Thank you so much for this elegant solution, Jeff.
Philip
On 2020-11-12 02:20, Jeff Newmiller wrote:
I am not a data.table afficiando, but here is how I would do it with
dplyr/tidyr:
library(dplyr)
library(tidyr)
do_per_REL <- function( DF ) {
rng <- range( DF$REF1 ) # watch out for
I am not a data.table afficiando, but here is how I would do it with
dplyr/tidyr:
library(dplyr)
library(tidyr)
do_per_REL <- function( DF ) {
rng <- range( DF$REF1 ) # watch out for missing months?
DF <- ( data.frame( REF1 = seq( rng[ 1 ], rng[ 2 ], by = "month" ) )
%>%
I am stuck on a data transformation problem. I have a data frame, df1 in
my example, with some original "levels" data. The data pertain to some
variable, such as GDP, in various reference periods, REF, as estimated
and released in various release periods, REL. The release periods follow
after
There is no "perhaps" about it. Nonsense phrases like "similar to logit, where
I dont [sic] lose normality of the data" that lead into off-topic discussions
of why one introduces transformations in the first place are perfect examples
of why questions like this belong on a statistical theory
this might work for you
newy <- sign(oldy)*f(abs(oldy))
where f() is a monotonic transformation, perhaps a power function.
On Sun, Jan 20, 2019 at 11:08 AM Adrian Johnson
wrote:
>
> I apologize, I forgot to mention another key operation.
> in my matrix -1 to <0 has a different meaning while
of Anthropology
Texas A University
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Adrian Johnson
Sent: Sunday, January 20, 2019 8:02 AM
To: r-help
Subject: [R] data transformation
Dear group,
My question, perhaps is more of a statistical question using R
I have
I apologize, I forgot to mention another key operation.
in my matrix -1 to <0 has a different meaning while values between >0
to 1 has a different set of meaning. So If I do logit transformation
some of the positives becomes negative (values < 0.5 etc.). In such
case, the resulting transformed
Dear group,
My question, perhaps is more of a statistical question using R
I have a data matrix ( 400 x 400 normally distributed) with data
points ranging from -1 to +1..
For certain clustering algorithms, I suspect the tight data range is
not helping resolving the clusters.
Is there a way to
) which(!!x[,3]))
}
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of arun
Sent: Tuesday, November 12, 2013 2:13 PM
To: R help
Subject: Re: [R] Data transformation to list
Hi,
Say I have a following data
ID WeekEvent_Occurence
A 1 0
A 2 0
A 3 1
A 4 0
B 1 1
B 2 0
B 3 0
B 4 1
that whether an individual experienced an event in a particular week.
I wish to create list such as the first element of the list will be a
vector listing the week number when the event
Hi Anindya,
You may try:
dat1 - read.table(text=ID Week Event_Occurence
A 1 0
A 2 0
A 3 1
A 4 0
B 1 1
B 2 0
B 3 0
B 4 1,sep=,header=TRUE,stringsAsFactors=FALSE)
with(dat1,tapply(as.logical(Event_Occurence),ID,FUN=which ))
#or
lapply(split(dat1,dat1$ID),function(x) which(!!x[,3]))
A.K.
Seems your questions belong to rule mining for frequent item sets.
check arules package
Weidong Gu
On Tue, Sep 27, 2011 at 11:13 PM, pip56789 pd...@virginia.edu wrote:
Hi,
I have a few methodological and implementation questions for ya'll. Thank
you in advance for your help. I have a dataset
On 09/28/2011 01:13 PM, pip56789 wrote:
Hi,
I have a few methodological and implementation questions for ya'll. Thank
you in advance for your help. I have a dataset that reflects people's
preference choices. I want to see if there's any kind of clustering effect
among certain preference choices
Hi,
I have a few methodological and implementation questions for ya'll. Thank
you in advance for your help. I have a dataset that reflects people's
preference choices. I want to see if there's any kind of clustering effect
among certain preference choices (e.g. do people who pick choice A also
On a methodological level, if the choices do not correspond on a cardinal or
at least ordinal scale, you don't want to use correlations. Instead you
should probably use Cramer's V, in particular if the choices are
multinomial. Whether the wide format is necessary will depend on the format
the
Hi
Could any one please help how I can trnasform data based on Box-Cox
Transformations. I have massive data set with many variables. If
possible someone can write few lines so I can read in all data set
once and transform it.
g1 g2 g2
97.03703704 89.25925926 4.4
Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Stuart
Sent: Tuesday, May 03, 2011 9:37 AM
To: r-help@r-project.org
Subject: [R] data transformation Box-Cox Transformations
Hi
of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Stuart
Sent: May-03-11 11:37 AM
To: r-help@r-project.org
Subject: [R] data transformation
Dear Group,
Need to do the following transformation:
I have the dataset
structure(list(Date = structure(1L, .Label = 2010-06-16, class =
factor),
ACC.returns1Day = -0.018524832, ACC.returns5Day = 0.000863931,
ACC.returns7Day = -0.019795222, BCC.returns1Day = -0.009861859,
Dear all,
I have a dataset that looks like this:
x - read.table(textConnection(col1 col2
3 1
2 2
4 7
8 6
5 10), header=TRUE)
I want to rewrite it as below:
var1 var2 var3 var4 var5 var6 var7 var8 var9 var10
1 0 1 0 0 0 0 0 0 0
0 2
Well, I have no idea how to get from one to the other. There's
col1 and col2 but no var1 var2 var3, etc. I thought perhaps col1
was the row index and col2 was the column index, but that doesn't
match up either, and not all the cell values are 1.
