How to calculate a (co-)occurrence matrix from a data frame with several columns using R?

Question

I\'m a rookie in R and currently working with collaboration data in the form of an edge list with 32 columns and around 200.000 rows. I want to create a (co-)occurrence matr

Answer 1

Here is a way using dplyr and tidyr packages, the whole idea lies in creating a dataframe with row-wise occurrence of each country then joining it on itself.

library(dplyr)

# Create dataframe sammple
df <- data.frame(ID = c(1,2,3,4), 
                 V1 = c("England", "England", "China", "England"),
                 V2 = c("Greece", "England", "Greece", "England"),
                 V32 = c("USA", "China", "Greece", "England"),
                 stringsAsFactors = FALSE)

# Get the occurance of each country in every row.
row_occurance <- 
  df %>%
  tidyr::gather(key = "identifier", value = "country", -ID) %>%
  group_by(ID, country) %>%
  count()

row_occurance %>%
  # Join row_occurance on itself to simulate the matrix
  left_join(row_occurance, by = "ID") %>%
  # Get the highest occurance row wise, this to handle when country
  # name is repeated within same row
  mutate(Occurance = pmax(n.x, n.y)) %>%
  # Group by 2 countries
  group_by(country.x, country.y) %>%
  # Sum the occurance of 2 countries together
  summarise(Occurance = sum(Occurance)) %>%
  # Spread the data to make it in matrix format
  tidyr::spread(key = "country.y", value = "Occurance", fill = 0)

# # A tibble: 4 x 5
# # Groups:   country.x [4]
# country.x China England Greece   USA
# <chr>     <dbl>   <dbl>  <dbl> <dbl>
# China         2       2      2     0
# England       2       6      1     1
# Greece        2       1      3     1
# USA           0       1      1     1

Answer 2

An option using base::table:

df <- data.frame(ID = c(1,2,3,4), 
    V1 = c("England", "England", "China", "England"),
    V2 = c("Greece", "England", "Greece", "England"),
    V3 = c("USA", "China", "Greece", "England"))

#get paired combi and remove those from same country
pairs <- as.data.frame(do.call(rbind, 
    by(df, df$ID, function(x) t(combn(as.character(x[-1L]), 2L)))))
pairs <- pairs[pairs$V1!=pairs$V2, ]

#repeat data frame with columns swap so that 
#upper and lower tri have same numbers and all countries are shown
pairs <- rbind(pairs, data.frame(V1=pairs$V2, V2=pairs$V1))

#tabulate pairs
tab <- table(pairs)

#set diagonals to be the count of countries
cnt <- c(table(unlist(df[-1L])))
diag(tab) <- cnt[names(diag(tab))]

tab

output:

         V2
V1        China England Greece USA
  China       2       2      2   0
  England     2       6      1   1
  Greece      2       1      3   1
  USA         0       1      1   1

Answer 3

There may be better ways to do this, but try:

library(tidyverse)

df1 <- df %>%
pivot_longer(-ID, names_to = "Category", values_to = "Country") %>%
xtabs(~ID + Country, data = ., sparse = FALSE) %>% 
crossprod(., .) 

df_diag <- df %>% 
pivot_longer(-ID, names_to = "Category", values_to = "Country") %>%
mutate(Country2 = Country) %>%
xtabs(~Country + Country2, data = ., sparse = FALSE) %>% 
diag()

diag(df1) <- df_diag 

df1

Country   China England Greece USA
  China       2       2      2   0
  England     2       6      1   1
  Greece      2       1      3   1
  USA         0       1      1   1