[R] extracting characters from a string
Dear All,
I have a data frame of vectors of publication names such as 'pub':
pub1 - c('Brown DK, Santos R, Rome DF, Don Juan X')
pub2 - c('Benigni D')
pub3 - c('Arstra SD, Van den Hoops DD, lamarque D')
pub - rbind(pub1, pub2, pub3)
I would like to construct a dataframe with only author's last name and each
last name in columns and the publication in rows. Basically I want to get rid
of the initials (max 2, always before a comma) and spaces surounding last name.
I would like to avoid a loop.
ps: If I could have even a short explanation of the code that extract the
values of the character string that would also be great!
David
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Re: [R] extracting characters from a string
thanks, it works well. I have to work on Arun's previous answer to make it work
too.
David
De : Rui Barradas ruipbarra...@sapo.pt
À : Biau David djmb...@yahoo.fr
Cc : r help list r-help@r-project.org
Envoyé le : Mercredi 23 janvier 2013 19h57
Objet : Re: [R] extracting characters from a string
Hello,
I've just noticed that my first solution would only return the first set
of alphabetic characters, such as Van, not Van den Hoops.
The following will solve that problem.
fun2 - function(x, sep = , ){
x - strsplit(x, sep)
m - lapply(x, function(y) gregexpr( [[:alpha:]]*$, y))
res - lapply(seq_along(x), function(i)
regmatches(x[[i]], m[[i]], invert = TRUE))
res - lapply(res, unlist)
lapply(res, function(y) y[nchar(y) 0])
}
fun2(pub)
Hope this helps,
Rui Barradas
Em 23-01-2013 18:33, Rui Barradas escreveu:
Hello,
Try the following.
fun - function(x, sep = , ){
s - unlist(strsplit(x, sep))
regmatches(s, regexpr([[:alpha:]]*, s))
}
fun(pub)
Hope this helps,
Rui Barradas
Em 23-01-2013 17:38, Biau David escreveu:
Dear All,
I have a data frame of vectors of publication names such as 'pub':
pub1 - c('Brown DK, Santos R, Rome DF, Don Juan X')
pub2 - c('Benigni D')
pub3 - c('Arstra SD, Van den Hoops DD, lamarque D')
pub - rbind(pub1, pub2, pub3)
I would like to construct a dataframe with only author's last name and
each last name in columns and the publication in rows. Basically I
want to get rid of the initials (max 2, always before a comma) and
spaces surounding last name. I would like to avoid a loop.
ps: If I could have even a short explanation of the code that extract
the values of the character string that would also be great!
David
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Re: [R] removing loops from code in making data.frame
thanks, it goes a lot faster. Just one thing though, when I apply the code to
my data, both data.frames end up differente. Or at least identical(df1, df2)
if false
however when i do which(df1!=df2) it says 'integer (0)'.
Could that be due to the class of the vectors or some thing of the sort?
thanks,
David Biau
De : arun smartpink...@yahoo.com
À : Biau David djmb...@yahoo.fr
Cc : R help r-help@r-project.org
Envoyé le : Mardi 15 janvier 2013 21h54
Objet : Re: [R] removing loops from code in making data.frame
Hi,
You could also do this:
res1-do.call(rbind,lapply(xaulist,function(x)
as.numeric(apply(t(mapply(`==`,tata,x)),2,any
identical(res1,tutu)
#[1] TRUE
A.K.
- Original Message -
From: Biau David djmb...@yahoo.fr
To: r help list r-help@r-project.org
Cc:
Sent: Tuesday, January 15, 2013 2:41 PM
Subject: [R] removing loops from code in making data.frame
Dear all,
I am working on an author network and to do so I have to arrange a data.frame
(tutu) crossing author names (rows) per publication number (column). The
participation of the author to a study is indicated by a 1 and 0 otherwise.
I have a vector (xaulist) of all the names of authors and a data.frame (tata)
with all the publications in row and the authors in columns. I have writen a
loop to obtain my data.frame but it takes a long time when the number of
studies increases. I was looking for a more efficient code.
Here is a minimal working example (my code is terrible i know...):
#-
au1 - c('deb', 'art', 'deb', 'seb', 'deb', 'deb', 'mar', 'mar', 'joy', 'deb')
au2 - c('art', 'deb', 'soy', 'deb', 'joy', 'ani', 'deb', 'deb', 'nem', 'mar')
au3 - c('mar', 'lio', 'mil', 'mar', 'ani', 'lul', 'nem', 'art', 'deb', 'tat')
tata - data.frame(au1, au2, au3)
xaulist2 - levels(factor(unlist(tata[,])))
xaulist - levels(as.factor(xaulist2))
tutu - matrix(NA, nrow=length(xaulist), ncol=dim(tata)[1]) # row are authors
and col are papers
for (i in 1:length(xaulist))
{
for (j in 1:dim(tata)[1])
{
ifelse('TRUE' %in% as.character(tata[j,]==xaulist[i]), tutu[i,j] - 1,
tutu[i,j] - 0)
}
}
tutu[is.na(tutu)] - 0
#-
I am looking at some more efficient way to build 'tutu'.
Thank you very much,
David
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[R] removing loops from code in making data.frame
Dear all,
I am working on an author network and to do so I have to arrange a data.frame
(tutu) crossing author names (rows) per publication number (column). The
participation of the author to a study is indicated by a 1 and 0 otherwise.
I have a vector (xaulist) of all the names of authors and a data.frame (tata)
with all the publications in row and the authors in columns. I have writen a
loop to obtain my data.frame but it takes a long time when the number of
studies increases. I was looking for a more efficient code.
Here is a minimal working example (my code is terrible i know...):
#-
au1 - c('deb', 'art', 'deb', 'seb', 'deb', 'deb', 'mar', 'mar', 'joy', 'deb')
au2 - c('art', 'deb', 'soy', 'deb', 'joy', 'ani', 'deb', 'deb', 'nem', 'mar')
au3 - c('mar', 'lio', 'mil', 'mar', 'ani', 'lul', 'nem', 'art', 'deb', 'tat')
tata - data.frame(au1, au2, au3)
xaulist2 - levels(factor(unlist(tata[,])))
xaulist - levels(as.factor(xaulist2))
tutu - matrix(NA, nrow=length(xaulist), ncol=dim(tata)[1]) # row are authors
and col are papers
for (i in 1:length(xaulist))
{
for (j in 1:dim(tata)[1])
{
ifelse('TRUE' %in% as.character(tata[j,]==xaulist[i]), tutu[i,j] - 1,
tutu[i,j] - 0)
}
}
tutu[is.na(tutu)] - 0
#-
I am looking at some more efficient way to build 'tutu'.
