I've written a script to count the total size of SAN storage LUNs, and also display the LUN sizes.
From server to server, the LUNs sizes differ.
What I want to do is count the occurances as they occur and change.
These are the LUN sizes:
49.95
49.95
49.95
49.95
49.95
49.95
49.95
49.95... (2 Replies)
Hi all,
I have a pattern like this in a file:
123 4 56 789
234 5 67 789
121 3 56 789
222 4 65 789
321 6 90 100
478 8 40 789
243 7 80 789
How can I count the number of occurences of '789' (4th column) in this set...?
Thanks for all your help!
K (7 Replies)
I need to count the number of occurrences of a pattern, say 'key', between each occurrence of a different pattern, say 'lu'.
Here's a portion of the text I'm trying to parse:
lu S1234L_149_m1_vg.6, part-att 1, vdp-att 1 p-reserver IID 0xdb
registrations:
key 4156 4353 0000 0000
... (3 Replies)
how to count the total number of lines of all the files under a directory using perl script..
I mean if I have 10 files under a directory then I want to count the total number of lines of all the 10 files contain. Please help me in writing a perl script on this. (5 Replies)
Please advice how can we search for a string say (abc) in multiple files and to get total occurrence of that searched string. (Need number of records that exits in period of time).
File look like this (read as filename.yyyymmdd)
a.20100101
b.20100108
c.20100115
d.20100122
e.20100129... (2 Replies)
Hi!
In our current directory there are around 35000 files.
Out of these a few thousands(around 20000) start with, "testfiles9842323879838".
I want to count the number of files that have filenames starting with the above pattern. Please help me with the command i could use.
Thank... (7 Replies)
Hi ,
Below is my sample data,I have this 8 column(A,B,C,D,E,F,G,H) in csv file.
A , B ,C ,D ,E ,F,G ,H
4141,127337,24,15,20,69,72.0,-3
4141,128864,24,15,20,65,66.0,-1
4141,910053,24,15,4,4,5.0,-1
4141,910383,24,15,22,3,4.0,-1
4141,496969,24,15,14,6,-24.0,-18... (7 Replies)
How to find total number of special character in a column?
I am using awk -f "," '$col_number "*$" {print $col_number}' file.csv|wc -l but its not giving correct output. It's giving output as 1 even though i give no special character?
Please use code tags next time for your code and... (4 Replies)
Discussion started by: AjitKumar
4 Replies
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)