Reputation: 21
I'm trying to create an heatmap on the base of spearman correlation and with dendrogramm corresponding to spearman correlation values. My input file is composed as follow:
> data[1:6,1:6]
group EG PN C0 C10 C10.1
1 Patients 24 729 352.66598 43.80707 75.16226
2 Patients 24 729 195.48486 17.15763 33.60365
3 Patients 24 729 106.85937 15.13400 34.47340
4 Patients 27 1060 76.70645 14.98315 22.09885
5 Patients 27 1060 354.07169 50.61995 98.36765
6 Patients 27 1060 331.84956 92.00343 125.46658
> data[150:160,1:6]
group EG PN C0 C10 C10.1
150 Controls 27 1011 99.94756 9.018773 20.207498
151 Controls 30 616 300.20203 25.667548 37.363280
152 Controls 30 616 190.38030 18.811198 46.417332
153 Controls 26 930 79.44666 7.801935 4.569444
154 Controls 24 724 381.74026 39.842241 42.144842
155 Controls 24 724 191.39962 19.008729 31.064398
I'm able to made up a simple correlation plot but i would like to create an unique heatmap with both protein and subjects dendrogram on the base on spearman correlation. Does anyone know how to do? thanks in advance
Upvotes: 1
Views: 722
Reputation: 3038
The following code displays an interactive heatmap using Spearman's rank correlation to cluster both rows and columns (in this case for the mtcars
dataset).
heatmaply(mtcars,
distfun = function(x) as.dist(1 - cor(t(x), method="spearman")))
Upvotes: 1