Thank you very much,
David
[[alternative HTML version deleted]]
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[R] extracting character values
Dear all,
I have a dataframe of names (netw), with each cell including last name and
initials of an author; some cells have NA. I would like to extract only the
last name from each cell; this new dataframe is calle 'res'
Here is what I do:
res - data.frame(matrix(NA, nrow=dim(netw)[1], ncol=dim(netw)[2]))
for (i in 1:x)
{
wh - regexpr('[a-z]{3,}', as.character(netw[,i]))
res[i] - substring(as.character(netw[,i]), wh, wh + attr(wh,'match.length')-1)
}
the problem is that I cannot manage to extract 'complex' names properly such as
' van der hoops bf ': here I only get 'van', the real last name is 'van der
hoops' and 'bf' are the initials. Basically the last name has always a minimum
of 3 consecutive letters, but may have 3 or more letters separated by one or
more space; the cell may start by a space too; initials never have more than 2
letters.
Someone would have a nice idea for that? Thanks,
David
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] count combined occurrences of categories
OK thanks for the tips. I have abandonned the use of cbidn in dataframe. I've
used the obth dcast() and melt() and they both work fine. Thanks again
David Biau
De : David Winsemius dwinsem...@comcast.net
À : arun smartpink...@yahoo.com
Cc : R help r-help@r-project.org; Biau David djmb...@yahoo.fr
Envoyé le : Vendredi 11 janvier 2013 18h54
Objet : Re: [R] count combined occurrences of categories
On Jan 11, 2013, at 9:47 AM, arun wrote:
HI David,
I get different results with dcast()
library(reshape2)
dcast(melt(tutu,nam),nam~value,length)
# nam art deb joy mar seb lio nem tat
#1 da 2 3 1 4 1 1 0 0
#2 fr 2 2 2 3 0 1 1 1
#3 ya 1 2 1 0 0 1 1 0
tutus - data.frame(nam=tutu$nam, au=with(tutu, c(au1,au2,au3)))
with(tutus,table(nam,au))
# au
#nam 1 2 3 4 5 6 7
# da 2 3 1 2 4 0 0 #some numbers don't match the previous result
#fr 2 2 2 2 2 1 1
#ya 1 2 1 1 0 1 0
#If I convert to as.character(), it matched with the dcast() results
Probably due to the fact I used c() on factors:
tutu - data.frame(nam, au1, au2, au3, stringsAsFactors=FALSE)
tutus - data.frame(nam=tutu$nam, au=with(tutu, c(au1,au2,au3)))
tutab - with(tutus, table(nam, au) )
tutab
au
nam art deb joy lio mar nem seb tat
da 2 3 1 1 4 0 1 0
fr 2 2 2 1 3 1 0 1
ya 1 2 1 1 0 1 0 0
-- David.
tutunew-data.frame(nam=tutu$nam,au=with(tutu,c(as.character(au1),as.character(au2),as.character(au3
with(tutunew,table(nam,au))
# au
#nam art deb joy lio mar nem seb tat
# da 2 3 1 1 4 0 1 0
#fr 2 2 2 1 3 1 0 1
#ya 1 2 1 1 0 1 0 0
A.K.
- Original Message -
From: David Winsemius dwinsem...@comcast.net
To: Biau David djmb...@yahoo.fr
Cc: r help list r-help@r-project.org
Sent: Friday, January 11, 2013 12:20 PM
Subject: Re: [R] count combined occurrences of categories
On Jan 11, 2013, at 2:54 AM, Biau David wrote:
Dear all,
i would like to count the number of times where I have combined occurrences
of the categories of 2 variables.
For instance, in the dataframe below, i would like to know how many times
each author (au1, au2, au3 represent the first, second, third author) is
associated with each of the category of the variable 'nam'. The position of
the author does not matter.
nam - c('da', 'ya', 'da', 'da', 'fr', 'fr', 'fr', 'da', 'ya', 'fr')
au1 - c('deb', 'art', 'deb', 'seb', 'deb', 'deb', 'mar', 'mar', 'joy',
'joy')
au2 - c('art', 'deb', 'mar', 'deb', 'joy', 'mar', 'art', 'lio', 'nem',
'mar')
au3 - c('mar', 'lio', 'joy', 'mar', 'art', 'lio', 'nem', 'art', 'deb',
'tat')
tutu - data.frame(cbind(nam, au1, au2, au3))
You should first abandon the practice of using `cbind` inside `data.frame`.
Obscure errors will plague your R experience until you do so.
Bas solution:
tutus - data.frame(nam=tutu$nam, au=with(tutu, c(au1,au2,au3)))
tutab - with(tutus, table(nam, au) )
tutab
au
nam 1 2 3 4 5 6 7
da 2 3 1 2 4 0 0
fr 2 2 2 2 2 1 1
ya 1 2 1 1 0 1 0
--
David Winsemius, MD
Alameda, CA, USA
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David Winsemius, MD
Alameda, CA, USA
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Re: [R] extracting character values
OK,
here is a minimal working example:
au1 - c('biau dj', 'jones kb', 'van den hoofs j', ' biau dj', 'biau dj',
'campagna r', 'biau dj', 'weiss kr', 'verdegaal sh', 'riad s')
au2 - c('weiss kr', 'ferguson pc', ' greidanus nv', ' porcher r', 'ferguson
pc', 'pessis e', 'leclerc p', 'biau dj', 'bovee jv', 'biau d')
au3 - c('bhumbra rs', 'lam b', 'garbuz ds', NA, 'chung p', ' biau dj', 'marmor
s', 'bhumbra r', 'pansuriya tc', NA)
netw - data.frame(au1, au2, au3)
res - data.frame(matrix(NA, nrow=dim(netw)[1], ncol=dim(netw)[2]))
for (i in 1:dim(netw)[2])
{
wh - regexpr('[a-z]{3,}', as.character(netw[,i]))
res[i] - substring(as.character(netw[,i]), wh, wh + attr(wh,'match.length')-1)
}
problem is for author van den hoofs j who is only retrieved as 'van'
thanks,
David Biau
De : arun smartpink...@yahoo.com
À : Biau David djmb...@yahoo.fr
Envoyé le : Dimanche 13 janvier 2013 17h38
Objet : Re: [R] extracting character values
HI,
res - data.frame(matrix(NA, nrow=dim(netw)[1], ncol=dim(netw)[2]))
#Error in matrix(NA, nrow = dim(netw)[1], ncol = dim(netw)[2]) :
# object 'netw' not found
Can you provide an example dataset of netw?
