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Top Forums Shell Programming and Scripting awk to get values greater than Post 302862729 by SkySmart on Friday 11th of October 2013 10:54:47 AM
Old 10-11-2013
awk to get values greater than

data.txt
Code:
August 09 17:16 2013
August 09 17:17 2013
August 09 17:19 2013
August 09 17:20 2013
August 09 17:21 2013
August 09 17:22 2013
August 09 17:23 2013
August 09 17:24 2013

to print from a point in this file, to the end of the file, i type:

Code:
awk '/August 09 17:22/,0' data.txt.

the value "August 09 17:22" is a value supplied by the user to the script, from the command line. So since the user supplied that value, and that value actually existed in the data.txt file, when i typed the awk command, i got an output back.

now, what happens if the user supplied a value that doesn't exist?

Code:
awk '/August 09 17:18 2013/,0' data.txt

You'll get no response.

what i've been stuck on is, when the date supplied by the user does not exist in the data file, i want awk to grab the date that is the closest to the date the user supplied.

In this case, since "August 09 17:18 2013" was not found, the date that should have been grabbed instead should be "August 09 17:17 2013". And awk should print from that line to the end of the file.
 

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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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