertices$id[vertices$label == x]})
from0 <- sapply(from2, function(x) strsplit(x, ",")) %>% unlist() %>% as.character()
lengths0 <- lapply(sapply(from2, function(x) strsplit(x, ",")), length) %>% unlist()

to0 <- c()
for( i in 1:length(lengths0)) to0 <- c(to0, rep(to[i], lengths0[i]))

from <- sapply(from0, function(x){vertices$id[vertices$label == x]})
edges <- data.frame(from = from, to = to0, stringsAsFactors = FALSE)
edges

根据要求给出这个结果：

  from to
1    3  1
2    1  2
3    1  3
4    2  3
5    3  3

这个想法是用逗号分隔符分割from，并存储每个元素的大小以便“拉伸”每个节点。这里使用for 循环完成