Thanks.
A.K.
- Original Message -
From: Biau David djmb...@yahoo.fr
To: r help list r-help@r-project.org
Cc:
Sent: Sunday, January 13, 2013 3:53 AM
Subject: [R] extracting character values
Dear all,
I have a dataframe of names (netw), with each cell including last name and
initials of an author; some cells have NA. I would like to extract only the
last name from each cell; this new dataframe is calle 'res'
Here is what I do:
res - data.frame(matrix(NA, nrow=dim(netw)[1], ncol=dim(netw)[2]))
for (i in 1:x)
{
wh - regexpr('[a-z]{3,}', as.character(netw[,i]))
res[i] - substring(as.character(netw[,i]), wh, wh + attr(wh,'match.length')-1)
}
the problem is that I cannot manage to extract 'complex' names properly such
as ' van der hoops bf ': here I only get 'van', the real last name is 'van
der hoops' and 'bf' are the initials. Basically the last name has always a
minimum of 3 consecutive letters, but may have 3 or more letters separated by
one or more space; the cell may start by a space too; initials never have more
than 2 letters.
Someone would have a nice idea for that? Thanks,
David
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] extracting character values
works great thanks. And you cut off my code a lot and removed the loop.
David Biau
De : Uwe Ligges lig...@statistik.tu-dortmund.de
À : Biau David djmb...@yahoo.fr
Cc : arun smartpink...@yahoo.com; r help list r-help@r-project.org
Envoyé le : Dimanche 13 janvier 2013 18h22
Objet : Re: [R] extracting character values
On 13.01.2013 18:02, Biau David wrote:
OK,
here is a minimal working example:
au1 - c('biau dj', 'jones kb', 'van den hoofs j', ' biau dj', 'biau dj',
'campagna r', 'biau dj', 'weiss kr', 'verdegaal sh', 'riad s')
au2 - c('weiss kr', 'ferguson pc', ' greidanus nv', ' porcher r', 'ferguson
pc', 'pessis e', 'leclerc p', 'biau dj', 'bovee jv', 'biau d')
au3 - c('bhumbra rs', 'lam b', 'garbuz ds', NA, 'chung p', ' biau dj',
'marmor s', 'bhumbra r', 'pansuriya tc', NA)
netw - data.frame(au1, au2, au3)
res - data.frame(matrix(NA, nrow=dim(netw)[1], ncol=dim(netw)[2]))
for (i in 1:dim(netw)[2])
{
wh - regexpr('[a-z]{3,}', as.character(netw[,i]))
res[i] - substring(as.character(netw[,i]), wh, wh +
attr(wh,'match.length')-1)
}
There may be an easier solution, but this should do:
res - data.frame(lapply(netw,
function(x)
gsub(^ *([[:alpha:] ]*) +[[:alpha:]]+$, \\1, x)))
Uwe Ligges
problem is for author van den hoofs j who is only retrieved as 'van'
thanks,
David Biau
De : arun smartpink...@yahoo.com
À : Biau David djmb...@yahoo.fr
Envoyé le : Dimanche 13 janvier 2013 17h38
Objet : Re: [R] extracting character values
HI,
res - data.frame(matrix(NA, nrow=dim(netw)[1], ncol=dim(netw)[2]))
#Error in matrix(NA, nrow = dim(netw)[1], ncol = dim(netw)[2]) :
# object 'netw' not found
Can you provide an example dataset of netw?
Thanks.
A.K.
- Original Message -
From: Biau David djmb...@yahoo.fr
To: r help list r-help@r-project.org
Cc:
Sent: Sunday, January 13, 2013 3:53 AM
Subject: [R] extracting character values
Dear all,
I have a dataframe of names (netw), with each cell including last name and
initials of an author; some cells have NA. I would like to extract only the
last name from each cell; this new dataframe is calle 'res'
Here is what I do:
res - data.frame(matrix(NA, nrow=dim(netw)[1], ncol=dim(netw)[2]))
for (i in 1:x)
{
wh - regexpr('[a-z]{3,}', as.character(netw[,i]))
res[i] - substring(as.character(netw[,i]), wh, wh +
attr(wh,'match.length')-1)
}
the problem is that I cannot manage to extract 'complex' names properly
such as ' van der hoops bf ': here I only get 'van', the real last name is
'van der hoops' and 'bf' are the initials. Basically the last name has
always a minimum of 3 consecutive letters, but may have 3 or more letters
separated by one or more space; the cell may start by a space too; initials
never have more than 2 letters.
Someone would have a nice idea for that? Thanks,
David
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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[[alternative HTML version deleted]]
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Re: [R] extracting character values
thanks too. It works also perfect. Not sure I understand all the code though:
will have to look into it!
David Biau
De : arun smartpink...@yahoo.com
À : Biau David djmb...@yahoo.fr
Cc : R help r-help@r-project.org; Uwe Ligges
lig...@statistik.tu-dortmund.de
Envoyé le : Dimanche 13 janvier 2013 18h36
Objet : Re: [R] extracting character values
Hi,
This should also work:
do.call(data.frame,lapply(netw,function(x) gsub(^ *(\\D+) \\w+$,\\1,x)))
A.K.
