about transforming a data.frame
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Bogdan ▴ 670
@bogdan-2367
Last seen 13 months ago
Palo Alto, CA, USA

Dear all,
 

I would appreciate a suggestion on the following : I am working with a data.frame (below) :

  EXP    CT   row_names   col_names                 
1   test -5    B4:B5:B6    B1:B2:B3              
2   test -2    B7:B8:B9    B1:B2:B3              
3   test -2    D4:D5:D6    H4:H5:H6              
4   test -2    D10:D11:D12 F10:F11:F12             
5   test -2    D10:D11:D12    H1:H2:H3              
6   test -2    E10:E11:E12    G7:G8:G9             
7   test -4     A1:A2:A3    D1:D2:D3                
8   test -4   B10:B11:B12    B1:B2:B3              

what would be the easiest way to consider UNIQUE elements in the ROW_NAMES or the UNIQUE elements in the COL_NAMES and :

print how many times these UNIQUE ELEMENTS associate with the numbers -5, -2, or -4 (these numbers are on the column names CT) ..

thanks,

bogdan

data • 1.4k views
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@steve-lianoglou-2771
Last seen 20 months ago
United States

This is really an R question as opposed to a Bioconductor question, but I'll play a little code golf.

Not sure that I understand what you want, actually, but here is one way to get the result that I think you are after using data.table:

x <- read.table(textConnection("
  EXP    CT   row_names   col_names                 
1   test -5    B4:B5:B6    B1:B2:B3              
2   test -2    B7:B8:B9    B1:B2:B3              
3   test -2    D4:D5:D6    H4:H5:H6              
4   test -2    D10:D11:D12 F10:F11:F12             
5   test -2    D10:D11:D12    H1:H2:H3              
6   test -2    E10:E11:E12    G7:G8:G9             
7   test -4     A1:A2:A3    D1:D2:D3                
8   test -4   B10:B11:B12    B1:B2:B3              
"))

x <- transform(as.data.table(x), CT=factor(CT))
x[, as.list(table(CT)), by='row_names']

##         row_names -5 -4 -2
##    1:    B4:B5:B6  1  0  0
##    2:    B7:B8:B9  0  0  1
##    3:    D4:D5:D6  0  0  1
##    4: D10:D11:D12  0  0  2
##    5: E10:E11:E12  0  0  1
##    6:    A1:A2:A3  0  1  0
##    7: B10:B11:B12  0  1  0

I would provide the "competing" dplyr code, but I'm not exactly sure what the idiomatic dplyr way to do that is.

 

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Bogdan ▴ 670
@bogdan-2367
Last seen 13 months ago
Palo Alto, CA, USA

Thanks Steve, it helps a lot !

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