thanks for the tips! your answer will certainly help me a lot!
ishida
2013/3/13 Robert Baer
> On 3/13/2013 12:05 AM, ishi soichi wrote:
>
>> I am not sure if I should ask this question in this list. But I'll try.
>>
>> Currently I am trying to analyze images using
I am not sure if I should ask this question in this list. But I'll try.
Currently I am trying to analyze images using EBImage and biOps.
One of the features that I need to extract from various images is the color
spectrum, namely, which colors each image consists of.
So, each image hopefully can
yes. thank you.
ishida
2013/3/11 Jorge I Velez
> Is the following that you are looking for?
>
> unlist(lapply(x.list, "[", 2))
>
> HTH,
> Jorge.-
>
>
> On Mon, Mar 11, 2013 at 9:52 PM, ishi soichi <> wrote:
>
>> say I have a matrix and l
say I have a matrix and lists like
x <- matrix(c(12.1, 3.44, 0.1, 3, 12, 33.1, 1.1, 23), nrow=2)
x.list <- lapply(seq_len(nrow(x)), function(i) x[i,])
if I want a column of the matrix x, I write
x[, 2]
for example.
But how can I do something similar for a set of lists, x.list, above?
> x.lis
the title of this thread was wrong. it's misleading sorry.
ishida
2013/3/11 ishi soichi
>
> say I have a matrix and lists like
>
> x <- matrix(c(12.1, 3.44, 0.1, 3, 12, 33.1, 1.1, 23), nrow=2)
>
> x.list <- lapply(seq_len(nrow(x)), function(i) x[i,])
>
> if
t;
>
> For the second case:
> lst1<-as.list(data.frame(t(apply(x,1,calcnorm2.const,x
> names(lst1)<- NULL
> lst1
> #[[1]]
> #[1] 0.0 31.75257
>
> #[[2]]
> #[1] 31.75257 0.0
>
>
>
>
> A.K.
> - Original Message -
> From:
there is a typo.
lapply(x.list, calcnorm, x.list)
should be
lapply(x.list, calcnorm2, x.list)
sorry.
2013/3/10 ishi soichi
> I need to develop a simple list manipulation. Although it seems easier to
> do it in matrix form, but I need it in list form.
>
> I have a matrix
>
I need to develop a simple list manipulation. Although it seems easier to
do it in matrix form, but I need it in list form.
I have a matrix
x <- matrix(c(12.1, 3.44, 0.1, 3, 12, 33.1, 1.1, 23), nrow=2)
for list form example, the conversion is
x.list <- lapply(seq_len(nrow(x)), function(i) x[i,]
t; function (i) sapply(lis, "[", i))
>
>
> I hope it helps.
>
> Best,
> Dimitris
>
>
> On 3/8/2013 11:06 AM, ishi soichi wrote:
> > Thanks. The result should be a list of lists like...
> >
> >> x
> > [[1]]
> > [1] 12.10 3.44
>
help-boun...@r-project.org [mailto:r-help-bounces@r-
> > project.org] On Behalf Of ishi soichi
> > Sent: Friday, March 08, 2013 10:50 AM
> > To: r-help
> > Subject: [R] transpose lists
> >
> > Can you think of a function that transposes a list like
>
> What shall
Can you think of a function that transposes a list like
> x
[[1]]
[1] 12.1 0.1 12.0 1.1
[[2]]
[1] 3.44 3.00 33.10 23.00
?
ishida
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thanks! It works.
I couldn't possibly figure out such trick...
soichi
2013/3/8 Jorge I Velez
> If I understood correctly,
>
> lapply(x, "[", 1:3)
>
> will do what you want.
>
> HTH,
> Jorge.-
>
>
> On Fri, Mar 8, 2013 at 5:05 PM, ishi soic
hi. I have a list like
x <- list(1:10,11:20,21:30)
It's a sort of a 3 x 10 matrix in list form.
I would like to reduce the dimension of this list.
it would be something like
list(1:3, 11:13, 21,23)
I tried
x[,1:3]
does not work of course. Neither
lapply(x, [1:3])
works...
Any suggestions?
2013/3/4 Prof Brian Ripley
> See the recent discussion on R-sig-mac (the place to ask questions about
> OS X).
>
>
> On 04/03/2013 11:00, ishi soichi wrote:
>
>> version.string R version 2.15.2 (2012-10-26)
>>
>> I am trying to install biOps on MacOS X 10.8.2
&g
version.string R version 2.15.2 (2012-10-26)
I am trying to install biOps on MacOS X 10.8.2
First, I have tiff, fftw-3, jpeg
and set paths like
cd /usr/include
sudo ln -s /usr/local/include/fftw3.h
for x in /usr/local/include/j*.h; do sudo ln -s $x; done
for x in /usr/local/include/tiff*.h; do
-%m-%d %H:%M:%S")
> library(xts)
> dat3<-xts(dat2[,-1],order.by=dat2[,1])
> plot(dat3)
> A.K.
>
>
>
>
> - Original Message -
> From: ishi soichi
> To: r-help
> Cc:
> Sent: Monday, January 7, 2013 3:55 AM
> Subject:
R-64 latest
Hi. I am trying to plot a set of csv data, which looks like
> head(interval)
date inteval
1 2012-07-01 00:57:54 +0900 156
2 2012-07-01 01:07:41 +0900 587
3 2012-07-01 01:09:31 +0900 110
4 2012-07-01 01:18:42 +0900 551
5 2012-07-01 01:39:01 +0900
platform x86_64-apple-darwin9.8.0
arch x86_64
os darwin9.8.0
system x86_64, darwin9.8.0
version.string R version 2.13.1 (2011-07-08)
I am trying to write a function that takes a few objects as input.
test <- function(directory, num = 1:100) {
> }
the argument
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