From: Biau David djmb...@yahoo.fr
To: arun smartpink...@yahoo.com; r help list r-help@r-project.org
Sent: Sunday, January 13, 2013 12:02 PM
Subject: Re: [R] extracting character values
OK,
here is a minimal working example:
au1 - c('biau dj', 'jones kb', 'van den hoofs j', ' biau dj', 'biau dj',
'campagna r', 'biau dj', 'weiss kr', 'verdegaal sh', 'riad s')
au2 - c('weiss kr', 'ferguson pc', ' greidanus nv', ' porcher r', 'ferguson
pc', 'pessis e', 'leclerc p', 'biau dj', 'bovee jv', 'biau d')
au3 - c('bhumbra rs', 'lam b', 'garbuz ds', NA, 'chung p', ' biau dj',
'marmor s', 'bhumbra r', 'pansuriya tc', NA)
netw - data.frame(au1, au2, au3)
res - data.frame(matrix(NA, nrow=dim(netw)[1], ncol=dim(netw)[2]))
for (i in 1:dim(netw)[2])
{
wh - regexpr('[a-z]{3,}', as.character(netw[,i]))
res[i] - substring(as.character(netw[,i]), wh, wh + attr(wh,'match.length')-1)
}
problem is for author van den hoofs j who is only retrieved as 'van'
thanks,
David Biau
De : arun smartpink...@yahoo.com
À : Biau David djmb...@yahoo.fr
Envoyé le : Dimanche 13 janvier 2013 17h38
Objet : Re: [R] extracting character values
HI,
res - data.frame(matrix(NA, nrow=dim(netw)[1], ncol=dim(netw)[2]))
#Error in matrix(NA, nrow = dim(netw)[1], ncol = dim(netw)[2]) :
# object 'netw' not found
Can you provide an example dataset of netw?
Thanks.
A.K.
- Original Message -
From: Biau David djmb...@yahoo.fr
To: r help list r-help@r-project.org
Cc:
Sent: Sunday, January 13, 2013 3:53 AM
Subject: [R] extracting character values
Dear all,
I have a dataframe of names (netw), with each cell including last name and
initials of an author; some cells have NA. I would like to extract only the
last name from each cell; this new dataframe is calle 'res'
Here is what I do:
res - data.frame(matrix(NA, nrow=dim(netw)[1], ncol=dim(netw)[2]))
for (i in 1:x)
{
wh - regexpr('[a-z]{3,}', as.character(netw[,i]))
res[i] - substring(as.character(netw[,i]), wh, wh +
attr(wh,'match.length')-1)
}
the problem is that I cannot manage to extract 'complex' names properly such
as ' van der hoops bf ': here I only get 'van', the real last name is
'van der hoops' and 'bf' are the initials. Basically the last name has always
a minimum of 3 consecutive letters, but may have 3 or more letters separated
by one or more space; the cell may start by a space too; initials never have
more than 2 letters.
Someone would have a nice idea for that? Thanks,
David
[[alternative HTML version deleted]]
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[R] count combined occurrences of categories
Dear all,
i would like to count the number of times where I have combined occurrences of
the categories of 2 variables.
For instance, in the dataframe below, i would like to know how many times each
author (au1, au2, au3 represent the first, second, third author) is associated
with each of the category of the variable 'nam'. The position of the author
does not matter.
nam - c('da', 'ya', 'da', 'da', 'fr', 'fr', 'fr', 'da', 'ya', 'fr')
au1 - c('deb', 'art', 'deb', 'seb', 'deb', 'deb', 'mar', 'mar', 'joy', 'joy')
au2 - c('art', 'deb', 'mar', 'deb', 'joy', 'mar', 'art', 'lio', 'nem', 'mar')
au3 - c('mar', 'lio', 'joy', 'mar', 'art', 'lio', 'nem', 'art', 'deb', 'tat')
tutu - data.frame(cbind(nam, au1, au2, au3))
thanks,
David
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[R] Re : interpretation of coefficients in survreg AND obtaining the hazard function
Dear Prof Therneau,
thank yo for this information: this is going to be most useful for what I want
to do. I will look into the ACF model.
Yours,
David Biau.
De : Terry Therneau thern...@mayo.edu
Cc : r-help@r-project.org
Envoyé le : Lun 15 novembre 2010, 15h 33min 23s
Objet : Re: interpretation of coefficients in survreg AND obtaining the hazard
function
1. The weibull is the only distribution that can be written in both a
proportional hazazrds for and an accelerated failure time form. Survreg
uses the latter.
In an ACF model, we model the time to failure. Positive coefficients
are good (longer time to death).
In a PH model, we model the death rate. Positive coefficients are
bad (higher death rate).
You are not the first to be confused by the change in sign between the
two models.
2. There are about 5 different ways to parameterize a Weibull
distribution, 1-4 appear in various texts and the acf form is #5. This
is a second common issue with survreg that strikes only the more
sophisticated users: to understand the output they look up the Weibull
in a textbook, and become even more confused!
Kalbfliesch and Prentice is a good reference for the acf form. The
manual page for psurvreg has some information on this, as does the very
end of ?survreg. The psurvreg page also has an example of how to
extract the hazard function for a Weibull fit.
Begin included message
Dear R help list,
I am modeling some survival data with coxph and survreg (dist='weibull')
using
package survival. I have 2 problems:
1) I do not understand how to interpret the regression coefficients in
the
survreg output and it is not clear, for me, from ?survreg.objects how
to.
Here is an example of the codes that points out my problem:
- data is stc1
- the factor is dichotomous with 'low' and 'high' categories
slr - Surv(stc1$ti_lr, stc1$ev_lr==1)
mca - coxph(slr~as.factor(grade2=='high'), data=stc1)
mcb - coxph(slr~as.factor(grade2), data=stc1)
mwa - survreg(slr~as.factor(grade2=='high'), data=stc1,
dist='weibull',
scale=0)
mwb - survreg(slr~as.factor(grade2), data=stc1, dist='weibull',
scale=0)
summary(mca)$coef
coef
exp(coef) se(coef) z Pr(|z|)
as.factor(grade2 == high)TRUE 0.2416562 1.273356 0.2456232
0.9838494 0.3251896
summary(mcb)$coef
coef exp(coef)
se(coef) z Pr(|z|)
as.factor(grade2)low -0.2416562 0.7853261 0.2456232
-0.9838494
0.3251896
summary(mwa)$coef
(Intercept) as.factor(grade2 == high)TRUE
7.9068380 -0.4035245
summary(mwb)$coef
(Intercept) as.factor(grade2)low
7.5033135 0.4035245
No problem with the interpretation of the coefs in the cox model.
However, i do
not understand why
a) the coefficients in the survreg model are the opposite (negative when
the
other is positive) of what I have in the cox model? are these not the
log(HR)
given the categories of these variable?
b) how come the intercept coefficient changes (the scale parameter does
not
change)?
2) My second question relates to the first.
a) given a model from survreg, say mwa above, how should i do to extract
the
base hazard and the hazard of each patient given a set of predictors?
