[R] Replace missing values in lapply
I have some matrices stored as elements in a list that I am working with. On example is provided below as TP[[18]] TP[[18]] level2 level1 1 2 3 4 1 79 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 Now, using prop.table on this gives prop.table(TP[[18]],1) level2 level1 1 2 3 4 1 1 0 0 0 2 3 4 It is important for the zero's to retain their position as this matrix will subsequently be used in some matrix multiplication and hence, must be of dimension 4 by 4 so that is it conformable for multiplcation with another matrix. In looking at the structure of the object resulting from prop.table I see NaNs, and so I can do this rr - TP[[18]] rr[is.na(rr)] - 0 rr level2 level1 1 2 3 4 1 79 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 This is exactly what I want for each matrix. But, I have multiple matrices stored within the list that need to be changed and so I am trying to resolve this via lapply, but something is awry (namely the user), but I could use a little help. I was thinking the following function should work, but it doesn't. It reduces each matrix within the list to a 0. PP - lapply(TP, function(x) x[is.na(x)] - 0) Am I missing something obvious? Harold [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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.
Re: [R] Replace missing values in lapply
you need to return x in the function within lapply(), e.g., something
like
lapply(TP, function(x) { x[is.na(x)] - 0; x })
I hope it works.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Doran, Harold [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Wednesday, January 24, 2007 4:40 PM
Subject: [R] Replace missing values in lapply
I have some matrices stored as elements in a list that I am working
with. On example is provided below as TP[[18]]
TP[[18]]
level2
level1 1 2 3 4
1 79 0 0 0
2 0 0 0 0
3 0 0 0 0
4 0 0 0 0
Now, using prop.table on this gives
prop.table(TP[[18]],1)
level2
level1 1 2 3 4
1 1 0 0 0
2
3
4
It is important for the zero's to retain their position as this
matrix
will subsequently be used in some matrix multiplication and hence,
must
be of dimension 4 by 4 so that is it conformable for multiplcation
with
another matrix.
In looking at the structure of the object resulting from prop.table
I
see NaNs, and so I can do this
rr - TP[[18]]
rr[is.na(rr)] - 0
rr
level2
level1 1 2 3 4
1 79 0 0 0
2 0 0 0 0
3 0 0 0 0
4 0 0 0 0
This is exactly what I want for each matrix. But, I have multiple
matrices stored within the list that need to be changed and so I am
trying to resolve this via lapply, but something is awry (namely the
user), but I could use a little help.
I was thinking the following function should work, but it doesn't.
It
reduces each matrix within the list to a 0.
PP - lapply(TP, function(x) x[is.na(x)] - 0)
Am I missing something obvious?
Harold
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Replace missing values in lapply
Perfect, thxs
-Original Message-
From: Dimitris Rizopoulos
[mailto:[EMAIL PROTECTED]
Sent: Wednesday, January 24, 2007 10:49 AM
To: Doran, Harold
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Replace missing values in lapply
you need to return x in the function within lapply(), e.g.,
something like
lapply(TP, function(x) { x[is.na(x)] - 0; x })
I hope it works.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Doran, Harold [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Wednesday, January 24, 2007 4:40 PM
Subject: [R] Replace missing values in lapply
I have some matrices stored as elements in a list that I am working
with. On example is provided below as TP[[18]]
TP[[18]]
level2
level1 1 2 3 4
1 79 0 0 0
2 0 0 0 0
3 0 0 0 0
4 0 0 0 0
Now, using prop.table on this gives
prop.table(TP[[18]],1)
level2
level1 1 2 3 4
1 1 0 0 0
2
3
4
It is important for the zero's to retain their position as this
matrix
will subsequently be used in some matrix multiplication and hence,
must
be of dimension 4 by 4 so that is it conformable for multiplcation
with
another matrix.
In looking at the structure of the object resulting from prop.table
I
see NaNs, and so I can do this
rr - TP[[18]]
rr[is.na(rr)] - 0
rr
level2
level1 1 2 3 4
1 79 0 0 0
2 0 0 0 0
3 0 0 0 0
4 0 0 0 0
This is exactly what I want for each matrix. But, I have multiple
matrices stored within the list that need to be changed and so I am
trying to resolve this via lapply, but something is awry (namely the
user), but I could use a little help.
