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How to cbind DF:s with differing number of rows?

4 messages · Lauri Nikkinen, jim holtman, Moshe Olshansky

Lauri Nikkinen
# 
#Hi R-users,
#Suppose that I have a data.frame like this:

y1 <- rnorm(10) + 6.8
y2 <- rnorm(10) + (1:10*1.7 + 1)
y3 <- rnorm(10) + (1:10*6.7 + 3.7)
y <- c(y1,y2,y3)
x <- rep(1:3,10)
f <- gl(2,15, labels=paste("lev", 1:2, sep=""))
g <- seq(as.Date("2000/1/1"), by="day", length=30)
DF <- data.frame(x=x,y=y, f=f, g=g)
DF$g[DF$x == 1] <- NA
DF$x[3:6] <- NA
DF$wdays <- weekdays(DF$g)

DF

#For EDA purposes, I would like to calculate frequences in each variable
g <- lapply(DF, function(x) as.data.frame(table(x)))

#After this, I would like to cbind these data.frames (in g) into a
single data.frame (which to export to MS Excel)
#do.call(cbind, g) does not seem to work because of the different
number of rows in each data.frame.
#The resulting data.frame shoul look like this (only two variables
printed here):

Rowid;x;Freq.x;y;Freq.y; # etc...
1;1;9;1.69151845313816;1;
2;2;9;5.03748767699799;1;
3;3;8;5.37387749444247;1;
4;Empty;Empty;6.83926626214299;1;
5;Empty;Empty;6.97484558968873;1;
6;Empty;Empty;7.11023821708323;1;
7;Empty;Empty;7.1348316549091;1;
8;Empty;Empty;7.16727166992407;1;
9;Empty;Empty;7.35983428577469;1;
10;Empty;Empty;7.7596470136235;1;
11;Empty;Empty;7.86369414967578;1;
12;Empty;Empty;7.97164674771006;1;
13;Empty;Empty;8.0787295301318;1;
14;Empty;Empty;8.14161030348166;1;
15;Empty;Empty;8.20134832959661;1;
16;Empty;Empty;10.1469115339016;1
17;Empty;Empty;12.7442067301746;1
18;Empty;Empty;14.0865167751202;1
19;Empty;Empty;15.8280312307450;1
20;Empty;Empty;16.0484499360756;1
21;Empty;Empty;17.0795222149999;1
22;Empty;Empty;18.1254057823357;1
23;Empty;Empty;22.7169729331525;1
24;Empty;Empty;30.7237748005358;1
25;Empty;Empty;37.2141271786934;1
26;Empty;Empty;44.4954633229803;1
27;Empty;Empty;50.2302409305761;1
28;Empty;Empty;57.8913405112114;1
29;Empty;Empty;64.849897477945;1
30;Empty;Empty;71.4205263353053;1


#Anyone have an idea how to do this?