With the
hazard function for the ith individual in the study given by h_i(t) =
exp(\beta'x_i)*\lambda*\gamma*t^{\gamma-1}, it doesn't look like to me
that
predict(mwa, type='linear') is \beta'x_i.
b) since I need the coefficient intercept from the model to obtain the
scale
parameter to obtain the base hazard function as defined in Collett
(h_0(t)=\lambda*\gamma*t^{\gamma-1}), I am concerned that this
coefficient
intercept changes depending on the reference level of the factor entered
in the
model. The change is very important when I have more than one predictor
in the
model.
Any help would be greatly appreciated,
David Biau.
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[R] Re : interpretation of coefficients in survreg AND obtaining the hazard function for an individual given a set of predictors
Dear Prof Lumley,
This is a very clear, precise, and useful answer to all my questions.
Thank you very much.
David Biau.
De : Thomas Lumley tlum...@uw.edu
Cc : r help list r-help@r-project.org
Envoyé le : Dim 14 novembre 2010, 23h 54min 23s
Objet : Re: [R] interpretation of coefficients in survreg AND obtaining the
hazard function for an individual given a set of predictors
Dear R help list,
I am modeling some survival data with coxph and survreg (dist='weibull') using
package survival. I have 2 problems:
1) I do not understand how to interpret the regression coefficients in the
survreg output and it is not clear, for me, from ?survreg.objects how to.
Here is an example of the codes that points out my problem:
- data is stc1
- the factor is dichotomous with 'low' and 'high' categories
slr - Surv(stc1$ti_lr, stc1$ev_lr==1)
mca - coxph(slr~as.factor(grade2=='high'), data=stc1)
mcb - coxph(slr~as.factor(grade2), data=stc1)
mwa - survreg(slr~as.factor(grade2=='high'), data=stc1, dist='weibull',
scale=0)
mwb - survreg(slr~as.factor(grade2), data=stc1, dist='weibull', scale=0)
summary(mca)$coef
coef
exp(coef) se(coef) z Pr(|z|)
as.factor(grade2 == high)TRUE 0.2416562 1.273356 0.2456232
0.9838494 0.3251896
summary(mcb)$coef
coef exp(coef)
se(coef) z Pr(|z|)
as.factor(grade2)low -0.2416562 0.7853261 0.2456232 -0.9838494
0.3251896
summary(mwa)$coef
(Intercept) as.factor(grade2 == high)TRUE
7.9068380 -0.4035245
summary(mwb)$coef
(Intercept) as.factor(grade2)low
7.5033135 0.4035245
No problem with the interpretation of the coefs in the cox model. However, i
do
not understand why
a) the coefficients in the survreg model are the opposite (negative when the
other is positive) of what I have in the cox model? are these not the log(HR)
given the categories of these variable?
No. survreg() fits accelerated failure models, not proportional
hazards models. The coefficients are logarithms of ratios of
survival times, so a positive coefficient means longer survival.
b) how come the intercept coefficient changes (the scale parameter does not
change)?
Because you have reversed the order of the factor levels. The
coefficient of that variable changes sign and the intercept changes to
compensate.
2) My second question relates to the first.
a) given a model from survreg, say mwa above, how should i do to extract the
base hazard and the hazard of each patient given a set of predictors? With the
hazard function for the ith individual in the study given by h_i(t) =
exp(\beta'x_i)*\lambda*\gamma*t^{\gamma-1}, it doesn't look like to me that
predict(mwa, type='linear') is \beta'x_i.
No, it's beta'x_i for the accelerated failure parametrization of the
Weibull. In terms of the CDF
F_i(t) = F_0( exp((t+beta'x_i)/scale) )
So you need to multiply by the scale parameter and change sign to get
the log hazard ratios.
b) since I need the coefficient intercept from the model to obtain the scale
parameter to obtain the base hazard function as defined in Collett
(h_0(t)=\lambda*\gamma*t^{\gamma-1}), I am concerned that this coefficient
intercept changes depending on the reference level of the factor entered in
the
model. The change is very important when I have more than one predictor in the
model.
As Terry Therneau pointed out recently in the context of the Cox
model, there is no such thing as the baseline hazard. The baseline
hazard is the hazard when all your covariates are equal to zero, and
this depends on how you parametrize. In mwa, zero is grade2=low, in
mwb, zero is grade2=high, so the hazard at zero has to be different
in the two cases.
-thomas
--
Thomas Lumley
Professor of Biostatistics
University of Auckland
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[R] interpretation of coefficients in survreg AND obtaining the hazard function for an individual given a set of predictors
Dear R help list,
I am modeling some survival data with coxph and survreg (dist='weibull') using
package survival. I have 2 problems:
1) I do not understand how to interpret the regression coefficients in the
survreg output and it is not clear, for me, from ?survreg.objects how to.
Here is an example of the codes that points out my problem:
- data is stc1
- the factor is dichotomous with 'low' and 'high' categories
slr - Surv(stc1$ti_lr, stc1$ev_lr==1)
mca - coxph(slr~as.factor(grade2=='high'), data=stc1)
mcb - coxph(slr~as.factor(grade2), data=stc1)
mwa - survreg(slr~as.factor(grade2=='high'), data=stc1, dist='weibull',
scale=0)
mwb - survreg(slr~as.factor(grade2), data=stc1, dist='weibull', scale=0)
summary(mca)$coef
coef
exp(coef) se(coef) z Pr(|z|)
as.factor(grade2 == high)TRUE 0.2416562 1.273356 0.2456232
0.9838494 0.3251896
summary(mcb)$coef
coef exp(coef)
se(coef) z Pr(|z|)
as.factor(grade2)low -0.2416562 0.7853261 0.2456232 -0.9838494
0.3251896
summary(mwa)$coef
(Intercept) as.factor(grade2 == high)TRUE
7.9068380 -0.4035245
summary(mwb)$coef
(Intercept) as.factor(grade2)low
7.5033135 0.4035245
No problem with the interpretation of the coefs in the cox model. However, i do
not understand why
a) the coefficients in the survreg model are the opposite (negative when the
other is positive) of what I have in the cox model? are these not the log(HR)
given the categories of these variable?
b) how come the intercept coefficient changes (the scale parameter does not
change)?
