CODEWITHSUNDEEP
rapplysapply
symbolrush
symbolrush

Reputation: 7457

Using rep inside sapply to strech a vector according to another vector

I want to generate a data.frame of edges. Problems arise when many edges end on one node. Edges are defined in vectors from and to.

# 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")

What I tried:

# 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)
}

This works for example create_edges_1(from1, to), the output is: 

  from to
c    3  1
a    1  2
b    2  3

However for example from2 this attempt fails.

So I tried the following:

# 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)
}

The idea was to "stretch" to for every node where more than one edge ends. However create_edges_2(from1, to) and create_edges_2(from2, to) both throw an error

Error in rep(x, sapply(strsplit(from2, ","), length)) : invalid 'times' argument

What am I doing wrong in my sapply statements?

The expected output for create_edges_2(from2, to) is:

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

Upvotes: 1

Views: 63

Answers (2)

IceCreamToucan
IceCreamToucan

Reputation: 28685

You could use joins or match for this

f2 <- strsplit(from2, ',')

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

With 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

With Joins

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

Upvotes: 2

gaut
gaut

Reputation: 5958

Here is a way:

# 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

Giving this result as requested: 

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

The idea is to split from with comma separators, and to store the size of each element in order to "stretch" every node. Here done with a for loop

Upvotes: 1

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