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shift by one column given rows in a dataframe

10 messages · Massimo Bressan, Bert Gunter, Sarah Goslee +2 more

Original
Massimo Bressan
# 
by considering the following reproducible example:

v0<-c("a","xxx","c",rep("xxx",2))
v1<-c(1,"b",3,"d","e")
v2<-c(6,2,8,4,5)
v3<-c("xxx",7,"xxx",9,10)

df_start<-data.frame(v0,v1,v2,v3)
df_start

v0<-letters[1:5]
v1<-1:5
v2<-6:10

df_end<-data.frame(v0,v1,v2)
df_end

I need  to shift by one column some given rows in the initial data frame 
called "df_start" so that to get the final structure as in "df_end";
please consider that the value "xxx" in the rows of "df_start" can be 
anything so that I necessarly need to apply by row index position (in my 
reproducible example rows: 2, 3, 5);

I'm really stuck with that problem and I can not conceive any viable 
solution up to now

any hints?

best regards

m
Sarah Goslee
# 
With one minor change to your reproducible example (thank you!):

df_start <- data.frame(v0,v1,v2,v3, stringsAsFactors=FALSE)
data.frame(t(apply(df_start, 1, function(i)i[!grepl("xxx", i)])),
stringsAsFactors=FALSE)

I'll leave it to you to deal with columns that you'd like to have
numeric. (You might also try str(df_start)).

Sarah

On Thu, Jul 23, 2015 at 12:14 PM, Massimo Bressan
<mbressan at arpa.veneto.it> wrote:
Massimo Bressan
# 
Hi

thank you for your reply: it's a neat solution but unfortunately not
applicable to my specific case;

in fact as I specified in my first post (I may have been not enough clear,
sorry for that!) I can not rely on any search method grep-like because the
value "xxx" in the rows of "df_start" can be anything (string or numeric and
always different) so that I necessarely need to apply by row index position
(i.e. in my reproducible example rows: 2, 3, 5); 

thank you again for your kind help but still searching for a solution...

best



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Bert Gunter
# 
Ah, so apparently you require some sort of psychic abilities...

For how else would one choose which three values to keep in a row that was:

a  2   b  5

based on your specification that "xxx could be anything."

Cheers,

Bert
Bert Gunter

"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
   -- Clifford Stoll
On Thu, Jul 23, 2015 at 12:56 PM, maxbre <mbressan at arpa.veneto.it> wrote:
Sarah Goslee
#
On Thu, Jul 23, 2015 at 3:56 PM, maxbre <mbressan at arpa.veneto.it> wrote:
I'm going to assume you're replying to me, although there's no context
whatsoever in your response (this is the R-help email list, not
Nabble).
Then how do you know which positions? Do you have another R object
that specifies row and column number? Or do you guess? I can easily
remove a random element from each row... If you aren't removing based
on value, then you didn't provide a reproducible example after all,
and you need to supply the index for removal.

Always different? Different within a single data frame? Different
between data frames? (Then you can simply change "xxx" to "whatever".)
Is telepathy required? Or perhaps precognition?

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Massimo Bressan
    

    

      
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sorry but honestly I do not get your point

I need to shift to left by one position (i.e. one column) the entire rows
2,4,5 of "df_start" so that to obtain as final result the structure
indicated in "df_end"

I know in advance the rows that I need to shift

hope it clears a bit, now




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      William Dunlap
    

    

      
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You could do something like the following
  > rowsToShiftLeft <- c(2,4,5) # 4, not the 3 that was in the original post
  > mat <- as.matrix(df_start)
  > mat[rowsToShiftLeft, 1:3] <- mat[rowsToShiftLeft, 2:4]
  > result <- data.frame(mat[, 1:3], stringsAsFactors=FALSE)
  > str(result)
  'data.frame':   5 obs. of  3 variables:
   $ v0: chr  "a" "b" "c" "d" ...
   $ v1: chr  "1" "2" "3" "4" ...
   $ v2: chr  "6" "7" "8" "9" ...
You will then have to convert the columns which ought to be numeric
to numeric.  (All the columns in df_start were factors because of the
extra xxx that offset some of them.)


Bill Dunlap
TIBCO Software
wdunlap tibco.com

        
On Thu, Jul 23, 2015 at 1:19 PM, maxbre <mbressan at arpa.veneto.it> wrote:

        

            

      
      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Bert Gunter
    

    

      
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Oops, Bill's reply and mine crossed in the email. His is essentially
the same as mine except probably more efficient.


-- Bert
Bert Gunter

"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
   -- Clifford Stoll

        
On Thu, Jul 23, 2015 at 2:44 PM, William Dunlap <wdunlap at tibco.com> wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Marissa Fahlberg
    

    

      
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Turns out my column and row names were too long to fit in the console and
when I widened the window the table didn't automatically adjust and widen
as well. I had to widen the console first and then retype the same command
so it fit. Silly mistake. I usually keep the console pretty narrow on my
laptop and never considered that would change the format.

Thanks for all the help - I'm sure I'll be back again very soon with more
questions!!

        
On Jul 23, 2015 4:53 PM, "Bert Gunter" <bgunter.4567 at gmail.com> wrote:

        

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

        

      
      
    

        

  
  


  


      

    

    

      
      

        


  

        

      Massimo Bressan
    

    

      
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hi

thank you all for the great replies, very useful indeed even if some of them
a bit too aggressive (which is never, ever a good approach in my very humble
opinion... but that's a matter of taste and style I do not want to discuss
here); sorry again for bothering someone with such a trivial and
ill-conceived question

finally, I'm posting here my solution as a reference to the problem so that
to close this long and winding thread; 

hoping this code might be somehow useful for someonelse, sometime,
somewhere...

## start code

v0<-c("a","xxx","c",rep("xxx",2))
v1<-c(1,"b",3,"d","e")
v2<-c(6,2,8,4,5)
v3<-c("xxx",7,"xxx",9,10)
 
df_start<-data.frame(v0,v1,v2,v3, stringsAsFactors = FALSE)

df_start

# set vector of rows to be shifted left
shIftLeft<-c(2,4,5)

# shift selected rows
df_start[shIftLeft,1:3]<-df_start[shIftLeft,2:4]

# final result
df_end<-df_start[,1:3]

df_end

## end code

thanks




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