Hi:
There are two good reasons why the loop solution is not efficient in
this (and related) problem(s):
(i) There is more code and less transparency;
(ii) the vectorized solution is four times faster.
Here are the two proposed functions:
# Vectorized version
m1 - function(v) paste(v, ' to ', v
out - vector(list)
Ylab - for(i in 1:length(BndY))
{
out[i] - paste(BndY[i], to ,BndY[i],mN)
}
Ylab - do.call(c, out)
markm0705 wrote
Dear R helpers
I'm trying to make up some labels for plot from this vector
BndY-seq(from = 18900,to= 19700, by = 50)
using
Ylab-for(i in
Don't do this! paste() is vectorized.
paste (BndY,to,50+seq_len(BndY), mN, sep = )
Cheers,
Bert
On Wed, Nov 23, 2011 at 3:31 PM, B77S bps0...@auburn.edu wrote:
out - vector(list)
Ylab - for(i in 1:length(BndY))
{
out[i] - paste(BndY[i], to ,BndY[i],mN)
}
Ylab - do.call(c, out)
... and you can of course do the assignment:
Bndy - paste (BndY,to,50+seq_len(BndY), mN, sep = )
An Introduction to R tells you about such fundamentals and should be
a first read for anyone learning R.
--- Bert
On Wed, Nov 23, 2011 at 4:10 PM, Bert Gunter bgun...@gene.com wrote:
Don't do
Dear R helpers
I'm trying to make up some labels for plot from this vector
BndY-seq(from = 18900,to= 19700, by = 50)
using
Ylab-for(i in BndY) {c((paste(i, to ,i+50,mN)))}
but the vector created is NULL
However if i use
for(i in BndY) {print(c(paste(i, to ,i+50,mN)))}
I can see the for
Try this instead:
Ylab - paste(BndY, BndY+50, mN)
Michael
On Wed, Nov 23, 2011 at 5:26 PM, markm0705 markm0...@gmail.com wrote:
Dear R helpers
I'm trying to make up some labels for plot from this vector
BndY-seq(from = 18900,to= 19700, by = 50)
using
Ylab-for(i in BndY) {c((paste(i, to
Thank you
On Thu, Nov 24, 2011 at 7:31 AM, B77S [via R]
ml-node+s789695n4102066...@n4.nabble.com wrote:
out - vector(list)
Ylab - for(i in 1:length(BndY))
{
out[i] - paste(BndY[i], to ,BndY[i],mN)
}
Ylab - do.call(c, out)
markm0705 wrote
Dear R helpers
I'm trying to make up
And thanks fo rthe pointer to the R introduction book as well
On Thu, Nov 24, 2011 at 11:00 AM, Mark Murphy markm0...@gmail.com wrote:
Thank you
On Thu, Nov 24, 2011 at 7:31 AM, B77S [via R]
ml-node+s789695n4102066...@n4.nabble.com wrote:
out - vector(list)
Ylab - for(i in
: Re: [R] looping with paste
As Sarah has said, you probably don't need to use paste() at all. However
if command is a text string containing a (syntactically correct) R command
you can execute it via
eval(parse(text=command))
E.g.:
command - x - 42
eval(parse(text=command
Dear list,
I have a spacialPolygonDataFrame where variables were unnecessarily imported as
factors. So I am trying to unfactor variables from spatialPolygonDataFrame@data
with a loop
for (i in (1:length(names( spatialPolygonDataFrame{
Juta,
On Mon, Aug 22, 2011 at 4:29 PM, Juta Kawalerowicz
juta.kawalerow...@stx.ox.ac.uk wrote:
Dear list,
I have a spacialPolygonDataFrame where variables were unnecessarily imported
as factors. So I am trying to unfactor variables from
spatialPolygonDataFrame@data with a loop
for (i in
As Sarah has said, you probably don't need to use paste() at all. However
if command is a text string containing a (syntactically correct) R command
you can execute it via
eval(parse(text=command))
E.g.:
command - x - 42
eval(parse(text=command))
x
[1] 42
I find this to
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