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# 8  
Old 10-19-2007
man paste

(message too short)
# 9  
Old 10-19-2007
Using man paste, I have only the option to use:

Code:
paste -d "\t" name_*.dat > final_file.dat

But when I use it the fina_final.dat is empty.
I guess that using the option -d I can decalre the delimiter:

Code:
$man paste

 -d, --delimiters=LIST
              reuse characters from LIST instead of TABs

So I have declared it such tabulation, but does not work. Smilie

Thanks in advance again.Smilie
tonet
# 10  
Old 10-19-2007
Code:
paste <(for f in name{1..30}.dat;do [ -f "$f" ]&&printf "%s" "$f";done)

# 11  
Old 10-19-2007
thanks again.I tried to do this but I do not know how create this new file. I guess it should be:

Code:
paste <(for f in name{1..30}.dat;do [ -f "$f" ]&&printf "%s" "$f";done) > final_file.dat

But it does not work. Best regards Smilie
tonet
# 12  
Old 10-19-2007
Do you want an output like this:

Code:
$ for i in $(seq 2 3 14);do echo col$i>name$i.dat;done
$ ls
name11.dat  name14.dat  name2.dat  name5.dat  name8.dat
$ cat name*
col11
col14
col2
col5
col8
$ { for f in name{1..15}.dat;do [ -f "$f" ]&&printf "%s " "$(<$f)";done;printf "\n";}>final_file.dat
$ cat final_file.dat 
col2 col5 col8 col11 col14

Last edited by radoulov; 10-19-2007 at 08:05 AM.. Reason: corrected
# 13  
Old 10-19-2007
Yes Radoulov.
I tried but only paste the name of files, and I would like the content.
Then I guess that I do not explain correctly. Each file it has one column, like:

Code:
2.378E-04
1.020E-03
1.287E-03
1.643E-03
1.285E-03
1.073E-03
1.473E-03
2.994E-03
2.090E-03
1.884E-03
....
....

for example this is the namefile_1.dat. Then I would like to create a final file which it has all the columns for each file, and ordered by number.
Smilie
tonet
# 14  
Old 10-19-2007
...

with bash, sort and paste:

Code:
$ cat name1.dat 
2.378E-01
1.020E-01
1.287E-01
1.643E-01
1.285E-01
1.073E-01
1.473E-01
2.994E-01
2.090E-01
1.884E-01
$ cat name2.dat
2.378E-02
1.020E-02
1.287E-02
1.643E-02
1.285E-02
1.073E-02
1.473E-02
2.994E-02
2.090E-02
1.884E-02
$ cat name12.dat
2.378E-12
1.020E-12
1.287E-12
1.643E-12
1.285E-12
1.073E-12
1.473E-12
2.994E-12
2.090E-12
1.884E-12
$ set -- $(printf "%s\n" name*|sort -k1.5n)
$ paste "$@">namefile_1.dat
$ cat namefile_1.dat 
2.378E-01       2.378E-02       2.378E-12
1.020E-01       1.020E-02       1.020E-12
1.287E-01       1.287E-02       1.287E-12
1.643E-01       1.643E-02       1.643E-12
1.285E-01       1.285E-02       1.285E-12
1.073E-01       1.073E-02       1.073E-12
1.473E-01       1.473E-02       1.473E-12
2.994E-01       2.994E-02       2.994E-12
2.090E-01       2.090E-02       2.090E-12
1.884E-01       1.884E-02       1.884E-12


Without sort:
Code:
$ set -- $(for f in name{1..12}.dat;do [ -f "$f" ]&&printf "%s\n" "$f";done) 
$ paste "$@"
2.378E-01       2.378E-02       2.378E-12
1.020E-01       1.020E-02       1.020E-12
1.287E-01       1.287E-02       1.287E-12
1.643E-01       1.643E-02       1.643E-12
1.285E-01       1.285E-02       1.285E-12
1.073E-01       1.073E-02       1.073E-12
1.473E-01       1.473E-02       1.473E-12
2.994E-01       2.994E-02       2.994E-12
2.090E-01       2.090E-02       2.090E-12
1.884E-01       1.884E-02       1.884E-12

Last edited by radoulov; 10-19-2007 at 08:33 AM..
 
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