Finding a paired values from the row throughout the dataframe in R - r

Question

I am trying to polishing my R skills and sort of hit my limit.
The issue I am trying to find the solution as follows.
Suppose my dataframe is as below ,
n = c(2, 15, 31 , 33)
n2 = c( 10 , 9, 10 , 40)
n3 = c( 11 , 10 , 11 , 42)
df = data.frame(n , n2 , n3)
> df
n n2 n3
1 2 10 11
2 15 9 10
3 31 10 11
4 33 40 42
if I would like to go through each row , and generate a random pair :eg 2,10 and go through each of the rest of the rows to find a repeated pair and print out the common pairs and number of occurrence , how can I do that?
In the above example , the only pair that repeat is 10 and 11 at rows 1 and 3.
So far I have thought about the pseudo code as follows
for(each row in the dataframe)
{
for (each of the values in the row)
{
for every pair
}
find a repeated pair
if found store in a dataframe
}
and to find the random pair using combn function.
But I am a little lost at the part on iteration through out the dataframe rows.
Pls help.
Thanks a lot!

Answer 1

I think this is what you want. Instead of thinking about selecting every combination of two values for each row, we'll get every combination of two column numbers - which will be the same for every row. Then we use plyr::count as a convenience function to count rows with the same values for an entire data frame at once. This way we can loop over the combinations of column indices rather than over rows. I use apply, but you could write it as a for loop instead.
pairs = combn(ncol(df), m = 2)
result = apply(pairs, MAR = 2, FUN = function(p) {
plyr::count(df[p])
})
names(result) = apply(pairs, MAR = 2, FUN = paste, collapse = "_")
The result is a list where each item is is a data frame with two columns and a freq column giving the number of rows in the original data each value-pair occurred.
result
# $`1_2`
# n n2 freq
# 1 2 10 1
# 2 15 9 1
# 3 31 10 1
# 4 33 40 1
#
# $`1_3`
# n n3 freq
# 1 2 11 1
# 2 15 10 1
# 3 31 11 1
# 4 33 42 1
#
# $`2_3`
# n2 n3 freq
# 1 9 10 1
# 2 10 11 2
# 3 40 42 1
If you want to omit the values that aren't repeated, we can just subset them out:
lapply(result, subset, freq > 1)
# $`1_2`
# [1] n n2 freq
# <0 rows> (or 0-length row.names)
#
# $`1_3`
# [1] n n3 freq
# <0 rows> (or 0-length row.names)
#
# $`2_3`
# n2 n3 freq
# 2 10 11 2

Answer 2

Slightly different method
n = c(2, 15, 31 , 15) # changed the dataset to have some common pairs in n and n2 too
n2 = c( 10 , 9, 10 , 9)
n3 = c( 11 , 10 , 11 , 42)
df = data.frame(n , n2 , n3)
library(dplyr)
library(rlang)
library(utils)
cols<-colnames(df) # define the columns that you want to do the pair checking for
combinations<- as.data.frame(combn(cols,2),stringsAsFactors = FALSE)
# picks up all combinations of columns
#iterates over each pair of columns
all_combs<- lapply(names(combinations[cols1]), function(x){
df %>%
group_by(!! sym( combinations[[x]][1]),!! sym( combinations[[x]][2])) %>%
filter(n()>1) # groups by the two columns, and filters out pairs that occur more than once. You can add a distinct command below if you
#dont want them repeated
})
all_combs_df <- do.call("rbind", all_combs)# all_combs is in a list format, use rbind to convert into a dataframe
all_combs_df
the output is this
n n2 n3
<dbl> <dbl> <dbl>
1 15. 9. 10.
2 15. 9. 42.
3 2. 10. 11.
4 31. 10. 11.