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Sashi Challa

2010-Dec-09 17:44 UTC

[R] lapply getting names of the list

Hello All,

I have a toy dataframe like this. It has 8 columns separated by tab.

Name    SampleID        Al1     Al2     X       Y       R       Th
rs191191        A1      A       B       0.999   0.09    0.78    0.090
abc928291       A1      B       J       0.3838  0.3839  0.028   0.888
abcnab  A1      H       K       0.3939  0.939   0.3939  0.77
rx82922 B1      J       K       0.3838  0.393   0.393   0.00
rcn3939 B1      M       O       0.000   0.000   0.000   0.77
tcn39399        B1      P       I       0.393   0.393   0.393   0.56

Note that the SampleID is repeating. So I want to be able to split the dataset
based on the SampleID and write the splitted dataset of every SampleID into a
new file.
I tried split followed by lapply to do this.

infile <- read.csv("test.txt", sep="\t", as.is = TRUE,
header = TRUE)
infile.split  <- split(infile, infile$SampleID)
names(infile.split[1])  ## outputs “A1”
## now A1, B1 are two lists in infile.split as I understand it. Correct me if I
am wrong.

lapply(infile.split,function(x){
              filename <- names(x) #### here I expect to see A1 or B1, I
didn’t, I tried (names(x)[1]) and that gave me “Name” and not A1 or B1.
              final_filename <- paste(filename,”toy_set.txt”,sep=”_”)
              write.table(x, file = paste(path, final_filename,sep=”/”,
row.names=FALSE, quote=FALSE,sep=”\t”)
  } )

In lapply I wanted to give a unique filename to all the split Sample Ids, i.e.
name them here as A1_toy_set.txt, B1_toy_set_txt.
How do I get those names, i.e. A1, B1 to a create a filename like above.
When I write each of the element in the list obtained after split into a file,
the column names would have names like A1.Name, A1.SampleID, A1.Al1, ….. Can I
get rid of “A1” in the column names within the lapply (other than reading in the
file again and changing the names) ?

Thanks for your time,

Regards
Sashi


	[[alternative HTML version deleted]]

Joshua Wiley

2010-Dec-09 18:06 UTC

head link

[R] lapply getting names of the list

Hi Sashi,

On Thu, Dec 9, 2010 at 9:44 AM, Sashi Challa <challa at ohsu.edu>
wrote:> Hello All,
>
> I have a toy dataframe like this. It has 8 columns separated by tab.
>
> Name ? ?SampleID ? ? ? ?Al1 ? ? Al2 ? ? X ? ? ? Y ? ? ? R ? ? ? Th
> rs191191 ? ? ? ?A1 ? ? ?A ? ? ? B ? ? ? 0.999 ? 0.09 ? ?0.78 ? ?0.090
> abc928291 ? ? ? A1 ? ? ?B ? ? ? J ? ? ? 0.3838 ?0.3839 ?0.028 ? 0.888
> abcnab ?A1 ? ? ?H ? ? ? K ? ? ? 0.3939 ?0.939 ? 0.3939 ?0.77
> rx82922 B1 ? ? ?J ? ? ? K ? ? ? 0.3838 ?0.393 ? 0.393 ? 0.00
> rcn3939 B1 ? ? ?M ? ? ? O ? ? ? 0.000 ? 0.000 ? 0.000 ? 0.77
> tcn39399 ? ? ? ?B1 ? ? ?P ? ? ? I ? ? ? 0.393 ? 0.393 ? 0.393 ? 0.56
>
> Note that the SampleID is repeating. So I want to be able to split the
dataset based on the SampleID and write the splitted dataset of every SampleID
into a new file.
> I tried split followed by lapply to do this.
>
> infile <- read.csv("test.txt", sep="\t", as.is =
TRUE, header = TRUE)
> infile.split ?<- split(infile, infile$SampleID)
> names(infile.split[1]) ?## outputs ?A1?
correct, names() returns the top level names of infile.split (i.e.,
the two data frame names)
> ## now A1, B1 are two lists in infile.split as I understand it. Correct me
if I am wrong.
It is a single, named list containing two data frames (A1 and B1)
(though data frames are built from lists, I think so I suppose in a
way it contains two lists, but that is not really the point).
>
> lapply(infile.split,function(x){
> ? ? ? ? ? ? ?filename <- names(x) #### here I expect to see A1 or B1, I
didn?t, I tried (names(x)[1]) and that gave me ?Name? and not A1 or B1.
by using lapply() on the actual object, your function is getting each
element of the list.  That is:

