How to plot a binary matrix without using additional packages? - r

I created a binary matrix and I wanna plot 1's as black square.
How can I write it without using any package?
For example, my matrix is:
m <- matrix(c(0,1,1,0,0,1,0,1,1),nrow=3, ncol=3)


Do you want this?
m <- matrix(c(0,1,1,0,0,1,0,1,1), nrow=3, ncol=3)
image(m, main = "My binary matrix plot", col = c("white", "black"))


If image doesn't suffice, we could write a generalized function using mapply like this one.
chessplot <- function(m, col=1, border=NA) {
stopifnot(dim(m)[1] == dim(m)[2]) ## allows only square matrices
n <- nrow(m)
plot(n, n, type='n', xlim=c(0, n), ylim=c(0, n))
mapply(\(i, j, m) {
rect(-1 + i, n - j, 0 + i, n - j + 1, col=m, border=border)
}, seq(n), rep(seq(n), each=n), t(m)) |> invisible()
}
Gives:
chessplot(m3)
chessplot(m4)
chessplot(m8)
Data:
m3 <- structure(c(0, 1, 1, 0, 0, 1, 0, 1, 1), .Dim = c(3L, 3L))
m4 <- structure(c(0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0), .Dim = c(4L,
4L))
m8 <- structure(c(0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0,
1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1,
0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1,
0, 1, 0, 1, 0), .Dim = c(8L, 8L))

Related

R function to change value after a condition has been fulfilled

Participants in an experiment took a test that has a rule that says "once a participant has gotten 6 items wrong in a window of 8 items, you stop running the test". However, some experimenters kept testing past this point. I now need to find a way in which I can automatically see where the test should have been stopped, and change all values following the end to 0 (= item wrong). I am not even sure if this is something that can be done in R.
To be clear, I would like to go row by row (which are the participants) and once there are six 0s in a given window of 8 columns (items), I would need all values after the sixth 0 to be 0 too.
While the reproducible data is below, here is a visualization of what I would need, where the blue cells are the ones that should change to 0:
Pre-changes
Post-changes
Reproducible data:
structure(list(Participant_ID = c("E01P01", "E01P02", "E01P03",
"E01P04", "E01P05", "E01P06", "E01P07", "E01P08", "E02P01", "E02P02"
), A2 = c(1, 1, 1, 0, 0, 1, 1, 1, 1, 1), A3 = c(1, 1, 0, 0, 0,
1, 0, 0, 0, 0), B1 = c(1, 1, 1, 0, 0, 1, 0, 0, 1, 1), B2 = c(1,
1, 1, 1, 1, 1, 0, 0, 0, 1), C3 = c(1, 0, 0, 1, 0, 1, 0, 0, 0,
1), C4 = c(1, 0, 0, 0, 0, 1, 0, 0, 1, 1), D1 = c(1, 0, 0, 0,
0, 1, 0, 0, 0, 0), D3 = c(1, 1, 1, 1, 0, 0, 1, 0, 0, 1), E1 = c(1,
0, 0, 0, 0, 1, 0, 0, 0, 1), E3 = c(1, 1, 0, 1, 0, 1, 0, 0, 0,
0), F1 = c(1, 0, 0, 0, 1, 0, 0, 1, 0, 0), F4 = c(1, 1, 1, 1,
0, 1, 0, 1, 1, 0), G1 = c(1, 0, 0, 0, 0, 1, 0, 0, 0, 1), G2 = c(0,
0, 0, 0, 1, 1, 1, 0, 1, 1)), row.names = c(NA, -10L), class = c("tbl_df",
"tbl", "data.frame"))
Any help is highly appreciated!

