How can I create dummy variables from a numeric variable in R? - r

How can I create dummy variables from a numeric variable in R?
I want to create N dummy variables. In such a way the numeric variable means how many zeros will come, counting from the first column. Imagine N=6. Like this:
x
a 5
b 2
c 4
d 1
e 9
It must become:
1 2 3 4 5 6
a 0 0 0 0 0 1
b 0 0 1 1 1 1
c 0 0 0 0 1 1
d 0 1 1 1 1 1
e 0 0 0 0 0 0
Thank you!


Here's a hacky solution for you
x = c(5,2,4,1,9)
N = 6
out = matrix(1, length(x), N)
for (i in 1:length(x))
out[i,1:min(x[i], N)] = 0
> out
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 0 0 0 0 0 1
[2,] 0 0 1 1 1 1
[3,] 0 0 0 0 1 1
[4,] 0 1 1 1 1 1
[5,] 0 0 0 0 0 0


We could do this in a vectorized manner by creating row/column index and assigning an already created matrix of 1s to 0 based on the index
m1 <- matrix(1, ncol = N, nrow = length(x),
dimnames = list(letters[seq_along(x)], seq_len(N)))
x1 <- pmin(x, ncol(m1))
m1[cbind(rep(seq_len(nrow(m1)), x1), sequence(x1))] <- 0
m1
# 1 2 3 4 5 6
#a 0 0 0 0 0 1
#b 0 0 1 1 1 1
#c 0 0 0 0 1 1
#d 0 1 1 1 1 1
#e 0 0 0 0 0 0
data
x <- c(5,2,4,1,9)
N <- 6

Related

Replicate rows by value in column, change values to 1 or 0, in R

I have data structured as:
A B C D
3 2 1 1
I want it restructured as
A B C D
1 0 0 0
1 0 0 0
1 0 0 0
0 1 0 0
0 1 0 0
0 0 1 0
0 0 0 1
Any thoughts on how to do this in R? Many thanks.

If the input is a data.frame, you could do the following:
coln <- seq_along(df)
m = do.call(rbind, lapply(coln, function(i) {t(replicate(df[1,i], coln == i))})) +0
This will result in a matrix like this:
# [,1] [,2] [,3] [,4]
#[1,] 1 0 0 0
#[2,] 1 0 0 0
#[3,] 1 0 0 0
#[4,] 0 1 0 0
#[5,] 0 1 0 0
#[6,] 0 0 1 0
#[7,] 0 0 0 1
You can then convert it to a data.frame or set column names if you like.

Here is an option using dcast
library(data.table)
nm1 <- rep(names(df1), unlist(df1))
dcast(data.table(nm1, v1 = seq_along(nm1)), v1 ~ nm1, length)[, v1 := NULL][]
# A B C D
#1: 1 0 0 0
#2: 1 0 0 0
#3: 1 0 0 0
#4: 0 1 0 0
#5: 0 1 0 0
#6: 0 0 1 0
#7: 0 0 0 1
Or after creating the 'nm1', use model.matrix from base R
model.matrix(~-1 + nm1)
or in a single line
model.matrix(~ -1 + rep(names(df1), unlist(df1)))
and change the column names
data
df1 <- data.frame(A = 3, B = 2, C = 1, D = 1)

How can I create this special sequence?

I would like to create the following vector sequence.
0 1 0 0 2 0 0 0 3 0 0 0 0 4
My thought was to create 0 first with rep() but not sure how to add the 1:4.

Create a diagonal matrix, take the upper triangle, and remove the first element:
d <- diag(0:4)
d[upper.tri(d, TRUE)][-1L]
# [1] 0 1 0 0 2 0 0 0 3 0 0 0 0 4
If you prefer a one-liner that makes no global assignments, wrap it up in a function:
(function() { d <- diag(0:4); d[upper.tri(d, TRUE)][-1L] })()
# [1] 0 1 0 0 2 0 0 0 3 0 0 0 0 4
And for code golf purposes, here's another variation using d from above:
d[!lower.tri(d)][-1L]
# [1] 0 1 0 0 2 0 0 0 3 0 0 0 0 4

rep and rbind up to their old tricks:
rep(rbind(0,1:4),rbind(1:4,1))
#[1] 0 1 0 0 2 0 0 0 3 0 0 0 0 4
This essentially creates 2 matrices, one for the value, and one for how many times the value is repeated. rep does not care if an input is a matrix, as it will just flatten it back to a vector going down each column in order.
rbind(0,1:4)
# [,1] [,2] [,3] [,4]
#[1,] 0 0 0 0
#[2,] 1 2 3 4
rbind(1:4,1)
# [,1] [,2] [,3] [,4]
#[1,] 1 2 3 4
#[2,] 1 1 1 1

