By default, R's heatmap will cluster rows and columns:
mtscaled = as.matrix(scale(mtcars))
heatmap(mtscaled, scale='none')
I can disable the clustering:
heatmap(mtscaled, Colv=NA, Rowv=NA, scale='none')
And then the dendrogram goes away:
But now the data is not clustered anymore.
I don't want the dendrograms to be shown, but I still want the rows and/or columns to be clustered. How can I do this?
Example of what I want:
You can do this with pheatmap:
mtscaled <- as.matrix(scale(mtcars))
pheatmap::pheatmap(mtscaled, treeheight_row = 0, treeheight_col = 0)
See pheatmap output here:
library(gplots)
heatmap.2(mtscaled,dendrogram='none', Rowv=TRUE, Colv=TRUE,trace='none')
Rowv -is TRUE, which implies dendrogram is computed and reordered based on row means.
Colv - columns should be treated identically to the rows.
I had similar issue with pheatmap, which has better visualisation and heatmap or heatmap.2. Though heatmap.2 is a choice for your solution, Here is the solution with pheatmap, by extracting the order of clustered data.
library(pheatmap)
mtscaled = as.matrix(scale(mtcars))
H = pheatmap(mtscaled)
Here is the output of pheatmap
pheatmap(mtscaled[H$tree_row$order,H$tree_col$order],cluster_rows = F,cluster_cols = F)
Here is the output of pheatmap after extracting the order of clusters
For ComplexHeatmap, there are function parameters to remove the dendrograms:
library(ComplexHeatmap)
Heatmap(as.matrix(iris[,1:4]), name = "mat", show_column_dend = FALSE, show_row_dend = FALSE)
You can rely on base R structures and consider following approach based on building the hclust trees by yourself.
mtscaled = as.matrix(scale(mtcars))
row_order = hclust(dist(mtscaled))$order
column_order = hclust(dist(t(mtscaled)))$order
heatmap(mtscaled[row_order,column_order], Colv=NA, Rowv=NA, scale="none")
No need to install additional junk.
Do the dendrogram twice using the basic R heatmap function. Take the output of the first run, which clusters but has mandatory drawing of the dendrogram and feed it into the heatmap function again. This time, without clustering, and without drawing the dendrogram.
#generate a random symmetrical matrix with a little bit of structure, and make a heatmap
M100s<-matrix(runif(10000),nrow=100)
M100s[2,]<-runif(100,min=0.1,max=0.2)
M100s[4,]<-runif(100,min=0.1,max=0.2)
M100s[6,]<-runif(100,min=0.1,max=0.2)
M100s[99,]<-runif(100,min=0.1,max=0.2)
M100s[37,]<-runif(100,min=0.1,max=0.2)
M100s[lower.tri(M100s)] <- t(M100s)[lower.tri(M100s)]
heatmap(M100s)
#save the output
OutputH <- heatmap(M100s)
#run it again without clustering or the dendrogram
M100c <- M100s
M100c1 <- M100c[,OutputH$rowInd]
M100c2 <- M100c1[OutputH$colInd,]
heatmap(M100c2,Rowv = NA, Colv = NA, labRow = NA, labCol = NA)
Related
I am trying to visualize a PCA that includes 87 variables.
prc <-prcomp(df[,1:87], center = TRUE, scale. = TRUE)
ggbiplot(prc, labels = rownames(df[,1:87]), var.axes = TRUE)
When I create the biplot, many of the vectors overlap with each other, making it impossible to read the labels. I was wondering if there is any way to only show some of the labels at a time. For example, I think it'd be useful if I could create a few separate biplots with each one showing only a subset of the labels on the vectors.
This question seems closely related, but I don't know if it translates to the latest version of ggbiplot. I'm also not sure how to modify the original functions.
A potential solution is to use the factoextra package to visualize your PCA results. The fviz_pca_biplot() function includes a repel argument. When repel = TRUE the plot labels are spread out to minimize overlap. There are also select.var options mentioned in the documentation, such as select.var = list(contrib=5) to display only the 5 most influential vectors. Also a select.var = list(name) option that seems to allow for the specification of a specific subset of variables that you want shown.
# read data
df <- mtcars[, c(1:7,10:11)]
# perform PCA
library("FactoMineR")
res.pca <- PCA(df, graph = FALSE)
# visualize
library(factoextra)
fviz_pca_biplot(res.pca, repel = TRUE, select.var = list(contrib = 5))
So I'm trying to generate a heatmap for my data using Bioconductor's ComplexHeatmap package, but I get slightly different results depending on whether I make the dendrogram myself, or tell Heatmap to make it.
Packages:
require(ComplexHeatmap)
require(dendextend)
Data:
a=rnorm(400,1)
b=as.matrix(a)
dim(b)=c(80,5)
If I make the dendrogram myself:
d=dist(b,method="euclidean")
d=as.dist(d)
h=hclust(d,method="ward.D")
dend=as.dendrogram(h)
Heatmap(b,
cluster_columns=FALSE,
cluster_rows = dend)
Versus having Heatmap do the clustering:
Heatmap(b,
cluster_columns=FALSE,
clustering_distance_rows = "euclidean",
clustering_method_rows = "ward.D")
They tend to look very similar, but they'll be very slightly different.
