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Top Forums Shell Programming and Scripting Read column values from previous and next line using awk Post 302708921 by Nishi_Licious on Tuesday 2nd of October 2012 11:57:17 AM
Old 10-02-2012
Read column values from previous and next line using awk

Hi,

I have a csv file which contains data that looks something like this:

Code:
Key1 Key2 Key3 New_Key1 New_Key2 New_Key3
102   30      0        -                 -                  -
102   40      1        30               40               50
102   50      2        40               50               30
103   12      0        -                 -                  -
103   15      1        12              15               12

I need to fill in New_Key1, New_Key2 and New_Key3 based on the existing data taken from column Key2 keeping Key1 as the reference. A set of values of Key2 can be mapped to Key1. Like we have 30, 40 and 50 are mapped to Key1=102. Using this I need to fill in for New_Key1, New_Key2 and New_Key3. Here for my first row the New_Key1, New_Key2 and New_Key3 need to be dashed out. The value of New_Key1 on the second row is same as the 1st value of Key2(old) and the value of New_Key3 will be same as the 3rd value of Key2(new future value). While the value of New_Key2 will be same as the current value of Key2. Similarly for the next set of rows mapping to the same Key1 the process needs to repeat. But there's one more thing to it...The last value of New_Key3 will be the same as the first value of Key2. I hope my explanation is not confusing. Smilie

I want to this the awk way..but i'm kind of stuck. It might need me to use many more commands than is required. Please help me out!! Smilie

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Last edited by Corona688; 10-02-2012 at 01:13 PM..
 

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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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