Hi,
I have a log file that has the date in this format "2006-05-30_13:14:04,256". I need to find the time difference between two log entries in milliseconds.
How to achieve this in AWK/GAWK script? :confused: (2 Replies)
Hi,
i am using the command
find . -exec grep -i "sometexttofoundinfiles" '{}' \; -print
The problem with this command is it prints the path of file i want twice. I did try the "-q" option of the grep command. It generates an error "invalid option". i guess the shell i am using... (1 Reply)
I was wondering if it was possible to tell awk to print the output of a command in the print.
.... | awk '{print $0}'
I would like it to print the date right before $0, so something like (this doesn't work though)
.... | awk '{print date $0}' (4 Replies)
Hi folk,
Hope you enjoy the summer.
I am stock after one day working, the problems are the following:
1) I want to write a for loop in a shell script code that has a double precision step as:
#!/bin/bash
START=0.00001
STOP=0.001
for((i = START ; i< STOP ; i=2*i)) do
echo "$i"... (1 Reply)
Hi,
I have data as
"01/22/97-"aaaaaaaaaaaaaaaaa""aaa""aabbbbbbbbcccccc""zbcd""dddddddddeeeeeeeeefffffff"
I want to remove only the Consequitive double quotes and not the one which occurs single.
My O/P must be ... (2 Replies)
Hi! all here is my code
which is working fine no errors but I want to know how to take result and input to other program
awk 'FNR==1{i++}{LC=NR}
{for(k=1; k<=NF; k++) A=$k}
END{for (i=1;i<=LC;i++)
{
for(j=1;j<=LC;j++)
if(A=='$UID' && A>='$MX'+A &&... (7 Replies)
Hello,
Giving those commands:
cat > myfile
1
2
3
^D
cat myfile | awk '{ s=s+$1 ; print s}'
The output is:
1
3
6
It seems like this command iterates each time on a different row so $1 is the first field of each row.. But what caused it to refer to each row ?.
What I mean... (3 Replies)
Dear forum members,
I want the script to count ALA as one (an example in quotes) and return an integer as 1 and not return 5 as an integer as it does now (look bash script). So how can I upgrade my script that it first checks or after finding all instances of ALA checks whether it is the same... (25 Replies)
Discussion started by: Aurimas
25 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)
05-10-2010