Here is a solution that involves some pivoting, rollsum, cumsum, if_else logic, then pivoting back. Let me know if it works.
library(tidyverse)
library(zoo)
structure(list(Participant_ID = c("E01P01", "E01P02", "E01P03",
"E01P04", "E01P05", "E01P06", "E01P07", "E01P08", "E02P01", "E02P02"
), A2 = c(1, 1, 1, 0, 0, 1, 1, 1, 1, 1), A3 = c(1, 1, 0, 0, 0,
1, 0, 0, 0, 0), B1 = c(1, 1, 1, 0, 0, 1, 0, 0, 1, 1), B2 = c(1,
1, 1, 1, 1, 1, 0, 0, 0, 1), C3 = c(1, 0, 0, 1, 0, 1, 0, 0, 0,
1), C4 = c(1, 0, 0, 0, 0, 1, 0, 0, 1, 1), D1 = c(1, 0, 0, 0,
0, 1, 0, 0, 0, 0), D3 = c(1, 1, 1, 1, 0, 0, 1, 0, 0, 1), E1 = c(1,
0, 0, 0, 0, 1, 0, 0, 0, 1), E3 = c(1, 1, 0, 1, 0, 1, 0, 0, 0,
0), F1 = c(1, 0, 0, 0, 1, 0, 0, 1, 0, 0), F4 = c(1, 1, 1, 1,
0, 1, 0, 1, 1, 0), G1 = c(1, 0, 0, 0, 0, 1, 0, 0, 0, 1), G2 = c(0,
0, 0, 0, 1, 1, 1, 0, 1, 1)), row.names = c(NA, -10L), class = c("tbl_df",
"tbl", "data.frame")) %>%
as_tibble() %>%
pivot_longer(-1) %>%
group_by(Participant_ID) %>%
mutate(running_total = zoo::rollsumr(value==0, k = 8, fill = 0),
should_terminate = cumsum(running_total >= 6),
value = if_else(should_terminate > 0, 0, value)) %>%
ungroup() %>%
select(Participant_ID, name, value) %>%
pivot_wider(names_from = name, values_from = value)

Chi Square Test of Independence of Whole Dataset

I have a 3185x90 dataset of binary values and want to do a chi-squared test of independence, comparing all column variables against each other.
I've been tried using different variations of code from google searches with chisq.test() and some for loops, but none of them have worked so far.
How do I do this?
This is the frame I've tinkered with. My dataset is oak.
chi_trial <- data.frame(a = c(0,1), b = c(0,1))
for(row in 1:nrow(oak)){
print(row)
print(chisq.test(c(oak[row,1],d[row,2])))
}
I also tried this:
apply(d, 1, chisq.test)
which gives me the error: Error in FUN(newX[, i], ...) :
all entries of 'x' must be nonnegative and finite
dput(oak[1:2],)
structure(list(post_flu = structure(c(1, 1, 1, 1, 1, 0, 0, 0,
0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0,
0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1,
1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0,
label = "Receipt of Flu Vaccine - Encounter Survey", format.stata = "%10.0g")), row.names = c(NA,
-3185L), class = c("tbl_df", "tbl", "data.frame"), label = "Main Oakland Clinic Analysis Dataset")
I added a sample of my data with the final lines of the output. The portion of the dataset is small, but it all looks like this.

You could use something like the code below, which is similar to R's cor function. I don't have your data, so I'm simulating some. Note that I get one significant p-value, using the traditional cut-off of 0.05.
set.seed(3)
nr=3185; nc=3
oak <- as.data.frame(matrix(sample(0:1, size=nr*nc, replace=TRUE), ncol=nc))
oak
mult.chi <- function(data){
nc <- ncol(data)
res <- matrix(0, nrow=nc, ncol=nc) # or NA
for(i in 1:(nc-1))
for(j in (i+1):nc)
res[i,j] <- suppressWarnings(chisq.test(oak[,i], oak[,j])$p.value)
rownames(res) <- colnames(data)
colnames(res) <- colnames(data)
res
}
mult.chi(oak)
# V1 V2 V3
# V1 0 0.7847063 0.32012466
# V2 0 0.0000000 0.01410326
# V3 0 0.0000000 0.00000000
So consider applying a multiple testing adjustment as mentioned in the comments.

Here is a solution with combn to get all combinations of column numbers 2 by 2. Tested with the data in #Edward's answer.
chisq2cols <- function(X){
y <- matrix(0, ncol(X), ncol(X))
cmb <- combn(ncol(X), 2)
y[upper.tri(y)] <- apply(cmb, 2, function(k){
tbl <- table(X[k])
chisq.test(tbl)$p.value
})
y
}
chisq2cols(oak)
# [,1] [,2] [,3]
#[1,] 0 0.7847063 0.32012466
#[2,] 0 0.0000000 0.01410326
#[3,] 0 0.0000000 0.00000000

Intersecting ranges of consecutive values in logical vectors in R

I have two logical vectors which look like this:
x = c(0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0)
y = c(0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0)
I would like to count the intersections between ranges of consecutive values. Meaning that consecutive values (of 1s) are handled as one range. So in the above example, each vector contains one range of 1s and these ranges intersect only once.
Is there any R package for range intersections which could help here?

