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Help with substitution in shell script

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# 1  
Old 01-13-2012
Help with substitution in shell script
Hello,

I Have a file like this, with five columns:
Code:
B D F T L
---------
0 0 0 0 0
0 0 0 1 0
0 2 0 0 1

I need, for each row, to append another five columns similar to the first ones, but with the field labeled as "T" (4th field of the column) substituted in this way:

(16 * value of Fifth field) + value of Fourth field = expr \( 16 \* ${L} \) + ${T} and leave the value of Fifth field as "0".

An example of the output I want:

Code:
B D F T L -----        B D F T  L
--------------------------------
0 0 0 0 0         ----- 0 0 0 0  0
0 0 0 1 0          ----- 0 0 0 1  0
0 2 0 0 1          ----- 0 2 0 16 0

I don't know how to do the substitution. Any help will be appreciated.
(The "-" are for separating columns; in the post, with spaces instead, it is seen all joined)


Best Regards,

Moderator's Comments:
Mod Comment Please use next time code tags for your code and data
# 2  
Old 01-13-2012
Welcome to the forum

With awk

Code:
awk '{printf $0 FS; $4=16*$5+$4;$5=0;printf $0 RS}' file

# 3  
Old 01-13-2012
Hi,

Thank you for your response, but it does not work:

Code:
host1:/:awk '{printf $0 FS; $4=16*$5+$4;$5=0;printf $0 RS}' file
B D F T L B D F 0 0
--------- ---------   0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 1 0
0 2 0 0 1 0 2 0 16 0
host1:/:

Best Regards,
# 4  
Old 01-13-2012
The first two lines are the header part? need to ignore that.

Code:
awk 'NR<=2 {printf $0 FS $0 RS} NR>2{printf $0 FS; $4=16*$5+$4;$5=0;printf $0 RS}' file

O/P
Code:
B D F T L B D F T L
--------- ---------
0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 1 0
0 2 0 0 1 0 2 0 16 0

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