matching and recording in R - r

Question

I have a data frame like this, called df:
a b c d e f
b c f a a a
d f a b c c
f e d f f d
The first row is actually the column name. Let's take an example to explain the meaning here: df[1,1] is b, which means there is relation from a to b, so the values in the column means there is relation from the 'column name' to that entry.
I want create a matrix(df1) with 6*6 dimension, column and row names are both column names of df. The (i,j) entry is 1, if there is relation from 'i' to 'j', otherwise, 0.
The output I want is:
a b c d e f
a 0 1 0 1 0 1
b 0 0 1 0 1 1
c 1 0 0 1 0 1
d 1 1 0 0 0 1
e 1 0 1 0 0 1
f 1 0 1 1 0 0
How to do this with a loop in R?
How to do this without a loop, and only use basic R?
How to do this using some fancy packages in R?

Answer 1

Using the reshape2 package, this is one way to go. My sample data has all columns as character. You use melt() to reshape your data in a long format. Then, you use dcast() from the same package.
library(magrittr)
library(reshape2)
melt(mydf, measure.vars = names(mydf)) %>%
dcast(variable ~ value, length)
variable a b c d e f
1 a 0 1 0 1 0 1
2 b 0 0 1 0 1 1
3 c 1 0 0 1 0 1
4 d 1 1 0 0 0 1
5 e 1 0 1 0 0 1
6 f 1 0 1 1 0 0
EDIT
As mentioned below by akrun, you can do all work using recast() in the reshape2 package.
recast(mydf, measure.var= names(mydf),variable~value, length)
DATA
mydf <- structure(list(a = c("b", "d", "f"), b = c("c", "f", "e"), c = c("f",
"a", "d"), d = c("a", "b", "f"), e = c("a", "c", "f"), f = c("a",
"c", "d")), .Names = c("a", "b", "c", "d", "e", "f"), class = "data.frame", row.names = c(NA,
-3L))

Answer 2

Just use table:
table(colnames(mydf)[col(mydf)], unlist(mydf) )
# a b c d e f
# a 0 1 0 1 0 1
# b 0 0 1 0 1 1
# c 1 0 0 1 0 1
# d 1 1 0 0 0 1
# e 1 0 1 0 0 1
# f 1 0 1 1 0 0
If you have multiple matches, then:
pmin(table(colnames(mydf)[col(mydf)], unlist(mydf) ), 1)

Answer 3

You can do this with reshaping.
library(dplyr)
library(tidyr)
data %>%
gather(from, to) %>%
distinct %>%
mutate(value = 1) %>%
spread(to, value, fill = 0)

Answer 4

The other solution using dplyr is really neat and smart. I recommend using that solution.
Here is an alternative solution to your problem using most basic functions in R.
Say your data frame has n columns and m rows i.e. n <- ncol(df) and m <- nrow(df).
output_matrix <- matrix(rep(0, n*n), ncol = n)
for(i in 1:n){
for(j in 1:m){
# UTF to integer conversion
# utf8ToInt("a") = 97
rowWithRelation <- utf8ToInt(df[j, i]) - 96
output_matrix[rowWithRelation, i] <- 1
}
}
rownames(output_matrix) <- letters[seq(from = 1, to = n)]
colnames(output_matrix) <- letters[seq(from = 1, to = n)]