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Top Forums Shell Programming and Scripting script to replace numbers on lines according to condition on the same line Post 302486524 by neemoze on Sunday 9th of January 2011 06:26:27 AM
Old 01-09-2011
script to replace numbers on lines according to condition on the same line

hello everyone

my file contains many records, the following is a sample:
Code:
BEGIN
ASX1500000050002010120000000308450201012000177
ASX1100002000000201012000000038450201012000220
ASX1600100005000201012000000038450020101200177
ASX1900100006000201067000000058450020101200177
ASX1100002000000201012300000008450201012000177
ASX1100002000000201012380000000450201012000277
ASX1800100004000201012380000000452010120000177
ASX1100002000000201012380000000450201012000177
ASX1900100006000201067500000008450201012000120
ASX1100005000000201012380000000450201012000377
ASX1100002000000201012380000000450201012000177
ASX1100002000000201012384000000052010120000477
ASX1900100006000201067584000000050201012000177
ASX1900100006000201067584500000000201012000177
ASX1100003000000201012384500000000002010120377
ASX1100003000000201012380000000450201012000177
END

what i need is to:
for each line:
1- search for 13 digits number after ASX and that number starts with 11 and contains 2 at digit 7 of this number
2- after finding this number, search in the same line for the blue three digits
if they = 001 or 002
3- if step 1 and 2 is true (lines with red and blue numbers meet this requirement) then:
replace the digits 7 to 0

the output should be
Code:
ASX1500000050002010120000000308450201012000177
ASX1100000000000201012000000038450201012000220
ASX1600100005000201012000000038450020101200177
ASX1900100006000201067000000058450020101200177
ASX1100000000000201012300000008450201012000177
ASX1100000000000201012380000000450201012000277
ASX1800100004000201012380000000452010120000177
ASX1100000000000201012380000000450201012000177
ASX1900100006000201067500000008450201012000120
ASX1100005000000201012380000000450201012000377
ASX1100000000000201012380000000450201012000177
ASX1100002000000201012384000000052010120000477
ASX1900100006000201067584000000050201012000177
ASX1900100006000201067584500000000201012000177
ASX1100003000000201012384500000000002010120377
ASX1100003000000201012380000000450201012000177
BEGIN
 
END

how to do it?
thank you
 

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bup-margin(1)						      General Commands Manual						     bup-margin(1)

NAME
bup-margin - figure out your deduplication safety margin SYNOPSIS
bup margin [options...] DESCRIPTION
bup margin iterates through all objects in your bup repository, calculating the largest number of prefix bits shared between any two entries. This number, n, identifies the longest subset of SHA-1 you could use and still encounter a collision between your object ids. For example, one system that was tested had a collection of 11 million objects (70 GB), and bup margin returned 45. That means a 46-bit hash would be sufficient to avoid all collisions among that set of objects; each object in that repository could be uniquely identified by its first 46 bits. The number of bits needed seems to increase by about 1 or 2 for every doubling of the number of objects. Since SHA-1 hashes have 160 bits, that leaves 115 bits of margin. Of course, because SHA-1 hashes are essentially random, it's theoretically possible to use many more bits with far fewer objects. If you're paranoid about the possibility of SHA-1 collisions, you can monitor your repository by running bup margin occasionally to see if you're getting dangerously close to 160 bits. OPTIONS
--predict Guess the offset into each index file where a particular object will appear, and report the maximum deviation of the correct answer from the guess. This is potentially useful for tuning an interpolation search algorithm. --ignore-midx don't use .midx files, use only .idx files. This is only really useful when used with --predict. EXAMPLE
$ bup margin Reading indexes: 100.00% (1612581/1612581), done. 40 40 matching prefix bits 1.94 bits per doubling 120 bits (61.86 doublings) remaining 4.19338e+18 times larger is possible Everyone on earth could have 625878182 data sets like yours, all in one repository, and we would expect 1 object collision. $ bup margin --predict PackIdxList: using 1 index. Reading indexes: 100.00% (1612581/1612581), done. 915 of 1612581 (0.057%) SEE ALSO
bup-midx(1), bup-save(1) BUP
Part of the bup(1) suite. AUTHORS
Avery Pennarun <apenwarr@gmail.com>. Bup unknown- bup-margin(1)
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