infile.split[[1]]
infile.split[[2]]

trying names() on those:

names(infile.split[[1]])

should show what you are getting
> ? ? ? ? ? ? ?final_filename <- paste(filename,?toy_set.txt?,sep=?_?)
> ? ? ? ? ? ? ?write.table(x, file = paste(path, final_filename,sep=?/?,
row.names=FALSE, quote=FALSE,sep=?\t?)
FYI I think you are missing a parenthesis in there
somewhere> ?} )
>
> In lapply I wanted to give a unique filename to all the split Sample Ids,
i.e. name them here as A1_toy_set.txt, B1_toy_set_txt.
> How do I get those names, i.e. A1, B1 to a create a filename like above.
Try this:

## read your data from the clipboard (obviously you do not need to)
infile <- read.table("clipboard", header = TRUE)
split.infile <- split(dat, dat$SampleID) #split data
path <- "~" # generic path

## rather than applying to the data itself, instead apply to the names
lapply(names(split.infile), function(x) {
  write.table(x = split.infile[[x]],
    file = paste(path, paste(x, "toy_set.txt", sep = "_"),
sep = "/"),
    row.names = FALSE, quote = FALSE, sep = "\t")
  cat("wrote ", x, fill = TRUE)
})

it will return two NULL lists, but that is fine because it should have
written the files.
> When I write each of the element in the list obtained after split into a
file, the column names would have names like A1.Name, A1.SampleID, A1.Al1, ?..
Can I get rid of ?A1? in the column names within the lapply (other than reading
in the file again and changing the names) ?
Can you report the results of str(yourdataframe) ?  I did not have
that issue just copying and pasting from your email and using the code
I showed above.

Cheers,

Josh
>
> Thanks for your time,
>
> Regards
> Sashi
>
>
> ? ? ? ?[[alternative HTML version deleted]]
>
>
> ______________________________________________
> R-help at 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.


-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

David Winsemius

2010-Dec-09 19:21 UTC

head link

[R] lapply getting names of the list

On Dec 9, 2010, at 12:44 PM, Sashi Challa wrote:
> Hello All,
>
> I have a toy dataframe like this. It has 8 columns separated by tab.
>
> Name    SampleID        Al1     Al2     X       Y       R       Th
> rs191191        A1      A       B       0.999   0.09    0.78    0.090
> abc928291       A1      B       J       0.3838  0.3839  0.028   0.888
> abcnab  A1      H       K       0.3939  0.939   0.3939  0.77
> rx82922 B1      J       K       0.3838  0.393   0.393   0.00
> rcn3939 B1      M       O       0.000   0.000   0.000   0.77
> tcn39399        B1      P       I       0.393   0.393   0.393   0.56
>
> Note that the SampleID is repeating. So I want to be able to split  
> the dataset based on the SampleID and write the splitted dataset of  
> every SampleID into a new file.
> I tried split followed by lapply to do this.
>
> infile <- read.csv("test.txt", sep="\t", as.is =
TRUE, header = TRUE)
> infile.split  <- split(infile, infile$SampleID)
> names(infile.split[1])  ## outputs ?A1?
> ## now A1, B1 are two lists in infile.split as I understand it.  
> Correct me if I am wrong.
>
> lapply(infile.split,function(x){
>              filename <- names(x) #### here I expect to see A1 or  
> B1, I didn?t, I tried (names(x)[1]) and that gave me ?Name? and not  
> A1 or B1.
>              final_filename <- paste(filename,?toy_set.txt?,sep=?_?)
>              write.table(x, file = paste(path,  
> final_filename,sep=?/?, row.names=FALSE, quote=FALSE,sep=?\t?)
>  } )
>
> In lapply I wanted to give a unique filename to all the split Sample  
> Ids, i.e. name them here as <dragged to the c() construct>.
> How do I get those names, i.e. A1, B1 to a create a filename like  
> above.
names(file.split) <- c("A1_toy_set.txt",
"B1_toy_set_txt")
> When I write each of the element in the list obtained after split  
> into a file,
How are you proposing do do this "writing"?
> the column names would have names like A1.Name, A1.SampleID, A1.Al1,  
> ?..
Are you sure? Why would you think that?

