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Generating an edge list from ID and grouping vectors

Tags:

r

sna

I have a data frame of 205,000+ rows formatted as follows:

df <- data.frame(project.id = c('SP001', 'SP001', 'SP001', 'SP017', 'SP018', 'SP017'),
                 supplier.id = c('1224', '5542', '7741', '1224', '2020', '9122'))

In the actual data frame there are 6700+ unique values of project.id. I would like to create an edge list that pairs suppliers who have worked on the same project.

Desired end result for project.id = SP001:

to     from
1224   5542
1224   7741
5542   7741

So far I've tried using split to create a list by project.id and then running lapply+combn to generate all possible combinations of supplier.id within each list/group:

try.list <- split(df, df$project.id)
try.output <- lapply(try.list, function(x) combn(x$supplier.id, 2))

Is there a more elegant/efficient (read "computed in less than 2hrs") way to generate something like this?

Any help would be much appreciated

like image 425
user3179350 Avatar asked Dec 19 '22 20:12

user3179350


2 Answers

Instead of using split and lapply, you can use the dplyr package.

df <- data.frame(project.id = c('SP001', 'SP001', 'SP001', 'SP017', 'SP018', 'SP017'),
                 supplier.id = c('1224', '5542', '7741', '1224', '2020', '9122'),
                 stringsAsFactors = FALSE)

library(dplyr)

df %>% group_by(project.id) %>%
  filter(n()>=2) %>% group_by(project.id) %>%
 do(data.frame(t(combn(.$supplier.id, 2)), stringsAsFactors=FALSE))
# Source: local data frame [4 x 3]
# Groups: project.id [2]

#   project.id    X1    X2
#        (chr) (chr) (chr)
# 1      SP001  1224  5542
# 2      SP001  1224  7741
# 3      SP001  5542  7741
# 4      SP017  1224  9122
like image 93
Ven Yao Avatar answered Jan 18 '23 20:01

Ven Yao


You can just merge it with itself which gets you all the Cartesian pairs:

 temp <- merge(df,df, by="project.id")
 res <- temp[ temp$supplier.id.x != temp$supplier.id.y , ]

> res

   project.id supplier.id.x supplier.id.y
2       SP001          1224          5542
3       SP001          1224          7741
4       SP001          5542          1224
6       SP001          5542          7741
7       SP001          7741          1224
8       SP001          7741          5542
11      SP017          1224          9122
12      SP017          9122          1224
like image 33
IRTFM Avatar answered Jan 18 '23 18:01

IRTFM