2) My second question relates to the first.
a) given a model from survreg, say mwa above, how should i do to extract the
base hazard and the hazard of each patient given a set of predictors? With the
hazard function for the ith individual in the study given by h_i(t) =
exp(\beta'x_i)*\lambda*\gamma*t^{\gamma-1}, it doesn't look like to me that
predict(mwa, type='linear') is \beta'x_i.
b) since I need the coefficient intercept from the model to obtain the scale
parameter to obtain the base hazard function as defined in Collett
(h_0(t)=\lambda*\gamma*t^{\gamma-1}), I am concerned that this coefficient
intercept changes depending on the reference level of the factor entered in the
model. The change is very important when I have more than one predictor in the
model.
Any help would be greatly appreciated,
David Biau.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] Re : interpretation of coefficients in survreg AND obtaining the hazard function for an individual given a set of predictors
Thank you David for your answer,
- grade2 is a factor with 2 categories: high and low
- yes as.factor is superfluous; it is just that it avoids warnings sometimes.
This can be overlooked.
- I will look into Terry Therneau answers; he gives a good explanation on how
to
obtain the hazard for an individual given a set of predictors for the Cox
model;
I will look to see if this works for survreg andlook into survreg.distributions
if it doesn't
- I'll come back if I can't figure it out.
Thanks again.
Best,
David Biau.
De : David Winsemius dwinsem...@comcast.net
Cc : r help list r-help@r-project.org
Envoyé le : Sam 13 novembre 2010, 19h 55min 10s
Objet : Re: [R] interpretation of coefficients in survreg AND obtaining the
hazard function for an individual given a set of predictors
On Nov 13, 2010, at 12:51 PM, Biau David wrote:
Dear R help list,
I am modeling some survival data with coxph and survreg (dist='weibull') using
package survival. I have 2 problems:
1) I do not understand how to interpret the regression coefficients in the
survreg output and it is not clear, for me, from ?survreg.objects how to.
Have you read:
?survreg.distributions # linked from survreg help
Here is an example of the codes that points out my problem:
- data is stc1
- the factor is dichotomous with 'low' and 'high' categories
Not an unambiguous description for the purposes of answering your many
questions. Please provide data or at the very least: str(stc1)
slr - Surv(stc1$ti_lr, stc1$ev_lr==1)
mca - coxph(slr~as.factor(grade2=='high'), data=stc1)
Not sure what that would be returning since we do not know the encoding of
grade2. If you want an estimate on a subset wouldn't you do the subsetting
outside of the formula? (You may be reversing the order by offering a logical
test for grade2.)
mcb - coxph(slr~as.factor(grade2), data=stc1)
You have not provided the data or str(stc1), so it is entirely possible that
as.factor is superfluous in this call.
mwa - survreg(slr~as.factor(grade2=='high'), data=stc1, dist='weibull',
scale=0)
mwb - survreg(slr~as.factor(grade2), data=stc1, dist='weibull', scale=0)
summary(mca)$coef
coef
exp(coef) se(coef) z Pr(|z|)
as.factor(grade2 == high)TRUE 0.2416562 1.273356 0.2456232
0.9838494 0.3251896
summary(mcb)$coef
coef exp(coef)
se(coef) z Pr(|z|)
as.factor(grade2)low -0.2416562 0.7853261 0.2456232 -0.9838494
0.3251896
summary(mwa)$coef
(Intercept) as.factor(grade2 == high)TRUE
7.9068380 -0.4035245
summary(mwb)$coef
(Intercept) as.factor(grade2)low
7.5033135 0.4035245
No problem with the interpretation of the coefs in the cox model. However, i
do
not understand why
a) the coefficients in the survreg model are the opposite (negative when the
other is positive) of what I have in the cox model? are these not the log(HR)
given the categories of these variable?
Probably because the order of the factor got reversed when you changed the
covariate to logical and them back to factor.
b) how come the intercept coefficient changes (the scale parameter does not
change)?
2) My second question relates to the first.
a) given a model from survreg, say mwa above, how should i do to extract the
base hazard
Answered by Therneau earlier this week and the next question last month:
https://stat.ethz.ch/pipermail/r-help/2010-November/259570.html
https://stat.ethz.ch/pipermail/r-help/2010-October/257941.html
and the hazard of each patient given a set of predictors? With the
hazard function for the ith individual in the study given by h_i(t) =
exp(\beta'x_i)*\lambda*\gamma*t^{\gamma-1}, it doesn't look like to me that
predict(mwa, type='linear') is \beta'x_i.
b) since I need the coefficient intercept from the model to obtain the scale
parameter to obtain the base hazard function as defined in Collett
(h_0(t)=\lambda*\gamma*t^{\gamma-1}), I am concerned that this coefficient
intercept changes depending on the reference level of the factor entered in
the
model. The change is very important when I have more than one predictor in the
model.
Any help would be greatly appreciated,
David Biau.
David Winsemius, MD
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
[R] How to compare the effect of a variable across regression models?
Hello, I would like, if it is possible, to compare the effect of a variable across regression models. I have looked around but I haven't found anything. Maybe someone could help? Here is the problem: I am studying the effect of a variable (age) on an outcome (local recurrence: lr). I have built 3 models: - model 1: lr ~ age y = \beta_(a1).age - model 2: lr ~ age + presentation variables (X_p)y = \beta_(a2).age + \BETA_(p2).X_p - model 3: lr ~ age + presentation variables + treatment variables( X_t) y = \beta_(a3).age + \BETA_(p3).X_(p) + \BETA_(t3).X_t Presentation variables include variables such as tumor grade, tumor size, etc... the physician cannot interfer with these variables. Treatment variables include variables such as chemotherapy, radiation, surgical margins (a surrogate for adequate surgery). I have used cph for the models and restricted cubic splines (Design library) for age. I have noted that the effect of age decreases from model 1 to 3. I would like to compare the effect of age on the outcome across the different models. A test of \beta_(a1) = \beta_(a2) = \beta_(a3) and then two by two comparisons or a global trend test maybe? Is that possible? Thank you for your help, David Biau. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Re : How to compare the effect of a variable across regression models?
OK,
thank you very much for the answer.I will look into that. Hopefully I'll find
smoething that will work out.
Best,
David Biau.
De : Frank Harrell f.harr...@vanderbilt.edu
Cc : r help list r-help@r-project.org
Envoyé le : Ven 13 août 2010, 15h 50min 18s
Objet : Re: [R] How to compare the effect of a variable across regression
models?