-- 
David.
> Can I get rid of ?A1? in the column names within the lapply (other  
> than reading in the file again and changing the names) ?
>
> Thanks for your time,
>
> Regards
> Sashi
>
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at 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.
David Winsemius, MD
West Hartford, CT

Sashi Challa

2010-Dec-09 19:24 UTC

head link

[R] lapply getting names of the list

Thanks a lot Joshua, that works perfectly fine. 
I could not think to lapply on the names instead of data itself.
I don't now notice SampleID names in the column names.
Thanks for your time,

-Sashi

-----Original Message-----
From: Joshua Wiley [mailto:jwiley.psych at gmail.com] 
Sent: Thursday, December 09, 2010 10:07 AM
To: Sashi Challa
Cc: r-help at R-project.org
Subject: Re: [R] lapply getting names of the list

Hi Sashi,

On Thu, Dec 9, 2010 at 9:44 AM, Sashi Challa <challa at ohsu.edu>
wrote:> Hello All,
>
> I have a toy dataframe like this. It has 8 columns separated by tab.
>
> Name ? ?SampleID ? ? ? ?Al1 ? ? Al2 ? ? X ? ? ? Y ? ? ? R ? ? ? Th
> rs191191 ? ? ? ?A1 ? ? ?A ? ? ? B ? ? ? 0.999 ? 0.09 ? ?0.78 ? ?0.090
> abc928291 ? ? ? A1 ? ? ?B ? ? ? J ? ? ? 0.3838 ?0.3839 ?0.028 ? 0.888
> abcnab ?A1 ? ? ?H ? ? ? K ? ? ? 0.3939 ?0.939 ? 0.3939 ?0.77
> rx82922 B1 ? ? ?J ? ? ? K ? ? ? 0.3838 ?0.393 ? 0.393 ? 0.00
> rcn3939 B1 ? ? ?M ? ? ? O ? ? ? 0.000 ? 0.000 ? 0.000 ? 0.77
> tcn39399 ? ? ? ?B1 ? ? ?P ? ? ? I ? ? ? 0.393 ? 0.393 ? 0.393 ? 0.56
>
> Note that the SampleID is repeating. So I want to be able to split the
dataset based on the SampleID and write the splitted dataset of every SampleID
into a new file.
> I tried split followed by lapply to do this.
>
> infile <- read.csv("test.txt", sep="\t", as.is =
TRUE, header = TRUE)
> infile.split ?<- split(infile, infile$SampleID)
> names(infile.split[1]) ?## outputs ?A1?
correct, names() returns the top level names of infile.split (i.e.,
the two data frame names)
> ## now A1, B1 are two lists in infile.split as I understand it. Correct me
if I am wrong.
It is a single, named list containing two data frames (A1 and B1)
(though data frames are built from lists, I think so I suppose in a
way it contains two lists, but that is not really the point).
>
> lapply(infile.split,function(x){
> ? ? ? ? ? ? ?filename <- names(x) #### here I expect to see A1 or B1, I
didn?t, I tried (names(x)[1]) and that gave me ?Name? and not A1 or B1.
by using lapply() on the actual object, your function is getting each
element of the list.  That is:

infile.split[[1]]
infile.split[[2]]

trying names() on those:

names(infile.split[[1]])

should show what you are getting
> ? ? ? ? ? ? ?final_filename <- paste(filename,?toy_set.txt?,sep=?_?)
> ? ? ? ? ? ? ?write.table(x, file = paste(path, final_filename,sep=?/?,
row.names=FALSE, quote=FALSE,sep=?\t?)
FYI I think you are missing a parenthesis in there
somewhere> ?} )
>
> In lapply I wanted to give a unique filename to all the split Sample Ids,
i.e. name them here as A1_toy_set.txt, B1_toy_set_txt.
> How do I get those names, i.e. A1, B1 to a create a filename like above.
Try this:

## read your data from the clipboard (obviously you do not need to)
infile <- read.table("clipboard", header = TRUE)
split.infile <- split(dat, dat$SampleID) #split data
path <- "~" # generic path

## rather than applying to the data itself, instead apply to the names
lapply(names(split.infile), function(x) {
  write.table(x = split.infile[[x]],
    file = paste(path, paste(x, "toy_set.txt", sep = "_"),
sep = "/"),
    row.names = FALSE, quote = FALSE, sep = "\t")
  cat("wrote ", x, fill = TRUE)
})

it will return two NULL lists, but that is fine because it should have
written the files.
> When I write each of the element in the list obtained after split into a
file, the column names would have names like A1.Name, A1.SampleID, A1.Al1, ?..
Can I get rid of ?A1? in the column names within the lapply (other than reading
in the file again and changing the names) ?
Can you report the results of str(yourdataframe) ?  I did not have
that issue just copying and pasting from your email and using the code
I showed above.

