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dividing values of each column in a dataframe

5 messages · Abhishek Pratap, David Winsemius, Rolf Turner +1 more

Original
Abhishek Pratap
# 
Hey Guys

I want to divide(numerically) all the columns of a data frame by
different numbers. Here is what I am  doing but getting a weird error.
The values in each column are not getting divided by the corresponding
value in the denominator vector instead by the alternative values. I
am sure I am messing up something.

Appreciate your help
-Abhi

head(counts)
               WT_CON WT_RB MU_CON MU_RB
row1       839   180    477   187


head(counts/c(2,3,4,5))
               WT_CON    WT_RB    MU_CON    MU_RB
row1      419.50000  45.0000  238.5000  46.7500
David Winsemius
#
On Feb 6, 2012, at 2:22 PM, Abhishek Pratap wrote:

            

      
      
      
    

        
The argument recycling is going to be done on a column basis. You  
apaear to expect it to be done on a row basis. For that you need apply

Probably:

t( apply(counts, 1, "/", c(2,3,4,5)) )


But untested in absence of example.

---
David Winsemius, MD
West Hartford, CT

  
  


  


      

    

    

      
      

        


  

        

      Berend Hasselman
    

    

      
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On 06-02-2012, at 20:22, Abhishek Pratap wrote:

        

            

      
      
      
    

        
I don't think you are getting an error.
You are getting an unexpected result (you think).

            

      
      
      
    

        
I tried this

df <- data.frame(a=runif(10),b=runif(10),c=runif(10))
scal <- c(2,3,4)

            

      
      
      
    

        
a          b          c
1  0.47661685 0.73457617 0.90045279
2  0.06916502 0.76374600 0.07630196
3  0.17029174 0.29450289 0.07416969
4  0.03126839 0.09864740 0.08230353
5  0.23713816 0.06342224 0.91241698
6  0.21970595 0.86890690 0.47316101
7  0.46380324 0.26142304 0.87823277
8  0.81256517 0.76097474 0.98956553
9  0.74425369 0.29228545 0.27496707
10 0.65425285 0.40166967 0.12231213

            

      
      
      
    

        
a          b          c
1  0.23830843 0.24485872 0.22511320
2  0.02305501 0.19093650 0.03815098
3  0.04257293 0.14725144 0.02472323
4  0.01563419 0.03288247 0.02057588
5  0.07904605 0.01585556 0.45620849
6  0.05492649 0.43445345 0.15772034
7  0.23190162 0.08714101 0.21955819
8  0.27085506 0.19024369 0.49478276
9  0.18606342 0.14614273 0.09165569
10 0.32712642 0.13388989 0.03057803

Checking:

            

      
      
      
    

        
a          b          c
4 0.01563419 0.03288247 0.02057588


This is what you want guessing from your description.
You haven't given us an example with desired output.

Berend

  
  


  


      

    

    

      
      

        


  

        

      Rolf Turner
    

    

      
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I believe your post is  misleading.  Your example "works"
purely by chance.

R uses "column ordering",  so entries 1 to 3 of column 1 in your
example get divided by 2, 3, and 4 respectively.  Then "scal" is
*recycled* and entries 4, 5, and 6, get divided by 2, 3, and 4
respectively, and so on.

It just *happens* that entry 4 of column 1 gets divided by 2,
entry 4 of column 2 gets divided by 3, and entry 4 of column 3
gets divided by 4, giving the impression that you are getting
what you want.  But if you look at row 5 of your "df" you'll see
something different.

# The same (due to serendipity --- or it's converse!):
 > (df/scal)[4,]
                   a                  b                  c
4 0.0156342 0.03288247 0.02057588
 > df[4,]/scal
                   a                  b                  c
4 0.0156342 0.03288247 0.02057588

# Not the same!
 > (df/scal)[5,]
                   a                  b                  c
5 0.07904605 0.01585556 0.4562085
 > df[5,]/scal
                   a                  b                  c
5 0.1185691 0.02114075 0.2281042

As has already been pointed out in this thread, to get what you think
you're getting, you need to use transpose t(t(df)/scal).

     cheers,

         Rolf Turner

        
On 07/02/12 08:54, Berend Hasselman wrote:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Berend Hasselman
    

    

      
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On 07-02-2012, at 00:09, Rolf Turner wrote:

        

            

      
      
      
    

        
You are completely correct.
I should have checked more thoroughly.
Using this dataframe makes the error glaring

df <- data.frame(a=rep(2,10),b=rep(3,10),c=rep(4,10))

My apologies to the OP and the list.

Berend Hasselman