David,
In the Cox and many other regression models, the effect of a variable is
context-dependent. There is an identifiability problem in what you are doing,
as discussed by
@ARTICLE{for95mod,
author = {Ford, Ian and Norrie, John and Ahmadi, Susan},
year = 1995,
title = {Model inconsistency, illustrated by the {Cox} proportional hazards
model},
journal = Stat in Med,
volume = 14,
pages = {735-746},
annote = {covariable adjustment; adjusted estimates; baseline imbalances;
RCT; model misspecification; model identification}
}
One possible remedy, which may not work for your goals, is to embed all models
in a grand model that is used for inference.
When coefficients ARE comparable in some sense, you can use the bootstrap to
get
confidence bands for differences in regressor effects between models.
Frank
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
On Fri, 13 Aug 2010, Biau David wrote:
Hello,
I would like, if it is possible, to compare the effect of a variable across
regression models. I have looked around but I haven't found anything. Maybe
someone could help? Here is the problem:
I am studying the effect of a variable (age) on an outcome (local recurrence:
lr). I have built 3 models:
- model 1: lr ~ age y = \beta_(a1).age
- model 2: lr ~ age + presentation variables (X_p)y = \beta_(a2).age
+
\BETA_(p2).X_p
- model 3: lr ~ age + presentation variables + treatment variables( X_t)
y = \beta_(a3).age + \BETA_(p3).X_(p) + \BETA_(t3).X_t
Presentation variables include variables such as tumor grade, tumor size,
etc...
the physician cannot interfer with these variables.
Treatment variables include variables such as chemotherapy, radiation,
surgical
margins (a surrogate for adequate surgery).
I have used cph for the models and restricted cubic splines (Design library)
for
age. I have noted that the effect of age decreases from model 1 to 3.
I would like to compare the effect of age on the outcome across the different
models. A test of \beta_(a1) = \beta_(a2) = \beta_(a3) and then two by two
comparisons or a global trend test maybe? Is that possible?
Thank you for your help,
David Biau.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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[R] Re : Re : How to compare the effect of a variable across regression models?
} proportional hazards
model},
journal = Stat in Med,
volume = 14,
pages = {735-746},
annote = {covariable adjustment; adjusted estimates; baseline imbalances;
RCT; model misspecification; model identification}
}
One possible remedy, which may not work for your goals, is to embed all models
in a grand model that is used for inference.
When coefficients ARE comparable in some sense, you can use the bootstrap to
get
confidence bands for differences in regressor effects between models.
Frank
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
On Fri, 13 Aug 2010, Biau David wrote:
Hello,
I would like, if it is possible, to compare the effect of a variable across
regression models. I have looked around but I haven't found anything. Maybe
someone could help? Here is the problem:
I am studying the effect of a variable (age) on an outcome (local recurrence:
lr). I have built 3 models:
- model 1: lr ~ age y = \beta_(a1).age
- model 2: lr ~ age + presentation variables (X_p)y = \beta_(a2).age
+
\BETA_(p2).X_p
- model 3: lr ~ age + presentation variables + treatment variables( X_t)
y = \beta_(a3).age + \BETA_(p3).X_(p) + \BETA_(t3).X_t
Presentation variables include variables such as tumor grade, tumor size,
etc...
the physician cannot interfer with these variables.
Treatment variables include variables such as chemotherapy, radiation,
surgical
margins (a surrogate for adequate surgery).
I have used cph for the models and restricted cubic splines (Design library)
for
age. I have noted that the effect of age decreases from model 1 to 3.
I would like to compare the effect of age on the outcome across the different
models. A test of \beta_(a1) = \beta_(a2) = \beta_(a3) and then two by two
comparisons or a global trend test maybe? Is that possible?
Thank you for your help,
David Biau.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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[R] How to extract se(coef) from cph?
Hello, I am modeling some survival data wih cph (Design). I have modeled a predictor which showed non linear effect with restricted cubic splines. I would like to retrieve the se(coef) for other, linear, predictors. This is just to make nice LateX tables automatically. I have the coefficients with coef(). How do I do that? Thanks, David Biau. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Re : How to extract se(coef) from cph?
Excellent! Yes, FH has a function to get LateX tables, but I not malleable enough. Thanks, David Biau. De : David Winsemius dwinsem...@comcast.net Cc : r help list r-help@r-project.org Envoyé le : Jeu 5 août 2010, 22h 11min 20s Objet : Re: [R] How to extract se(coef) from cph? On Aug 5, 2010, at 4:03 PM, Biau David wrote: Hello, I am modeling some survival data wih cph (Design). I have modeled a predictor which showed non linear effect with restricted cubic splines. I would like to retrieve the se(coef) for other, linear, predictors. The cph object has a var. The vcov function is an extractor function. You would probably be using something like: diag(vcov(fit))^(1/2) This is just to make nice LateX tables automatically. Are you sure Frank has not already programed that for you somewhere? Perhaps latex.cph? I have the coefficients with coef(). How do I do that? Thanks, David Biau. -- David Winsemius, MD West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Re : How to extract se(coef) from cph?
thanks, it works just great. David Biau. De : Abhijit Dasgupta, PhD aikidasgu...@gmail.com Cc : r help list r-help@r-project.org Envoyé le : Jeu 5 août 2010, 22h 15min 37s Objet : Re: [R] How to extract se(coef) from cph? if the cph model fit is m1, you can try sqrt(diag(m1$var)) This is coded in print.cph.fit (library(rms)) On 08/05/2010 04:03 PM, Biau David wrote: Hello, I am modeling some survival data wih cph (Design). I have modeled a predictor which showed non linear effect with restricted cubic splines. I would like to retrieve the se(coef) for other, linear, predictors. This is just to make nice LateX tables automatically. I have the coefficients with coef(). How do I do that? Thanks, David Biau. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Abhijit Dasgupta, PhD Director and Principal Statistician ARAASTAT Ph: 301.385.3067 E: adasgu...@araastat.com W: http://www.araastat.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] COXPH: how to get the score test and likelihood ratio test for a specific variable in a multivariate Coxph ?
Hello, I would like to get the likelihood ratio and score tests for specific variables in a multivariate coxph model. The default is Wald, so the tests for each separate variable is based on Wald's test. I have the other tests for the full model but I don't know how to get them for each variable. Any idea? David Biau. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Re : COXPH: how to get the score test and likelihood ratio test for a specific variable in a multivariate Coxph ?