I think this should work (calling your logical vectors x and y):
sum(rle(x & y)$values)
A few examples:
x = c(0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0)
y = c(0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0)
sum(rle(x & y)$values)
# [1] 1
x = c(1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0)
y = c(0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0)
sum(rle(x & y)$values)
# [1] 2
x = c(1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0)
y = c(0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0)
sum(rle(x & y)$values)
# [1] 3
By way of explanation, x & y gives the intersections on a per-element level, rle collapses runs of adjacent intersections, and sum counts.

Optimum algorithm to check various combinations of items when number of items is too large

I have a data frame which has 20 columns/items in it, and 593 rows (number of rows doesn't matter though) as shown below:
Using this the reliability of test is obtained as 0.94, with the help of alpha from psych package psych::alpha. The output also gives me the the new value of cronbach's alpha if I drop one of the items. However, I want to know how many items can I drop to retain an alpha of at least 0.8 I used a brute force approach for the purpose where I am creating the combination of all the items that exists in my data frame and check if their alpha is in the range (0.7,0.9). Is there a better way of doing this, as this is taking forever to run because number of items is too large to check for all the combination of items. Below is my current piece of code:
numberOfItems <- 20
for(i in 2:(2^numberOfItems)-1){
# ignoring the first case i.e. i=1, as it doesn't represent any model
# convert the value of i to binary, e.g. i=5 will give combination = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1
# using the binaryLogic package
combination <- as.binary(i, n=numberOfItems)
model <- c()
for(j in 1:length(combination)){
# choose which columns to consider depending on the combination
if(combination[j])
model <- c(model, j)
}
itemsToUse <- itemResponses[, c(model)]
#cat(model)
if(length(model) > 13){
alphaVal <- psych::alpha(itemsToUse)$total$raw_alpha
if(alphaVal > 0.7 && alphaVal < 0.9){
cat(alphaVal)
print(model)
}
}
}
A sample output from this code is as follows:
0.8989831 1 4 5 7 8 9 10 11 13 14 15 16 17 19 20
0.899768 1 4 5 7 8 9 10 11 12 13 15 17 18 19 20
0.899937 1 4 5 7 8 9 10 11 12 13 15 16 17 19 20
0.8980605 1 4 5 7 8 9 10 11 12 13 14 15 17 19 20
Here are the first 10 rows of data:
dput(itemResponses)
structure(list(CESD1 = c(1, 2, 2, 0, 1, 0, 0, 0, 0, 1), CESD2 = c(2,
3, 1, 0, 0, 1, 1, 1, 0, 1), CESD3 = c(0, 3, 0, 1, 1, 0, 0, 0,
0, 0), CESD4 = c(1, 2, 0, 1, 0, 1, 1, 1, 0, 0), CESD5 = c(0,
1, 0, 2, 1, 2, 2, 0, 0, 0), CESD6 = c(0, 3, 0, 1, 0, 0, 2, 0,
0, 0), CESD7 = c(1, 2, 1, 1, 2, 0, 1, 0, 1, 0), CESD8 = c(1,
3, 1, 1, 0, 1, 0, 0, 1, 0), CESD9 = c(0, 1, 0, 2, 0, 0, 1, 1,
0, 1), CESD10 = c(0, 1, 0, 2, 0, 0, 1, 1, 0, 1), CESD11 = c(0,
2, 1, 1, 1, 1, 2, 3, 0, 0), CESD12 = c(0, 3, 1, 1, 1, 0, 2, 0,
0, 0), CESD13 = c(0, 3, 0, 2, 1, 2, 1, 0, 1, 0), CESD14 = c(0,
3, 1, 2, 1, 1, 1, 0, 1, 1), CESD15 = c(0, 2, 0, 1, 0, 1, 0, 1,
1, 0), CESD16 = c(0, 2, 2, 0, 0, 1, 1, 0, 0, 0), CESD17 = c(0,
0, 0, 0, 0, 1, 1, 0, 0, 0), CESD18 = c(0, 2, 0, 0, 0, 0, 0, 0,
0, 1), CESD19 = c(0, 3, 0, 0, 0, 0, 0, 1, 1, 0), CESD20 = c(0,
3, 0, 1, 0, 0, 0, 0, 0, 0)), .Names = c("CESD1", "CESD2", "CESD3",
"CESD4", "CESD5", "CESD6", "CESD7", "CESD8", "CESD9", "CESD10",
"CESD11", "CESD12", "CESD13", "CESD14", "CESD15", "CESD16", "CESD17",
"CESD18", "CESD19", "CESD20"), row.names = c(NA, -10L), class = c("tbl_df",
"tbl", "data.frame"))

The idea is to replace the computation of alpha with the so-called discrimination for each item from classical test theory (CTT). The discrimination is the correlation of the item score with a "true score" (which we would assume to be the row sum).
Let the data be
dat <- structure(list(CESD1 = c(1, 2, 2, 0, 1, 0, 0, 0, 0, 1), CESD2 = c(2, 3, 1, 0, 0, 1, 1, 1, 0, 1),
CESD3 = c(0, 3, 0, 1, 1, 0, 0, 0, 0, 0), CESD4 = c(1, 2, 0, 1, 0, 1, 1, 1, 0, 0),
CESD5 = c(0, 1, 0, 2, 1, 2, 2, 0, 0, 0), CESD6 = c(0, 3, 0, 1, 0, 0, 2, 0, 0, 0),
CESD7 = c(1, 2, 1, 1, 2, 0, 1, 0, 1, 0), CESD8 = c(1, 3, 1, 1, 0, 1, 0, 0, 1, 0),
CESD9 = c(0, 1, 0, 2, 0, 0, 1, 1, 0, 1), CESD10 = c(0, 1, 0, 2, 0, 0, 1, 1, 0, 1),
CESD11 = c(0, 2, 1, 1, 1, 1, 2, 3, 0, 0), CESD12 = c(0, 3, 1, 1, 1, 0, 2, 0, 0, 0),
CESD13 = c(0, 3, 0, 2, 1, 2, 1, 0, 1, 0), CESD14 = c(0, 3, 1, 2, 1, 1, 1, 0, 1, 1),
CESD15 = c(0, 2, 0, 1, 0, 1, 0, 1, 1, 0), CESD16 = c(0, 2, 2, 0, 0, 1, 1, 0, 0, 0),
CESD17 = c(0, 0, 0, 0, 0, 1, 1, 0, 0, 0), CESD18 = c(0, 2, 0, 0, 0, 0, 0, 0, 0, 1),
CESD19 = c(0, 3, 0, 0, 0, 0, 0, 1, 1, 0), CESD20 = c(0, 3, 0, 1, 0, 0, 0, 0, 0, 0)),
.Names = c("CESD1", "CESD2", "CESD3", "CESD4", "CESD5", "CESD6", "CESD7", "CESD8", "CESD9",
"CESD10", "CESD11", "CESD12", "CESD13", "CESD14", "CESD15", "CESD16", "CESD17",
"CESD18", "CESD19", "CESD20"), row.names = c(NA, -10L),
class = c("tbl_df", "tbl", "data.frame"))
We compute (1) the discrimination and (2) the alpha coefficient.
stat <- t(sapply(1:ncol(dat), function(ii){
dd <- dat[, ii]
# discrimination is the correlation of the item to the rowsum
disc <- if(var(dd, na.rm = TRUE) > 0) cor(dd, rowSums(dat[, -ii]), use = "pairwise")
# alpha that would be obtained when we skip this item
alpha <- psych::alpha(dat[, -ii])$total$raw_alpha
c(disc, alpha)
}))
dimnames(stat) <- list(colnames(dat), c("disc", "alpha^I"))
stat <- data.frame(stat)
Observe that the discrimination (which is more efficient to compute) is inversely proportional to alpha that is obtained when deleting this item. In other words, alpha is highest when there are many high "discriminating" items (that correlate with each other).
plot(stat, pch = 19)
Use this information to select the sequence with which the items should be deleted to fall below a benchmark (say .9, since the toy data doesn't allow for a lower mark):
1) delete as many items as possible to stay above the benchmark; that is, start with the least discriminating items.
stat <- stat[order(stat$disc), ]
this <- sapply(1:(nrow(stat)-2), function(ii){
ind <- match(rownames(stat)[1:ii], colnames(dat))
alpha <- psych::alpha(dat[, -ind, drop = FALSE])$total$raw_alpha
})
delete_these <- rownames(stat)[which(this > .9)]
psych::alpha(dat[, -match(delete_these, colnames(dat)), drop = FALSE])$total$raw_alpha
length(delete_these)
2) delete as few items as possible to stay above the benchmark; that is, start with the highest discriminating items.
stat <- stat[order(stat$disc, decreasing = TRUE), ]
this <- sapply(1:(nrow(stat)-2), function(ii){
ind <- match(rownames(stat)[1:ii], colnames(dat))
alpha <- psych::alpha(dat[, -ind, drop = FALSE])$total$raw_alpha
})
delete_these <- rownames(stat)[which(this > .9)]
psych::alpha(dat[, -match(delete_these, colnames(dat)), drop = FALSE])$total$raw_alpha
length(delete_these)
Note, that 1) is coherent with classical item selection procedures in (psychological/educational) diagnostic/assessments: remove items from the assessment, that fall below a benchmark in terms of discriminatory power.

I changed the code as follows, now I am dropping a fixed number of items and changing the value of numberOfItemsToDrop from 1 to 20 manually. Although it is a lil better, but it still is taking too long to run :(
I hope there is some better way of doing this.
numberOfItemsToDrop <- 13
combinations <- combinat::combn(20, numberOfItemsToDrop)
timesToIterate <- length(combinations)/numberOfItemsToDrop
for(i in 1:timesToIterate){
model <- combinations[,i]
itemsToUse <- itemResponses[, -c(model)]
alphaVal <- psych::alpha(itemsToUse)$total$raw_alpha
if(alphaVal < 0.82){
cat("Cronbach's alpha =",alphaVal, ", number of items dropped = ", length(model), " :: ")
print(model)
}
}

Modification to heat map

I am making a heat map though would like to separate the columns and add a line between each row. I am well aware that doing so makes this well, not a heat map. But this is how my boss envisions it.
Below is my code for the current heat map. Any advice on separating the columns & adding a line between each "person" would be much appreciated.
x11 <- c(0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1)
x22 <- c(1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1)
x <- rbind(x11, x22)
hv <- heatmap(t(x), col = c("cornflowerblue", "hotpink"), margins = c(4, 12), Colv = NA, Rowv = NA, scale = "none", xlab ="", ylab ="", main = "", labCol=c("BP", "Cx"), cexCol =2)
legend("topright", c("No Osteomyelitis", "Osteomyelitis"), col=c("cornflowerblue", "hotpink"), bty="n", fill=c("cornflowerblue", "hotpink"))
Yes I ended up using the code below. Thank you for the answer. I used gplots & got rid of the color key & histogram & added my own legend instead.
hv <- heatmap.2(t(x), key=FALSE, trace="none", colsep = seq(1,nrow(x)-1),
rowsep = seq(1,ncol(x)-1),
sepcolor = "white",
sepwidth = c(0.1, 0.0005), col = c("cornflowerblue", "hotpink"), margins = c(4, 12), Colv = NA, Rowv = NA, scale = "none", xlab ="", ylab ="", main = "", labCol=c("BP", "Cx"), cexCol =2)
legend("topleft", c("No Osteomyelitis", "Osteomyelitis"), col=c("cornflowerblue", "hotpink"), bty="n", fill=c("cornflowerblue", "hotpink"))

I suggest you swap to gplots::heatmap.2() which allows greater control over plotting with mostly the same arguments.
Building on your good example (+1 btw) by adding the colsep, rowsep, sepcolor and sepwidth arguments to control the separation between the rows and columns (and trace = 'none' because I don't like it) gives:
x11 <- c(0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1)
x22 <- c(1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1)
x <- rbind(x11, x22)
hv <- heatmap(t(x), col = c("cornflowerblue", "hotpink"), margins = c(4, 12), Colv = NA, Rowv = NA, scale = "none", xlab ="", ylab ="", main = "", labCol=c("BP", "Cx"), cexCol =2)
legend("topright", c("No Osteomyelitis", "Osteomyelitis"), col=c("cornflowerblue", "hotpink"), bty="n", fill=c("cornflowerblue", "hotpink"))
library(gplots)
heatmap.2(t(x),
col = c("cornflowerblue", "hotpink"),
margins = c(4, 12),
Colv = NA, Rowv = NA,
scale = "none",
xlab ="",
ylab ="",
main = "",
labCol=c("BP", "Cx"),
cexCol = 2,
trace = 'none',
colsep = seq(1,nrow(x)-1),
rowsep = seq(1,ncol(x)-1),
sepcolor = "white",
sepwidth = c(0.1, 0.05))
To separate the columns more, increse the first element of sepwidth and similarly for the rows.

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