So you will need to explain more clearly what you
Hi,
On Mon, Jan 25, 2010 at 5:39 PM, Lisa lisa...@gmail.com wrote:
Dear all,
I have a dataset that looks like this:
x - read.table(textConnection(col1 col2
3 1
2 2
4 7
8 6
5 10), header=TRUE)
I want to rewrite it as below:
var1 var2 var3 var4 var5 var6 var7 var8 var9 var10
1
Thank you so much.
Lisa
--
View this message in context:
http://n4.nabble.com/Data-transformation-tp1289899p1289915.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
r-help-boun...@r-project.org wrote on 01/25/2010 02:39:32 PM:
x - read.table(textConnection(col1 col2
3 1
2 2
4 7
8 6
5 10), header=TRUE)
I want to rewrite it as below:
var1 var2 var3 var4 var5 var6 var7 var8 var9 var10
1 0 1 0 0 0 0 0 0
Try this:
t(apply(x, 1, function(r) table(factor(r, levels = seq_len(max(x))
1 2 3 4 5 6 7 8 9 10
[1,] 1 0 1 0 0 0 0 0 0 0
[2,] 0 2 0 0 0 0 0 0 0 0
[3,] 0 0 0 1 0 0 1 0 0 0
[4,] 0 0 0 0 0 1 0 1 0 0
[5,] 0 0 0 0 1 0 0 0 0 1
If you use aaply in the plyr package instead of apply then
Try this:
x - read.table(textConnection(idcode1code2 p
+ 148 0.1
+ 157 0.9
+ 218 0.4
+ 262 0.2
+ 243 0.6
+ 356 0.7
+
Your script works very well. Thank you very much.
Legen
Henrique Dallazuanna wrote:
Try this also:
xtabs(rep(p, 2) ~ rep(id, 2) + sprintf(var%d, c(code1, code2)), data =
x)
On Wed, Nov 11, 2009 at 2:10 AM, legen lege...@gmail.com wrote:
Thank you for your kind help. Your script
That's what I want. Many thanks for your help.
Legen
jholtman wrote:
Try this:
x - read.table(textConnection(idcode1code2 p
+ 148 0.1
+ 157 0.9
+ 218 0.4
+ 262
Dear all,
I have a dataset as below:
idcode1code2 p
148 0.1
157 0.9
218 0.4
262 0.2
243 0.6
356 0.7
37
Is this what you want:
x - read.table(textConnection(idcode1code2 p
+ 148 0.1
+ 157 0.9
+ 218 0.4
+ 262 0.2
+ 243 0.6
+ 356
Thank you for your kind help. Your script works very well. Would you please
show me how to change NaN to zero and column variables 1, 2, ..., 8 to var1,
var2, ..., var8? Thanks again.
Legen
jholtman wrote:
Is this what you want:
x - read.table(textConnection(idcode1code2
Hi R-users,
I have this code to uniformise the data using gamma:
length(dp1)
[1] 696
dim(dp1)
[1] 58 12
dim(ahall)
[1] 1 12
dim(bhall)
[1] 1 12
trans_dt - function(dt,a,b)
+ { n1 - ncol(dt)
+ n2 - length(dt)
+ trans - vector(mode='numeric', length=n2)
+ dim(trans) - dim(dt)
+
Roslina,
this code performs what you need:
dt = matrix((1:(58*12))/58/12,58) # some numbers
# if dt is a data.frame use dt = as.matrix(dt)
a = (1:12)/12 # some a coef
b = (12:1)/12 # some b coef
dtgam = matrix(pgamma(dt,a,b),58)
# dtgam is the transformation you're looking for
Dear all,
how can I, with R, transform a presence-only table (with the names of
the species (1st column), the lat information of the sites (2nd column)
and the lon information of the sites (3rd column)) into a
presence-absence (0/1) matrix of species occurrences across sites, as
given in the
on 07/22/2008 11:24 AM Christian Hof wrote:
Dear all,
how can I, with R, transform a presence-only table (with the names of
the species (1st column), the lat information of the sites (2nd column)
and the lon information of the sites (3rd column)) into a
presence-absence (0/1) matrix of
Dear all,
how can I, with R, transform a presence-absence (0/1) matrix of species
occurrences into a presence-only table (3 columns) with the names of the
species (1st column), the lat information of the sites (2nd column) and
the lon information of the sites (3rd column), as given in the
PM
To: r-help@r-project.org
Subject: [R] data transformation
Dear all,
how can I, with R, transform a presence-absence (0/1) matrix
of species occurrences into a presence-only table (3 columns)
with the names of the species (1st column), the lat
information of the sites (2nd column
Hi Christian,
Here's a way using the reshape package:
dfr
site lat lon spec1 spec2 spec3 spec4
1 site1 10 11 1 0 1 0
2 site2 20 21 1 1 1 0
3 site3 30 31 0 1 1 1
library(reshape)
dfr - melt(dfr[, -1], id=1:2, variable_name='species')
Try this:
newx - with(x, cbind(stack(x, select = grep(spec, names(x))), lat, lon))
newx[newx$values 0, -1]
On 5/2/08, Christian Hof [EMAIL PROTECTED] wrote:
Dear all,
how can I, with R, transform a presence-absence (0/1) matrix of species
occurrences into a presence-only table (3 columns)
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