Thx for the answer. I am using survival. I didn't know that the Wald and score tests were the same for individual variables in a coxph; I Thought the score test was the multivariate version of the Log-rank. However, say I have only one variable in the model, I should expect the test for the full model and the one for a single variable to be the same? Then it seems to me that the default test is the Wald and that the Wald and the Score are different. cox_lr_age - coxph(Surv(tilr, ev_lr==1)~age, data=tam) summary(cox_lr_age) Call: coxph(formula = Surv(tilr, ev_lr == 1) ~ age, data = tam) n=2156 (76 observations deleted due to missingness) coef exp(coef) se(coef) z Pr(|z|) age 0.019504 1.019696 0.004651 4.193 2.75e-05 *** --- Signif. codes: 0 â***â 0.001 â**â 0.01 â*â 0.05 â.â 0.1 â â 1 exp(coef) exp(-coef) lower .95 upper .95 age 1.020 0.9807 1.010 1.029 Rsquare= 0.008 (max possible= 0.669 ) Likelihood ratio test= 18.4 on 1 df, p=1.787e-05 Wald test= 17.58 on 1 df, p=2.751e-05 Score (logrank) test = 17.86 on 1 df, p=2.375e-05 David Biau. De : David Winsemius dwinsem...@comcast.net Cc : r help list r-help@r-project.org Envoyé le : Ven 30 juillet 2010, 17h 34min 28s Objet : Re: [R] COXPH: how to get the score test and likelihood ratio test for a specific variable in a multivariate Coxph ? On Jul 30, 2010, at 11:08 AM, Biau David wrote: Hello, I would like to get the likelihood ratio and score tests for specific variables in a multivariate coxph model. The default is Wald, so the tests for each separate variable is based on Wald's test. I have the other tests for the full model but I don't know how to get them for each variable. Any idea? The first idea would be to specify which function in which package you are asking questions about. In the case of coxph in the survival package, for instance, you do get a likelihood ratio test (== differences in log-likelihoods) by default. A score test is, at least as as I understand it for individual variables, equivalent to a Wald test, so I don't really understand your question, since youa re already getting all of that in the survival package. (You can extract a score value and loglik values from a coxph object by: (with the first example in the coxph help page) coxph(Surv(time, status) ~ x + strata(sex), test1)$score xoxph(Surv(time, status) ~ x + strata(sex), test1)$loglik But anova(coxph-object) would give you these values in a neater bundle. #Analysis of Deviance Table # Cox model: response is Surv(time, status) #Terms added sequentially (first to last) # loglik Chisq Df Pr(|Chi|) # NULL -3.8712 # x-3.3277 1.0871 1 0.2971 The question about getting them for each variable does not make a lot of sense to me, since likelihood tests are model comparisons. You can only make such statements about the consequences of adding or deleting a variable to/from an existing model. --David Winsemius, MD West Hartford, CT [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Re : Re : COXPH: how to get the score test and likelihood ratio test for a specific variable in a multivariate Coxph ?
Well thank you very much for these explanations. Unfortunately, I must admit the book I have for survival analysis seems less precise as to which test to use and why. Still, in coxph (survival), if I have multiple variables in a model, say X_1, X_2, and X_3, how do I test their respective coefficients \beta_1, \beta_2, and \beta_3 with the LR, score and Wald? I guess i can do it by comparing the model with all three variables to those without each of the variables, but is there not a more straightforward manner? David Biau. De : Therneau, Terry M., Ph.D. thern...@mayo.edu À : David Winsemius dwinsem...@comcast.net; Biau David djmb...@yahoo.fr Cc : r help list r-help@r-project.org Envoyé le : Ven 30 juillet 2010, 19h 07min 15s Objet : RE: Re : [R] COXPH: how to get the score test and likelihood ratio test for a specific variable in a multivariate Coxph ? The Wald, score, and LR tests are discussed in full in my book. They are not the same. The LR test is the difference between LR(beta=0) and LR(beta=final). The score test is a Taylor series approximation to this using an expansion around beta=0. The Wald test is a similar Taylor series approximation, but around beta=final. If there are no tied times the score test = Log-rank test. If there are ties, then they are just a tiny bit different: the paper using the log-rank has an n-1 in his variance term and the Cox model has an n. Neither is right or wrong, just a different choice. Terry Therneau [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] longitudinal tobit regression in R
Hi, I am trying to model a score over time. This score shows a ceiling effect. I was willing to use a longitudinal tobit model, such as the one described by Twisk et al. (Twisk_Longitudinal tobit regression: A new approach to analyze outcome variables with floor or ceiling effects_JCE_2009) but it is programmed for STATA. Has anyone used such models in R? Any other idea? David Biau. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] crosstabling multiple variables at once
Hi, I am trying to describe a data.frame by obtaining multiple crosstable summary statistics at once. I have tried table, xtab, crosstable, summaryBy and describe but none of these functions seems to allow muliple conparisons at once. Here, is what I would like to do: I have, for instance, age, sex (M and F), grade (1, 2, 3) and site (limb, trunk) and I want the, for instance, following summary statistics: - age (mean, SD) for males and age for females - age for grade 1, grade2, and grade 3 - age for site limb, site trunk - sex (count, proportions) for grade 1, grade2, and grade 3 - sex (count, proportions) site limb, site trunk (already have sex/age above) - grade (count, proportions) for site limb, site trunk (already have grade/sex and grade/age above) a lso, I want each of these not crossed by any others (mean overall age, numbers of males, etc) which could be seen as each crossed with its own. I have at least 10 variables, continuous, categorical ordered and non ordered. I don't want any tests. Any idea? David Biau. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] multiple 2 by 2 crosstabulations?
Hello, I have a dataframe (var_1, var_2, ..., var_n) and I would like to export summary statistics to Latex in the form of a table. I want specific summary statistics by crossing numerous variables 2x2 AT ONCE. In each cell I would like sometimes to have the median (Q1 - Q3), or frequency and proportion, etc. CrossTable, xtab, etc... do not allow for multiple 2 by 2 crosstabulation. The table would look like this: var_1 var_2 var3, ... var_1 a b c var_2 d e f var_3 .. ... ... with a, b, c, ... the results of each crosstabulation. I have continuous and categorical variables. Any idea? Thank you very much, David. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
