Re: [R] conditional Dataframe filling

Wed, 27 Mar 2013 13:13:04 -0700 

Thanks Arun,
Well that is interesting. My intention was to have a dataframe with the same number of rows in the original data, and for the rows with NAs, then return NA (If there are NAs, often the entire row has NAs). What is interesting is that in your code with NAs, the row that has NAs gets NAs in the output, which is what I am looking for. 


I guess a solution is to subset complete rows and then run your line  
of code. Unless there is an alternative, to tell cumsum to leave NAs  
as NAs?


Thanks again,

Camilo


Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:   Country code: 57
         Provider code: 313
         Phone 776 2282
         From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/



Quoting arun <smartpink...@yahoo.com>:


Dear Camilo,

How do you want to deal with the NAs?

If I remove the NAs:
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),

y =  
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),

z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")

dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat1<- as.data.frame(na.omit(dat1))
dat2<-dat1

dat2[]<-t(apply(!dat1,1,function(x)  
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))

 dat2
#  a b c d e f g h i j k l m
#w 0 0 0 0 0 1 2 3 4 0 0 0 0
#y 1 2 3 4 5 0 0 1 2 0 0 0 1
#z 0 0 0 0 1 0 0 0 1 0 0 0 1


 dat1
#      a     b     c     d     e     f     g     h     i    j    k    l     m
#w  TRUE  TRUE  TRUE  TRUE  TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE  TRUE
#y FALSE FALSE FALSE FALSE FALSE  TRUE  TRUE FALSE FALSE TRUE TRUE TRUE FALSE
#z  TRUE  TRUE  TRUE  TRUE FALSE  TRUE  TRUE  TRUE FALSE TRUE TRUE TRUE FALSE


A.K.





----- Original Message -----
From: Camilo Mora <cm...@dal.ca>
To: arun <smartpink...@yahoo.com>
Cc: R help <r-help@r-project.org>
Sent: Wednesday, March 27, 2013 3:27 PM
Subject: Re: [R] conditional Dataframe filling

Dear Arun,


Thank you very  much for your help with this.I did not know where to  
start looking to solve that problem, so I truly appreciate your input.



The line of code you sent seems to work but it duplicates the  
results. Do you know why that may happen?

Below is a larger database, to which I apply your line of code.

Thank you very much again,
Camilo


dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),

y =  
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),

z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")

dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")

dat2<-dat1


dat2[]<-t(apply(!dat1,1,function(x)  
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))














Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:   Country code: 57
         Provider code: 313
         Phone 776 2282
         From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/



Quoting arun <smartpink...@yahoo.com>:


HI,

Just a correction:

:


dat2[]<-t(apply(!dat1,1,function(x)  
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))   
#should also work

A.K.



----- Original Message -----
From: arun <smartpink...@yahoo.com>
To: Camilo Mora <cm...@dal.ca>
Cc: R help <r-help@r-project.org>
Sent: Wednesday, March 27, 2013 9:09 AM
Subject: Re: [R] conditional Dataframe filling



Hi,
You could try:
dat1<- read.table(text="
a    b    c    d
TRUE  TRUE  TRUE  TRUE
FALSE FALSE FALSE TRUE
FALSE  TRUE  FALSE  FALSE
",sep="",header=TRUE)
dat2<-dat1

 dat2[]<-t(apply(1*!dat1,1,function(x)  
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))

 dat2
#  a b c d
#1 0 0 0 0
#2 1 2 3 0
#3 1 0 1 2
A.K.


----- Original Message -----
From: Camilo Mora <cm...@dal.ca>
To: r-help@r-project.org
Cc:
Sent: Wednesday, March 27, 2013 4:31 AM
Subject: [R] conditional Dataframe filling

Hi everyone:

This may be trivial but I just have not been able to figure it out.

Imagine the following dataframe:
a     b     c     d
TRUE  TRUE  TRUE  TRUE
FALSE FALSE FALSE TRUE
FALSE  TRUE  FALSE  FALSE


I would like to create a new dataframe, in which TRUE gets 0 but if  
false then add 1 to the cell to the left. So the results for the  
example above should be something like:


a     b     c     d
0     0     0     0
1     2     3     0
1     0     1     2

I wonder if you may know?.

Thanks,

Camilo




Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:   Country code: 57
         Provider code: 313
         Phone 776 2282
         From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/

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