Cheers,

Josh
>
> Thanks for your time,
>
> Regards
> Sashi
>
>
> ? ? ? ?[[alternative HTML version deleted]]
>
>
> ______________________________________________
> R-help at 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.


-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

Jon Erik Ween

2010-Dec-16 14:43 UTC

head link

[R] Optimal method to scan cells and set flag

Hi

I am still working on my large dataset (sample attached) that contains a series
of binary variables (flags, yes/no) regarding affected brain areas
("Lica","LAtChor","LA1" ,etc).

I need to scan these columns, if value = Y for "Lxxx" set
"LesionSide" to "L", if Y for "Rxxx" set to
"R" and "B" if both.  There are >2500 records, so
for-loops would be inefficient. Any suggestions?

Much obliged.

Best

Jon

Soli Deo Gloria

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Jon Erik Ween, MD, MS
Scientist, Kunin-Lunenfeld Applied Research Unit 
Director, Stroke Clinic, Brain Health Clinic, Baycrest Centre
Assistant Professor, Dept. of Medicine, Div. of Neurology
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Kimel Family Building, 6th Floor, Room 644 
Baycrest Centre
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unauthorized use or disclosure of this information is strictly prohibited.



On 2010-12-09, at 2:21 PM, David Winsemius wrote:
> 
> On Dec 9, 2010, at 12:44 PM, Sashi Challa wrote:
> 
>> Hello All,
>> 
>> I have a toy dataframe like this. It has 8 columns separated by tab.
>> 
>> Name    SampleID        Al1     Al2     X       Y       R       Th
>> rs191191        A1      A       B       0.999   0.09    0.78    0.090
>> abc928291       A1      B       J       0.3838  0.3839  0.028   0.888
>> abcnab  A1      H       K       0.3939  0.939   0.3939  0.77
>> rx82922 B1      J       K       0.3838  0.393   0.393   0.00
>> rcn3939 B1      M       O       0.000   0.000   0.000   0.77
>> tcn39399        B1      P       I       0.393   0.393   0.393   0.56
>> 
>> Note that the SampleID is repeating. So I want to be able to split the
dataset based on the SampleID and write the splitted dataset of every SampleID
into a new file.
>> I tried split followed by lapply to do this.
>> 
>> infile <- read.csv("test.txt", sep="\t", as.is =
TRUE, header = TRUE)
>> infile.split  <- split(infile, infile$SampleID)
>> names(infile.split[1])  ## outputs ?A1?
>> ## now A1, B1 are two lists in infile.split as I understand it. Correct
me if I am wrong.
>> 
>> lapply(infile.split,function(x){
>>             filename <- names(x) #### here I expect to see A1 or B1,
I didn?t, I tried (names(x)[1]) and that gave me ?Name? and not A1 or B1.
>>             final_filename <- paste(filename,?toy_set.txt?,sep=?_?)
>>             write.table(x, file = paste(path, final_filename,sep=?/?,
row.names=FALSE, quote=FALSE,sep=?\t?)
>> } )
>> 
>> In lapply I wanted to give a unique filename to all the split Sample
Ids, i.e. name them here as <dragged to the c() construct>.
>> How do I get those names, i.e. A1, B1 to a create a filename like
above.
> 
> names(file.split) <- c("A1_toy_set.txt",
"B1_toy_set_txt")
> 
>> When I write each of the element in the list obtained after split into
a file,
> 
> How are you proposing do do this "writing"?
> 
>> the column names would have names like A1.Name, A1.SampleID, A1.Al1,
?..
> 
> Are you sure? Why would you think that?
> 
> -- 
> David.
> 
>> Can I get rid of ?A1? in the column names within the lapply (other than
reading in the file again and changing the names) ?
>> 
>> Thanks for your time,
>> 
>> Regards
>> Sashi
>> 
>> 
>> 	[[alternative HTML version deleted]]
>> 
>> ______________________________________________
>> R-help at 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.
> 
> David Winsemius, MD
> West Hartford, CT
> 
> ______________________________________________
> R-help at 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.

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