Plot grouped values of a dataframe in R

Question

I have the following dataframe:

          stat                     mTADs                    DE_genes
    5267 -5.452819  chr2:167337500-167447500  chr2:167318145-167341673:+
    5268  4.114012    chr6:41532500-41642500    chr6:41555481-41570508:+
    5269  9.812369   chr10:18157500-18262500   chr10:18259929-18265882:-
    5270  3.371969   chr17:40957500-41062500   chr17:41060000-41071996:-
    5271  4.576930   chr17:40957500-41062500   chr17:41012431-41017507:-
    5272  2.952151   chr11:72251250-72352500   chr11:72254857-72265270:+
    5273 -3.349795  chr1:174307500-174407500  chr1:174405489-174408706:+
    5274 -2.685897 chr13:100777500-100877500 chr13:100787949-100874025:-
    5275  2.865269 chr13:100777500-100877500 chr13:100718488-100785594:-
    5276  6.436959  chr4:150417500-150517500  chr4:150377761-150418774:-
    5277  2.622196      chr7:6072500-6162500      chr7:6123828-6142951:+
    5278 -5.605531   chr11:48597500-48682500   chr11:48675470-48685185:-
    5279  3.554733   chr11:48597500-48682500   chr11:48639642-48665711:+
    5280  4.399655   chr11:48597500-48682500   chr11:48638848-48640157:-

As you see some DE_genes fall within the same mTAD. I would like to plot for all the DE_genes their stat values and group them by mTAD. I thought of doing this as a horizontal barplot having on the yaxis the genes, and on the x axis the stat values and group them by TAD but first i do not know how to do it and second i thought that a heatmap might be a better option. Is there any way to do that in R? In total I have 1700 mTADs and I would like to see if there are any patterns in the data.

Many thanks, Dimitris

Heatmaps are not useful for the data that you present here. Heatmaps nearly always have two dimensions so that they make a grid of color. Your `stat` variable is one dimension for a heatmap. What is your second dimension, mTAD? That could be a very difficult heatmap to interpret. — David, May 06 '16 at 08:50

score 0 · Accepted Answer · answered May 06 '16 at 08:37

You might consider a sorted dotchart instead of a barchart.

> thing
     ID      stat                     mTADs                    DE_genes
1  5267 -5.452819  chr2:167337500-167447500  chr2:167318145-167341673:+
2  5268  4.114012    chr6:41532500-41642500    chr6:41555481-41570508:+
3  5269  9.812369   chr10:18157500-18262500   chr10:18259929-18265882:-
4  5270  3.371969   chr17:40957500-41062500   chr17:41060000-41071996:-
5  5271  4.576930   chr17:40957500-41062500   chr17:41012431-41017507:-
6  5272  2.952151   chr11:72251250-72352500   chr11:72254857-72265270:+
7  5273 -3.349795  chr1:174307500-174407500  chr1:174405489-174408706:+
8  5274 -2.685897 chr13:100777500-100877500 chr13:100787949-100874025:-
9  5275  2.865269 chr13:100777500-100877500 chr13:100718488-100785594:-
10 5276  6.436959  chr4:150417500-150517500  chr4:150377761-150418774:-
11 5277  2.622196      chr7:6072500-6162500      chr7:6123828-6142951:+
12 5278 -5.605531   chr11:48597500-48682500   chr11:48675470-48685185:-
13 5279  3.554733   chr11:48597500-48682500   chr11:48639642-48665711:+
14 5280  4.399655   chr11:48597500-48682500   chr11:48638848-48640157:-

First we will take medians by mTADs.

medians.of.stat.by.mTADs<-aggregate(stat~mTADs,data=thing,FUN=median)
names(medians.of.stat.by.mTADs)[2]<-"median stat for mTAD"

Now merge these medians with the original data frame and create a factor by ordering on the sorted stat values.

thing<-merge(thing,medians.of.stat.by.mTADs,all.x = T,by="mTADs")
thing$mTADs.reordered <-factor(thing$mTADs, levels=thing[order(thing$`median stat for mTAD`), "mTADs"])

This will throw a warning because of duplicated levels but it seems to work.

> thing
                       mTADs   ID      stat                    DE_genes median stat for mTAD           mTADs.reordered
1   chr1:174307500-174407500 5273 -3.349795  chr1:174405489-174408706:+            -3.349795  chr1:174307500-174407500
2    chr10:18157500-18262500 5269  9.812369   chr10:18259929-18265882:-             9.812369   chr10:18157500-18262500
3    chr11:48597500-48682500 5278 -5.605531   chr11:48675470-48685185:-             3.554733   chr11:48597500-48682500
4    chr11:48597500-48682500 5279  3.554733   chr11:48639642-48665711:+             3.554733   chr11:48597500-48682500
5    chr11:48597500-48682500 5280  4.399655   chr11:48638848-48640157:-             3.554733   chr11:48597500-48682500
6    chr11:72251250-72352500 5272  2.952151   chr11:72254857-72265270:+             2.952151   chr11:72251250-72352500
7  chr13:100777500-100877500 5274 -2.685897 chr13:100787949-100874025:-             0.089686 chr13:100777500-100877500
8  chr13:100777500-100877500 5275  2.865269 chr13:100718488-100785594:-             0.089686 chr13:100777500-100877500
9    chr17:40957500-41062500 5270  3.371969   chr17:41060000-41071996:-             3.974449   chr17:40957500-41062500
10   chr17:40957500-41062500 5271  4.576930   chr17:41012431-41017507:-             3.974449   chr17:40957500-41062500
11  chr2:167337500-167447500 5267 -5.452819  chr2:167318145-167341673:+            -5.452819  chr2:167337500-167447500
12  chr4:150417500-150517500 5276  6.436959  chr4:150377761-150418774:-             6.436959  chr4:150417500-150517500
13    chr6:41532500-41642500 5268  4.114012    chr6:41555481-41570508:+             4.114012    chr6:41532500-41642500
14      chr7:6072500-6162500 5277  2.622196      chr7:6123828-6142951:+             2.622196      chr7:6072500-6162500

Now do a simple dotchart.

ggplot() + geom_point(data=thing, aes(x=stat,y=mTADs.reordered), shape=20, cex=3.3)

Thank you very much David for suggesting this way of visualing. It worked! — user2931662, May 06 '16 at 12:42

Plot grouped values of a dataframe in R

1 Answers1