You can use rep() to create a sequence that has n + 1 of each value:
n <- 4
myseq <- rep(seq_len(n), seq_len(n) + 1)
# [1] 1 1 2 2 2 3 3 3 3 4 4 4 4 4
Then you can use diff() to find the elements you want. You need to append a 1 to the end of the diff() output, since you always want the last value.
c(diff(myseq), 1)
# [1] 0 1 0 0 1 0 0 0 1 0 0 0 0 1
Then you just need to multiply the original sequence with the diff() output.
myseq <- myseq * c(diff(myseq), 1)
myseq
# [1] 0 1 0 0 2 0 0 0 3 0 0 0 0 4

unlist(lapply(1:4, function(i) c(rep(0,i),i)))

# the sequence
s = 1:4
# create zeros vector
vec = rep(0, sum(s+1))
# assign the sequence to the corresponding position in the zeros vector
vec[cumsum(s+1)] <- s
vec
# [1] 0 1 0 0 2 0 0 0 3 0 0 0 0 4
Or to be more succinct, use replace:
replace(rep(0, sum(s+1)), cumsum(s+1), s)
# [1] 0 1 0 0 2 0 0 0 3 0 0 0 0 4

R: Matrix counting matches when 2 teams interacted from schedule with 3 participants per match

I'd like to make some calculations on FIRST robotics teams and need to build, for lack of better words, a binary interaction matrix. That is when two teams were on the same alliance. Each alliance has three teams, so there are 7 values from each match added to the matrix, when considering (i,j), (j,i), and (i,i).
The full data I'm using is here: http://frc-events.firstinspires.org/2016/MOKC/qualifications
But for simplicity, here is an example of 9 teams playing 1 match each.
> data.frame(Team.1=1:3,Team.2=4:6,Team.3=7:9)
Team.1 Team.2 Team.3
1 1 4 7
2 2 5 8
3 3 6 9
The matrix should count each binary interaction, (1,4),(4,7),(3,6),(6,3),(9,9), etc, and will be an N x N matrix, where in the above example N=9. Here's the matrix that represents the above lists:
> matrix(data=c(1,0,0,1,0,0,1,0,0,+
+ 0,1,0,0,1,0,0,1,0,+
+ 0,0,1,0,0,1,0,0,1,+
+ 1,0,0,1,0,0,1,0,0,+
+ 0,1,0,0,1,0,0,1,0,+
+ 0,0,1,0,0,1,0,0,1,+
+ 1,0,0,1,0,0,1,0,0,+
+ 0,1,0,0,1,0,0,1,0,+
+ 0,0,1,0,0,1,0,0,1),9,9)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 1 0 0 1 0 0 1 0 0
[2,] 0 1 0 0 1 0 0 1 0
[3,] 0 0 1 0 0 1 0 0 1
[4,] 1 0 0 1 0 0 1 0 0
[5,] 0 1 0 0 1 0 0 1 0
[6,] 0 0 1 0 0 1 0 0 1
[7,] 1 0 0 1 0 0 1 0 0
[8,] 0 1 0 0 1 0 0 1 0
[9,] 0 0 1 0 0 1 0 0 1
In the real data, the team number are not sequential, and are would be more like 5732,1345,3451,etc, and there are more matches per team meaning the matrix values would be between 0 and max number of matches any of the teams played. This can be seen in the real data.
Thanks to anyone that can help.