I was thinking the following function should work, but it doesn't.
It
reduces each matrix within the list to a 0.
PP - lapply(TP, function(x) x[is.na(x)] - 0)
Am I missing something obvious?
Harold
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch 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.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
__
R-help@stat.math.ethz.ch 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.
Re: [R] Replace missing values in lapply
I wonder if a list of matrices is the best representation? Do your matrices all have the same dimension as in: TP - list(matrix(c(1:3, NA), 2), matrix(c(NA, 1:3), 2)) # Then you could consider representing them as an array: TPa - array(unlist(TP), c(2,2,2)) # in which case its just TPa[is.na(TPa)] - 0 TPa On 1/24/07, Doran, Harold [EMAIL PROTECTED] wrote: I have some matrices stored as elements in a list that I am working with. On example is provided below as TP[[18]] TP[[18]] level2 level1 1 2 3 4 1 79 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 Now, using prop.table on this gives prop.table(TP[[18]],1) level2 level1 1 2 3 4 1 1 0 0 0 2 3 4 It is important for the zero's to retain their position as this matrix will subsequently be used in some matrix multiplication and hence, must be of dimension 4 by 4 so that is it conformable for multiplcation with another matrix. In looking at the structure of the object resulting from prop.table I see NaNs, and so I can do this rr - TP[[18]] rr[is.na(rr)] - 0 rr level2 level1 1 2 3 4 1 79 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 This is exactly what I want for each matrix. But, I have multiple matrices stored within the list that need to be changed and so I am trying to resolve this via lapply, but something is awry (namely the user), but I could use a little help. I was thinking the following function should work, but it doesn't. It reduces each matrix within the list to a 0. PP - lapply(TP, function(x) x[is.na(x)] - 0) Am I missing something obvious? Harold [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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-help@stat.math.ethz.ch 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.
Re: [R] Replace missing values in lapply
I hadn't thought of that. I use the following at one point in my program tmp - with(data, tapply(variable, index, table)) Which returns a list. So, I just went with it for the rest of my program. I'm changing code now to arrays, I think you're right and this may be a better representation. I need to walk through this and see what turns up. Thanks for the recommendation. -Original Message- From: Gabor Grothendieck [mailto:[EMAIL PROTECTED] Sent: Wednesday, January 24, 2007 11:06 AM To: Doran, Harold Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Replace missing values in lapply I wonder if a list of matrices is the best representation? Do your matrices all have the same dimension as in: TP - list(matrix(c(1:3, NA), 2), matrix(c(NA, 1:3), 2)) # Then you could consider representing them as an array: TPa - array(unlist(TP), c(2,2,2)) # in which case its just TPa[is.na(TPa)] - 0 TPa On 1/24/07, Doran, Harold [EMAIL PROTECTED] wrote: I have some matrices stored as elements in a list that I am working with. On example is provided below as TP[[18]] TP[[18]] level2 level1 1 2 3 4 1 79 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 Now, using prop.table on this gives prop.table(TP[[18]],1) level2 level1 1 2 3 4 1 1 0 0 0 2 3 4 It is important for the zero's to retain their position as this matrix will subsequently be used in some matrix multiplication and hence, must be of dimension 4 by 4 so that is it conformable for multiplcation with another matrix. In looking at the structure of the object resulting from prop.table I see NaNs, and so I can do this rr - TP[[18]] rr[is.na(rr)] - 0 rr level2 level1 1 2 3 4 1 79 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 This is exactly what I want for each matrix. But, I have multiple matrices stored within the list that need to be changed and so I am trying to resolve this via lapply, but something is awry (namely the user), but I could use a little help. I was thinking the following function should work, but it doesn't. It reduces each matrix within the list to a 0. PP - lapply(TP, function(x) x[is.na(x)] - 0) Am I missing something obvious? Harold [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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-help@stat.math.ethz.ch 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.
