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Deriving unique entries from multiple repeating patterns

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# 1  
Old 10-23-2012
Deriving unique entries from multiple repeating patterns
Dear All,

I have a below one column data.(example)
Code:
Col1
1
2
.
.
25
8
9
25
1
2
.
.
25

Where each entry(row) is a number from 1-25, but in place whereever mentioned with
Code:
.

we have all the entries 1-25, but some places where ever no
Code:
.

like in 8 9 25 I have only 3 entries.
No I need the below output for this kind of data.
(In reality I have 1 million rows in my data)
Code:
Col1
A1
A2
.
.
A25
B8
B9
B25
C1
C2
.
.
C25

I need an awk based automatic script.

Thanks
Sidda
# 2  
Old 10-23-2012
And what have you tried so far?
# 3  
Old 10-23-2012
Hi Radoulov,

I do not have any clue on the code an I am not an expert in awk.
Since I am in some urgency of preparing a data summary I am manipulating manually all the entries prefixing with A,B etc.

Regards
Sidda
# 4  
Old 10-23-2012
Quote:
Originally Posted by ks_reddy
(In reality I have 1 million rows in my data)
As you have mentioned you have 1 million rows..

what you want print after Z..?
# 5  
Old 10-23-2012
OK, instead of modifying the file manually, you could use Perl:

Code:
perl -i.orig -lpe'
  $l = "A" and next if $. == 1;
  next if /^\./;
  $_ = (/^25/ ? $l++ : $l).$_
  ' infile

Warning: the script will modify your original file (and back it up as filename.orig) ! Incrementing Z will wrap around to A. The script assumes that the first line contains the column header
and therefor it will leave it as is.
Last edited by radoulov; 10-23-2012 at 06:21 AM.. Reason: Corrected.
# 6  
Old 10-23-2012
with awk..

Code:
awk -v i="65" '
function get_word() {
if(i <= 90){
c=sprintf("%c",i)}
else{i=65;c=sprintf("%c",i)}
}
{if($0 !~ /\./){get_word();print c""$0}
else{print}}
/25/{i++;}' file

It will again start printing A after Z.
# 7  
Old 10-23-2012
Another awk
Code:
awk 'NR>1{if($1<p)c++; p=$1; $0=sprintf( "%c", 65+c) $0}1' file
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