There is probably a more elegant approach, but here is one using data.table.
library(data.table)
dat <- data.table(Team.1=1:3,Team.2=4:6,Team.3=7:9)
#add match ID
dat[,match:=1:.N]
#turn to long
mdat <- melt(dat,id="match",value.name="team")[,variable:=NULL]
#merge with itself
dat2 <- merge(mdat, mdat, by=c("match"),all=T, allow.cartesian = T)
# reshape
dcast(dat2, team.x~team.y, fun.agg=length)
team.x 1 2 3 4 5 6 7 8 9
1: 1 1 0 0 1 0 0 1 0 0
2: 2 0 1 0 0 1 0 0 1 0
3: 3 0 0 1 0 0 1 0 0 1
4: 4 1 0 0 1 0 0 1 0 0
5: 5 0 1 0 0 1 0 0 1 0
6: 6 0 0 1 0 0 1 0 0 1
7: 7 1 0 0 1 0 0 1 0 0
8: 8 0 1 0 0 1 0 0 1 0
9: 9 0 0 1 0 0 1 0 0 1
And, because I can, one in base-R. A case where I think the use of a for-loop is justified (as you keep modifying the same object).
#make matrix to put results in
nteams = length(unique(unlist(dat)))
res <- matrix(0,nrow=nteams, ncol=nteams)
#split data by row, generate combinations for each row and add to matrix
for(i in 1:nrow(dat)){
x=unlist(dat[i,])
coords=as.matrix(expand.grid(x,x))
res[coords] <- res[coords]+1
}

Here is my suggestion with base functions. I tried to create a matrix. My approach was to look for the position indexes for 1.
library(magrittr)
mydf <- data.frame(Team.1 = 1:3, Team.2 = 4:6,Team.3 = 7:9)
### Create a matrix with position indexes
lapply(1:nrow(mydf), function(x){
a <- t(combn(mydf[x, ], 2)) # Get some combination
b <- a[, 2:1] # Get other combination by reversing columns
foo <- rbind(a, b)
foo
}) %>%
do.call(rbind, .) -> ana
ana <- matrix(unlist(ana), nrow = nrow(ana))
### Another set: Get indexes for self (e.g., (1,1), (2,2), (3,3))
foo <- rep(1:max(mydf), times = 2)
matrix(foo, nrow = length(foo) / 2) -> bob
### A matric with all position indexes
cammy <- rbind(ana, bob)
### Create a plain matrix
mat <- matrix(0, nrow = max(mydf), ncol = max(mydf))
### Fill in the matrix with 1
mat[cammy] <- 1
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
# [1,] 1 0 0 1 0 0 1 0 0
# [2,] 0 1 0 0 1 0 0 1 0
# [3,] 0 0 1 0 0 1 0 0 1
# [4,] 1 0 0 1 0 0 1 0 0
# [5,] 0 1 0 0 1 0 0 1 0
# [6,] 0 0 1 0 0 1 0 0 1
# [7,] 1 0 0 1 0 0 1 0 0
# [8,] 0 1 0 0 1 0 0 1 0
# [9,] 0 0 1 0 0 1 0 0 1
EDIT
Here is a revised version based on the previous idea. This is not concise like Heroka's idea with base functions. In my modified data, team 1 and 4 had two matches. The idea here is that I counted how many times each pair appeared in the data set. The dplyr part is doing that. In the for loop, I filled in the matrix, mat by going through each row of cammy.
mydf <- data.frame(Team.1=c(1:3,1),Team.2=c(4:6,4),Team.3=c(7:9,5))
# Team.1 Team.2 Team.3
#1 1 4 7
#2 2 5 8
#3 3 6 9
#4 1 4 5
library(dplyr)
lapply(1:nrow(mydf), function(x){
a <- t(combn(mydf[x, ], 2)) # Get some combination
b <- a[, 2:1] # Get other combination by reversing columns
foo <- rbind(a, b)
foo
}) %>%
do.call(rbind, .) -> ana
ana <- data.frame(matrix(unlist(ana), nrow = nrow(ana)))
### Another set: Get indexes for self (e.g., (1,1), (2,2), (3,3))
foo <- rep(1:max(mydf), times = 2)
data.frame(matrix(foo, nrow = length(foo) / 2)) -> bob
cammy <- bind_rows(ana, bob) %>%
group_by(X1, X2) %>%
mutate(total = n()) %>%
as.matrix
### Create a plain matrix
mat <- matrix(0, nrow = max(mydf), ncol = max(mydf))
for(i in 1:nrow(cammy)){
mat[cammy[i, 1], cammy[i, 2]] <- cammy[i, 3]
}
print(mat)
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
# [1,] 1 0 0 2 1 0 1 0 0
# [2,] 0 1 0 0 1 0 0 1 0
# [3,] 0 0 1 0 0 1 0 0 1
# [4,] 2 0 0 1 1 0 1 0 0
# [5,] 1 1 0 1 1 0 0 1 0
# [6,] 0 0 1 0 0 1 0 0 1
# [7,] 1 0 0 1 0 0 1 0 0
# [8,] 0 1 0 0 1 0 0 1 0
# [9,] 0 0 1 0 0 1 0 0 1

