Hello all
I have data like below where the column with values (PRI, SEC ) is the char field and the rest are Numeric Fields.
200707,9580,58,7,2,1,PRI,1,1,137,205594,0,5,10,-45.51,-45.51
200707,9580,58,7,2,1,SEC,1,1,137,205594,0,5,10,-45.51,45.51... (1 Reply)
I want to sort alphabetically on the first field and sort in descending numerical order on the 2nd field. With a normal "sort -r -n" it does this:
abc ||| 5e-05 ||| bla
abc ||| 3 ||| ble
def ||| 1 ||| abc
def ||| 0.2 ||| def
As you can see it ignores the fact that 5e-05 is actually 0.00005... (1 Reply)
HI everyone,
I am trying to use the unix sort command to get a list of numbers sorted in ascending order but having trouble in getting it to work.
An example of this issue would be when i am trying to sort the following three
number each on a different line "1" , "2" and "116" the sort command... (3 Replies)
Hi,
I'm a learner of PERL programming.
I've a input file with the below data:
SWAT_5, 1703, 2010-09-21
SWAT_6, 2345, 2010-09-21
SWAT_7, 1792, 2010-09-21
SWAT_8, 1662, 2010-09-21
SWAT_9, 1888, 2010-09-21
VXHARP_1, 171, 2010-09-21
I need to sort this data based on the second... (6 Replies)
dear all,
i have .dat files named as:
34.dat
2.dat
16.dat
107.dat
i would like to sort them by their filenames as:
2.dat
16.dat
34.dat
107.dat
i have tried numerous combinations of sort and ls command (in vain) to obtain :
107.dat
16.dat
2.dat
34.dat (1 Reply)
Hi
I am using this
cat substitutionFeats.txt | gawk '{$0=gensub(/\t/,"blabla",1);print}' | gawk '{print length, $0}' | sort -n | sort -r
and the "sort -n" command doesn't work as expected: it leads to a wrong ordering:
64 Adjustable cuffs
64 Abrasion-
64 Abrasion pas
647 Sanitized 647... (4 Replies)
I have ran into a heavy case of PEBCAK*) and could need some advice on what i do wrong:
OS is Linux (kernel 2.6.35), sort --version reports "8.5" from 2010, shell is ksh.
Originally i had a file with with the following structure:
hdisk1 yyy
hdisk2 yyy
hdisk3 yyy
hdisk4 yyy
hdisk5 yyy... (2 Replies)
Input file:
100%ABC2 3.44E-12 USA
A2M%H02579 0E0 UK
100%ABC2 5.34E-8 UK
100%ABC2 3.25E-12 USA
A2M%H02579 5E-45 UK
Output file:
100%ABC2 3.44E-12 USA
100%ABC2 3.25E-12 USA
100%ABC2 5.34E-8 UK
A2M%H02579 0E0 UK
A2M%H02579 5E-45 UK
Code try:
sort -k1,1 -g -k2 -r input.txt... (2 Replies)
From googling and reading man pages I figured out this sorts the first column by numeric values.
sort -g -k 1,1
Why does the -n option not work? The man pages were a bit confusing.
And what if I want to sort the second column numerically? I haven't been able to figure that out. The file... (7 Replies)
Discussion started by: cokedude
7 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)