I'm using the below command to list files older than 2 hours but it returns redundant output, am I missing something.
# find . -mmin +120 -exec ls -l {} \;
total 0
-rw-r--r-- 1 root system 0 Oct 13 09:52 test1
-rw-r--r-- 1 root system 0 Oct 13 09:52 test2
-rw-r--r-- 1 root ... (5 Replies)
I am using the following command:
nawk -F"," 'NR==FNR {a=$1;next} a {print a,$1,$2,$3}' file1 file2
I am getting 40 records output.
But when i import file1 and file2 in MS Access i get 140 records.
And i know 140 is correct count.
Appreciate your help on correcting the above script (5 Replies)
Hello,
I couldn't find anything on the Forum that would help me to solve this problem. Could any body help me process below data using awk?
I have got two files:
file1:
Worker1: Thomas
Position: Manager
Department: Sales
Salary: $5,000
Worker2: Jason
Position: ... (5 Replies)
cat T|awk -v format=$format '{ SUM += $1} END { printf format,SUM}'
the file T has below data
usghrt45tf:hrguat:/home/hrguat $ cat T
-1363000.00123456789
-95000.00789456123
-986000.0045612378
-594000.0015978
-368939.54159753258415
-310259.0578945612
-133197.37123456789... (4 Replies)
Hi All,
I am looking to filter out filesystems which are greter than a specific value.
I use the command
df -h | awk '$4 >=70.00 {print $4,$5}'
But this results out as below, which also gives for lower values.
9% /u01
86% /home
8% /u01/data
82% /install
70% /u01/app
Looks... (3 Replies)
The awk below is supposed to count all the matching $5 strings and count how many $7 values is less than 20. I don't think I need the portion in bold as I do not need any decimal point or format, but can not seem to get the correct counts. Thank you :).
file
chr5 77316500 77316628 ... (6 Replies)
Hello scripting geeks,
I am new to scripting and facing some issues in writing the logic of the script. Request your kind help here
Actually when i run a command i get o/p as below
o/p :
0x0000
0x0000
0x0000
0x0000
0x0000
0x0000
these are hex values i guess...now i want to... (15 Replies)
I was wondering whether anyone has any idea what is happening here. I'm using simple code to compare 2 tab delimited files based on column 1 values. If the column1 value of file1 exists in file2, then I'm to print the column4 value in file2 in column3 of file1. Here is my code:
1st I have to... (6 Replies)
Discussion started by: Geneanalyst
6 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)