And this matters a lot for my data. Heatmap's clustering ends up organizing my data way, way better, however, I also want to extract the list of clustered items via like cutree(), but I don't think I can extract it from Heatmap's clustering.
Does anyone know what's going on?
the dendrograms are the same. The only thing that changes is the ordering. You can verify this using:
hmap1 <- Heatmap(b,
cluster_columns=FALSE,
cluster_rows = dend)
hmap2 <- Heatmap(b,
cluster_columns=FALSE,
clustering_distance_rows = "euclidean",
clustering_method_rows = "ward.D")
#Reorder both row dendrograms using the same weights:
rowdend1 <- reorder(row_dend(hmap1)[[1]], 1:80)
rowdend2 <- reorder(row_dend(hmap2)[[1]], 1:80)
#check that they are identical:
identical( rowdend1, rowdend2)
## [1] TRUE
The ComplexHeatmap::Heatmap function has an argument row_dend_reorder with default value TRUE that you should check.
I have been asked to obtain a correlation plot for a colaborator.
My choice is to use R for the task, specifically the corrplot package.
I have been researching on the internet and I found multiple ways to obtain such graphics, but not the specific graphic I was asked for (as you can see in the picture the significant values are highlighted by drawing a square around the significant tile), which is puzzling me.
Example of the correlation plot required
The closest result I achieve is using the code under this lines, but I do not seem to be able to find the option to draw line around the significant tiles (if exists).
#Insignificant correlations are leaved blank
corrplot(res3$r, type="upper", order="hclust",
p.mat = res3$P, sig.level = 0.01, insig = "blank")
I tried adding the "addrect" parameter but it didn't work.
#Insignificant correlation are crossed
corrplot(res3$r, type="upper", order="hclust", p.mat = res3$P,
addrect=2, sig.level = 0.01, insig = "blank")
Any help will be appreciated.
corrplot allows you to add new plots to an already existing one. Therefore, once you've created the plot of the initial correlation matrix, you can simply add those cells that you want to highlight in an iterative manner using corrplot(..., add = TRUE).
The only thing required to achieve your goal is an indices vecor (which I called 'ids') to tell R which cells to highlight. Note that for reasons of simplicity, I took a random sample of the initial correlation matrix, but things like ids <- which(p.value < 0.01) (assuming that you've stored your significance levels in a separate vector) would work similarly.
library(corrplot)
## create and visualize correlation matrix
data(mtcars)
M <- cor(mtcars)
corrplot(M, cl.pos = "n", na.label = " ")
## select cells to highlight (e.g., statistically significant values)
set.seed(10)
ids <- sample(1:length(M), 15L)
## duplicate correlation matrix and reject all irrelevant values
N <- M
N[-ids] <- NA
## add significant cells to the initial corrplot iteratively
for (i in ids) {
O <- N
O[-i] <- NA
corrplot(O, cl.pos = "n", na.label = " ", addgrid.col = "black", add = TRUE,
bg = "transparent", tl.col = "transparent")
}
Note that you could also add all values to highlight in one go (i.e., without requiring a for loop) using corrplot(N, ...), but in that case, an undesirable black margin is drawn all around the plotting area.
How to draw a heatmap from a matrix of 305 columns and 865 rows in R.
The code I have written for the matrix is
nba <- read.csv("mydata.csv", sep=",")
row.names(nba) <- nba[,1]
nba <- nba[,2:865]
nba_matrix <- data.matrix(nba)
nba_heatmap <- heatmap(nba_matrix, Rowv=NA, Colv=NA, col = brewer.pal(9, "Blues"), scale="column", margins=c(5,10))
Now the code gives me the heatmap as shown bellow, but the labels are not clear. Please help me to get a clear heatmap.
since you stated that you need all the labels, the only way I see is reducing the font size. You can do this by setting the cexCol and cexRow parameters in your call to heatmap(); for example like this:
heatmap(as.matrix(iris[,1:3]),cexRow = 0.1, cexCol = 0.1,)
I am using R to do a hierarchical cluster analysis using the Ward's squared euclidean distance. I have a matrix of x columns(stations) and y rows(numbers in float), the first row contain the header(stations' names). I want to have a good dendrogram where the name of the station appear at the bottom of the tree as i am not able to interprete my result. My aim is to find those stations which are similar. However using the following codes i am having numbers (100,101,102,...) for the lower branches.
Yu<-read.table("yu_s.txt",header = T, dec=",")
library(cluster)
agn1 <- agnes(Yu, metric = "euclidean", method="ward", stand = TRUE)
hcd<-as.dendrogram(agn1)
par(mfrow=c(3,1))
plot(hcd, main="Main")
plot(cut(hcd, h=25)$upper,
main="Upper tree of cut at h=25")
plot(cut(hcd, h=25)$lower[[2]],
main="Second branch of lower tree with cut at h=25")
A nice collection of examples are present here (http://gastonsanchez.com/blog/how-to/2012/10/03/Dendrograms.html)
Two methods:
with hclust from base R
hc<-hclust(dist(mtcars),method="ward")
plot(hc)
Default plot
ggplot
with ggplot and ggdendro
library(ggplot2)
library(ggdendro)
# basic option
ggdendrogram(hc, rotate = TRUE, size = 4, theme_dendro = FALSE)