Monica --
There has been a virtual population explosion of R books in recent years
and we all have our favorites. You may wish to pick one oriented toward
your specialty, but the absolute minimum lowest common denominator (by
which I mean that it has the ground zero essential information that all
users must share, not that it is minimal or incomplete) is the manual
An Introduction to R, available by download from the Cran website.
Beyond that, my favorite introduction is Peter Dalgaard's Introductory
Statistics with R. He has an elegance and clarity of style, as well as
a feel for what is necessary to include in an introduction, that some
others lack. Others may disagree, but I find myself returning to
Dalgaard again and again.
Ben
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Monica Pisica
Sent: Thursday, March 13, 2008 9:05 AM
To: Gabor Grothendieck; [EMAIL PROTECTED];
[EMAIL PROTECTED]
Cc: r-help@r-project.org
Subject: [PS] Re: [R] a more elegant way to get percentages?
Hi everybody,
I am amazed how quick i got my answer ;-) I have to recognize that
Gabor's code really puts to shame my skills in doing any programming in
R. Is there any book or documentation which really explains in details
all these neat tricks from {stats} like ave (i even didn't know this
function existed), apply and all its friends (sapply, tapply, etc) ? To
be honest it took me quite a while to come up with the fancy
subscripting to get my persantages ;-))
thank you so much, i really appreciate your help,
Monica
Date: Thu, 13 Mar 2008 09:45:05 -0400
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: Re: [R] a more elegant way to get percentages?
CC: r-help@r-project.org
Assuming your x is as follows:
x - data.frame(locat = c(a, b, b, c, c, c, c, d, d,
d),
val = c(5, 5, 15, 5, 20, 5, 10, 5, 15, 10))
Try this:
x$percent1 - ave(x$val, x$locat, FUN = function(x) 100*x/sum(x))
On Thu, Mar 13, 2008 at 9:36 AM, Monica Pisica wrote:
Hi,
I am trying to get percentages in a more elegant way. I have a
data.frame with locations and values (counts) of species at that
location. Each location is repeated for each species i have values for
and i would like to get percentages of each species at that location. I
am not sure if i am clear in my explanations so i will paste my code
below:
#
x
locat val
1 a 5
2 b 5
3 b 15
4 c 5
5 c 20
6 c 5
7 c 10
8 d 5
9 d 15
10 d 10
loc1 - x$locat
n - length(loc1)
locuniq1 - unique(loc1)
m - length(locuniq1)
counts - seq(1:m)
for (i in 1:m) {
+ count - 0
+ for (j in 1:n) {
+ if (loc1[j]==locuniq1[i]) count - count+1
+ counts[i] - count
+ }
+ }
percent1 - rep(0,n)
j - 0
for (i in 1:m) {
+
+ b - x[(j+1):(j+counts[i]),]
+ total - sum(b$val)
+ percent1[(j+1):(j+counts[i])] - round(apply(as.matrix(b$val), 1,
function(x) {x*100/total}),2)
+ j = j+counts[i]
+ }
x1 - cbind(x, percent1) # this is the result i want
x1
locat val percent1
1 a 5 100.00
2 b 5 25.00
3 b 15 75.00
4 c 5 12.50
5 c 20 50.00
6 c 5 12.50
7 c 10 25.00
8 d 5 16.67
9 d 15 50.00
10 d 10 33.33
I am wondering if there is any way to do it more efficiently, much
more that the first loop which gives how many times each location is
present in the data.frame is slow if you have a larger data.frame and
not only 10 rows.
Thanks for any input and sorry if the email is on the long side,
Monica
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and provide commented, minimal, self-contained, reproducible code.
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