How to transform a directed Dataset into a Matrix with R

I have a Dataset in R which looks like this:
ID LinkedTo
1 Null
2 1
3 1
4 3
5 4
I want transform it into a Matrix which looks similar to this:
0 0 0 0 0
1 0 0 0 0
1 0 0 0 0
0 0 1 0 0
0 0 0 1 0

Another option , is to modelize your directed dataset as a directed graph and extract adjacency matrix.
library(igraph)
dat <- read.table(text='ID LinkedTo
2 1
3 1
4 3
5 4',header=TRUE)
gg <- graph.data.frame(dat)
as.matrix(get.adjacency(gg))
2 3 4 5 1
2 0 0 0 0 1
3 0 0 0 0 1
4 0 1 0 0 0
5 0 0 1 0 0
1 0 0 0 0 0

It's more convenient if you replace "Null" by NA in your dataset. Something like
i <- structure(list(ID = c(1, 2, 3, 4, 5),
LinkedTo = c(NA, 1, 1, 3, 4)),
.Names = c("ID", "LinkedTo"),
row.names = c(NA, -5L), class = "data.frame")
i
# ID LinkedTo
# 1 1 NA
# 2 2 1
# 3 3 1
# 4 4 3
# 5 5 4
Then you can do
m <- matrix(0, nrow(i), nrow(i))
m[i$ID + (i$LinkedTo - 1) * nrow(i)] <- 1
(It would work the same way if i was a matrix, but you would have to change i$ID and i$LinkedTo to i[, 1] and i[, 2] resp)

you can start by replacing the null with zeros, i think .
Then you can do a little for loop:
data.frame(id=1:5, pos=sample(1:5))->df
matrix(nrow=max(nrow(df)),ncol= max(df$id),data=0)->m
for (i in 1:nrow(df)){
m[i,df$pos[i]]<-1
}

Using #konvas i dataset
i[,2][is.na(i[,2])] <- 0
m <- matrix(0, nrow(i), nrow(i))
m[as.matrix(i)] <- 1
m
# [,1] [,2] [,3] [,4] [,5]
#[1,] 0 0 0 0 0
#[2,] 1 0 0 0 0
#[3,] 1 0 0 0 0
#[4,] 0 0 1 0 0
#[5,] 0 0 0 1 0

table should also work if you combine it with factor. (I say "should" because your conditions aren't clearly specified and your sample data are not reproducible.)
Using #konvas's "i" sample data, try:
> table(i$ID, factor(i$LinkedTo, 1:5))
1 2 3 4 5
1 0 0 0 0 0
2 1 0 0 0 0
3 1 0 0 0 0
4 0 0 1 0 0
5 0 0 0 1 0

Convert a vector into logical matrix

Is there a native R function that will take an input vector and return the corresponding binary matrix where the matrix has the same number of columns as unique values in the input vector?
For example, given x <- 1:3, I want to return the following matrix:
[,1] [,2] [,3]
[1,] 1 0 0
[2,] 0 1 0
[3,] 0 0 1
The functions contrasts gets close, but I can't seem to get around the n-1 columns returned:
> contrasts(as.factor(x))
2 3
1 0 0
2 1 0
3 0 1

Actually, contrasts is what you want.
contrasts(as.factor(1:3), contrasts=FALSE)
1 2 3
1 1 0 0
2 0 1 0
3 0 0 1

model.matrix() might help here, but you need to suppress the intercept:
> model.matrix(~ factor(1:3) - 1)
factor(1:3)1 factor(1:3)2 factor(1:3)3
1 1 0 0
2 0 1 0
3 0 0 1
attr(,"assign")
[1] 1 1 1
attr(,"contrasts")
attr(,"contrasts")$`factor(1:3)`
[1] "contr.treatment"
Something slightly more complex:
> set.seed(1)
> fac <- factor(sample(1:3, 10, replace = TRUE))
> model.matrix(~ fac - 1)
fac1 fac2 fac3
1 1 0 0
2 0 1 0
3 0 1 0
4 0 0 1
5 1 0 0
6 0 0 1
7 0 0 1
8 0 1 0
9 0 1 0
10 1 0 0
attr(,"assign")
[1] 1 1 1
attr(,"contrasts")
attr(,"contrasts")$fac
[1] "contr.treatment"

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