#Thanks,
#Lauri
jim holtman
# 
This should do it for you by padding out the rows so they are the same length:
List of 5
 $ x    :'data.frame':  3 obs. of  2 variables:
  ..$ x   : Factor w/ 3 levels "1","2","3": 1 2 3
  ..$ Freq: int [1:3] 9 9 8
 $ y    :'data.frame':  30 obs. of  2 variables:
  ..$ x   : Factor w/ 30 levels "4.21178116845085",..: 1 2 3 4 5 6 7 8 9 10 ...
  ..$ Freq: int [1:30] 1 1 1 1 1 1 1 1 1 1 ...
 $ f    :'data.frame':  2 obs. of  2 variables:
  ..$ x   : Factor w/ 2 levels "lev1","lev2": 1 2
  ..$ Freq: int [1:2] 15 15
 $ g    :'data.frame':  20 obs. of  2 variables:
  ..$ x   : Factor w/ 20 levels "2000-01-02","2000-01-03",..: 1 2 3 4
5 6 7 8 9 10 ...
  ..$ Freq: int [1:20] 1 1 1 1 1 1 1 1 1 1 ...
 $ wdays:'data.frame':  7 obs. of  2 variables:
  ..$ x   : Factor w/ 7 levels "Friday","Monday",..: 1 2 3 4 5 6 7
  ..$ Freq: int [1:7] 2 3 3 4 3 2 3
+     if (nrow(.x) < max.rows) .x <- rbind(.x, matrix(NA, ncol=2,
nrow=max.rows - nrow(.x),
+         dimnames=list(NULL, c('x', 'Freq'))))
+     .x
+ })
x.x x.Freq              y.x y.Freq  f.x f.Freq        g.x g.Freq
wdays.x wdays.Freq
1     1      9 4.21178116845085      1 lev1     15 2000-01-02      1
 Friday          2
2     2      9 4.78984323641143      1 lev2     15 2000-01-03      1
 Monday          3
3     3      8  5.4787594194582      1 <NA>     NA 2000-01-05      1
Saturday          3
4  <NA>     NA  5.5853001128225      1 <NA>     NA 2000-01-06      1
 Sunday          4
5  <NA>     NA 5.96437138758995      1 <NA>     NA 2000-01-08      1
Thursday          3
6  <NA>     NA 5.97953161588198      1 <NA>     NA 2000-01-09      1
Tuesday          2
7  <NA>     NA 6.17354618925767      1 <NA>     NA 2000-01-11      1
Wednesday          3
8  <NA>     NA 6.49461161284364      1 <NA>     NA 2000-01-12      1
   <NA>         NA
9  <NA>     NA 6.98364332422208      1 <NA>     NA 2000-01-14      1
   <NA>         NA
10 <NA>     NA 7.12950777181536      1 <NA>     NA 2000-01-15      1
   <NA>         NA
11 <NA>     NA 7.28742905242849      1 <NA>     NA 2000-01-17      1
   <NA>         NA
12 <NA>     NA  7.3757813516535      1 <NA>     NA 2000-01-18      1
   <NA>         NA
13 <NA>     NA 7.53832470512922      1 <NA>     NA 2000-01-20      1
   <NA>         NA
14 <NA>     NA 8.39528080213779      1 <NA>     NA 2000-01-21      1
   <NA>         NA
15 <NA>     NA 10.6249309181431      1 <NA>     NA 2000-01-23      1
   <NA>         NA
16 <NA>     NA 11.1550663909848      1 <NA>     NA 2000-01-24      1
   <NA>         NA
17 <NA>     NA 11.3189773716082      1 <NA>     NA 2000-01-26      1
   <NA>         NA
18 <NA>     NA 12.8838097369011      1 <NA>     NA 2000-01-27      1
   <NA>         NA
19 <NA>     NA 15.5438362106853      1 <NA>     NA 2000-01-29      1
   <NA>         NA
20 <NA>     NA 17.1212211950981      1 <NA>     NA 2000-01-30      1
   <NA>         NA
21 <NA>     NA 17.8821363007311      1 <NA>     NA       <NA>     NA
   <NA>         NA
22 <NA>     NA 18.5939013212175      1 <NA>     NA       <NA>     NA
   <NA>         NA
On Dec 1, 2007 8:05 AM, Lauri Nikkinen <lauri.nikkinen at iki.fi> wrote:

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Lauri Nikkinen
    

    

      
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Thanks Jim! That helped me a lot!

Cheers,
Lauri

2007/12/1, jim holtman <jholtman at gmail.com>:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

  
  


  


      

    

    
1 day later

    

      
      

        


  

        

      Moshe Olshansky
    

    

      
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Hi Lauri,

I see two possibilities.

Let say that you have
a <-c(1:5)
b <- c(1:7)
c <- c(1:4)
l <- list(a,b,c)

and you want to create an Excel file with columnA (1)
containing a (5 rows), column B (2) containing b and
column C containing c.

One possibility would be to write each ROW of the
output file separately (7 rows in your case) writing
nothing (just \t ) when you reach the end of the
particular item in the list. 

Another possibility would be to use xlsReadWritePro
package which allows to write each item (a,b,c) (or
list element - l[[1]],l[[2]],... etc) to a given
column of the output file.

Regards,

Moshe.

        
--- Lauri Nikkinen <lauri.nikkinen at iki.fi> wrote: