fantastic Michael! The code "V = sort(unique(ave(x,cut(x,HistOut$breaks))))"
works good. I got what I need! that is 20 mean values. I calculate the first
value in excel, it's the same (79.61429).
> HistOut= hist(data$ DMP_m3.jaso..10, freq=F, breaks =20,
> xlim=c(0,3000),col="grey26", main="DMP_niger", plot=T)
V = sort(unique(ave(data$ DMP_m3.jaso..10,cut(data$
DMP_m3.jaso..10,HistOut$breaks))))
> write(V, "", sep = " ")
79.61429 149.9792 257.4969 347.8211 449.2063
552.9694 647.5132 749.23 859.0167 945.4912
1046.917 1147.703 1249.375 1359.794 1447.518
1553.838 1649.231 1735.217 1850.617 1957.85
2031.329 2156.8 2247.55 2340.6 2822.1
Thank you very much.
francesco
From: michael.weyla...@gmail.com
Date: Tue, 23 Aug 2011 11:27:13 -0400
Subject: Re: [R] [r] histogram with mean for every break
To: nutini.france...@gmail.com
CC: r-help@r-project.org
I do not believe your code (minimal as it is) would work: the correct argument
is "breaks". More generally, do you really mean to say that hist(x, breaks =
20) immediately returns the bin counts? It doesn't on my machine and if you
knew how to get the counts, you should be able to get the midpoints as well.
Try this
x = rnorm(5000)
HistOut = hist(x, breaks = 20)
HistOut$mids # will give midpoints for each bin of this histogram
If you need proper means, rather than midpoints:
# I *think* this works -- no guarantees: it assigns x to bins based on the
histogram breaks, averages those by bin, and then we pull out just the unique
values and sort them
V = sort(unique(ave(x,cut(x,HistOut$breaks))))
It does seem to be a somewhat strange task to use histogram breaks to get means
though: if I were you, I'd spend a minute or two contemplating if this really
makes the most sense, rather than just doing the breaks in some way yourself
directly.
There may well be a more elegant way but I think this gets the job done.
Hope it helps,
Michael
On Tue, Aug 23, 2011 at 9:46 AM, Francesco Nutini <nutini.france...@gmail.com>
wrote:
Dear R-users,
I need to produce a histogram where for every breaks there are the mean of the
data.
I tried tu use the function >hist(x, break=20 ... ) but this return the
numerosity for every breaks, not the mean.
Any hint?
Thanks in advance,
francesco
---------------------------------------
Francesco Nutini
CNR-IREA
Ist. per il Rilevamento Elettromagnetico dell'Ambiente
Via Bassini 15, 20133 Milano (Italy)
Tel: +39-02 23699 297
http://www.irea.cnr.it
nutin...@irea.cnr.it
Skype: ui...@hotmail.it
Univ. mail: francesco.nut...@unimi.it
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