Hash Function Speed Post: 302364236

9 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

Awk Hash Function.

I have a file with a format of A,2 B,2 G,3 A,2 A,3 A,2 D,7 A,2 E,2 A,2 I need to create a sum of each alphabet with the numbers assigned to it using awk.

2. Shell Programming and Scripting

Print Entire hash list (hash of hashes)

I have a script with dynamic hash of hashes , and I want to print the entire hash (with all other hashes). Itried to do it recursively by checking if the current key is a hash and if yes call the current function again with refference to the sub hash. Most of the printing seems to be OK but in...

3. Filesystems, Disks and Memory

data from blktrace: read speed V.S. write speed

I analysed disk performance with blktrace and get some data: read: 8,3 4 2141 2.882115217 3342 Q R 195732187 + 32 8,3 4 2142 2.882116411 3342 G R 195732187 + 32 8,3 4 2144 2.882117647 3342 I R 195732187 + 32 8,3 4 2145 ...

4. Shell Programming and Scripting

Assigning a hash to another hash key

Hello, I have a hash in hsh. I need to assign it to another hash globalHsh. I think the below statement does not work $globalHsh{$id} = %hsh; What is the right way to assign it? Thanks

5. Shell Programming and Scripting

Perl Hash:Can not keep hash data in the same order that it was inserted

Can Someone explain me why even using Tie::IxHash I can not get the output data in the same order that it was inserted? See code below. #!/usr/bin/perl use warnings; use Tie::IxHash; use strict; tie (my %programs, "Tie::IxHash"); while (my $line = <DATA>) { chomp $line; my(...

6. Shell Programming and Scripting

perl hash - using a range as a hash key.

Hi, In Perl, is it possible to use a range of numbers with '..' as a key in a hash? Something in like: %hash = ( '768..1536' => '1G', '1537..2560' => '2G' ); That is, the range operation is evaluated, and all members of the range are...

7. Shell Programming and Scripting

Compare values of hashes of hash for n number of hash in perl without sorting.

Hi, I have an hashes of hash, where hash is dynamic, it can be n number of hash. i need to compare data_count values of all . my %result ( $abc => { 'data_count' => '10', 'ID' => 'ABC122', } $def => { 'data_count' => '20', 'ID' => 'defASe', ...

8. Shell Programming and Scripting

Dynamically parse BibTeX and create hash of hash

Hello gurus, Iam trying to parse following BibTex file (bibliography.bib): @book{Lee2000a, abstract = {Abstract goes here}, author = {Lee, Wenke and Stolfo, Salvatore J}, title = {{Data mining approaches for intrusion detection}}, year = {2000} } @article{Forrest1996, abstract =...

9. Shell Programming and Scripting

Need to print hash of hash in table format

Hi, I have a hash of hash where it has name, activities and count i have data like this - $result->{$name}->{$activities} = $value; content of that are - name - robert tom cat peter activities - running, eating, sleeping , drinking, work i need to print output as below ...

LEARN ABOUT DEBIAN

bup-margin

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)