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Heatmap without levelplot

8 messages · Dieter Menne, Greg Snow, Antje +1 more

Antje
# 
Hi there,

as I'm not sure to understand the coloring levelplot uses, I'm looking for 
another easy way to create a heatmap like this:

library(lattice)
mat <- matrix(seq(1,5, length.out = 12), nrow = 3)
mat[1,2] <- 3.5

my.at <- seq(0.5,5.5, length.out = 6)
my.col.regions <- rainbow(5)

graph <- levelplot(t(mat[nrow(mat):1, ] ), at = my.at, col.regions = 
my.col.regions)
print(graph)

Can anybody help me with some hints or little examples?

Antje
Dieter Menne
# 
Antje <niederlein-rstat <at> yahoo.de> writes:
Looks fine to me, so what's wrong? Maybe have a look at RColorBrewer
to get more pleasing colors.

Dieter
Uwe Ligges
#
Antje wrote:
Dear Antje,

since you are asking the same question again now, maybe you can explain 
what you are going to get? In fact, I do not undertsand where your 
problem is. R places the colors according to the values in your matrix 
very well including the legend and I thought up to today that the plot 
is self explaining.

Best wishes,
Uwe Ligges
Antje
# 
Hi Uwe,

I tried to explain my problem with the given example.
I don't see any documentation which tells me that the length of "col.regions" 
should be one less than "at". (At least I assume now that it should be this way...)
If it's equal or longer some colors (in the middle of the color-vector) are 
simply not used.
Just try the example below with rainbow(5) and rainbow(6) and compare the 
results... both plot will use 5 colors!
Sorry, but this behaviour is not really self-explaining to me... maybe I'm to 
blind to find the documentation which says that only one color less will ensure 
the usage of all colors.

(It is so important for me because I need to display a heatmap with colors 
let's say
* all lower data outliers "green",
* all higher data outliers "blue" and
* everything else within the color range "yellow" to "red".
I've seen that some values do not get blue or green though they are outliers...
I've attached one graph, I've generated - maybe it helps to understand)

Any wrong assumption?

Ciao,
Antje


Uwe Ligges schrieb:
Greg Snow
# 
The image function in base graphics does the same type of plot, just different names and structure of the data (and the documentation says that the number of breaks should be 1 more than the number of colors).

Hope this helps,
Uwe Ligges
#
Antje wrote:
Well, of you have 5 at locations (i.e. breaks), then you have 4 
intervals in between and that's the amount of colors that is sensible.
Maybe:

Say you want everything below -1 be considered as a lower outlier and 
all above 1 is a higher outlier, then you can say:


levelplot(matrix(c(1,2,0,-2), nrow=2),
     at = c(-Inf, seq(-1, 1, length=10), Inf),
     col.regions = c(rgb(0,1,0),
          hcl(seq(20, 80, length=10), c=400),
          rgb(0,0,1)))

Then below -1 is green (rgb(0,1,0)), above 1 is blue (rgb(0,0,1)) and in 
between we have 10 regions from -1 to 1 each with a color between some 
kind of yellow and red in hcl() space.


Uwe Ligges
Antje
# 
Hi Uwe,

thanks a lot for your answer! And thanks a lot to all others helping me with 
this issue!

Uwe Ligges schrieb:
From the documentation this was not really clear to me (though it makes sense, 
I agree)
Thank you for this little example. Just two comments:

1) I was not aware of the possibility to use "Inf" - it just has the 
disadvantage that these colors are not displayed at the color vector (maybe 
this can be adjusted somehow)

2) if you replace one number of the matrix with -1, it will be displayed green.
So it would be considered as an outlier. From the documentation it was also not 
clear to me that the lower endpoint of the interval is always excluded (except 
for the very first value of the at-vector), while the upper endpoint will be 
included. (This also makes sense but in this case I have to slightly modify my 
data because I'd like to include both endpoints of my non-outlier-range...)

Anyway, I guess, I solved all problems and found a workable solution :-)

Ciao,
Antje
Uwe Ligges
#
Antje wrote:
You can add or substract .Machine$double.eps to the endpoint in order to 
in/exclude the "==" case.

Uwe Ligges