Hi,
This is my input file:
ali 5 usa abc
abu 4 uk bca
alan 6 brazil bac
pinky 10 utah sdc
My desired output:
pinky 10 utah sdc
alan 6 brazil bac
ali 5 usa abc
abu 4 uk bca
Based on the column two, I want to do the descending order and print out other related column at the... (3 Replies)
Hi,
My input file is
$cat samp
1 siva
1 raja
2 siva
1 siva
2 raja
4 venkat
i want sort this name wise...alos need to remove duplicate lines.
i am using
cat samp|awk '{print $2,$1}'|sort -u
it showing
raja 1 (3 Replies)
I have a column of numbers in the following format:
1.722e-05
2.018e-05
2.548e-05
2.747e-05
7.897e-05
4.016e-05
4.613e-05
4.613e-05
5.151e-05
5.151e-05
5.151e-05
6.1e-05
6.254e-05
7.04e-05
7.12e-05
7.12e-05 (6 Replies)
Hi Guru,
I need some help regarding awking the output so it only show the first line (based on column) of each row.
So If column has 1, three row, then it only show the first line of that row, based on similar character in column 1. So i am trying to achieve a sort, based on column one and... (3 Replies)
Hi,
I have a table to be imported for R as matrix or data.frame but I first need to edit it because I've got several lines with the same identifier (1st column), so I want to sum the each column (2nd -nth) of each identifier (1st column)
The input is for example, after sorted:
K00001 1 1 4 3... (8 Replies)
How to sort the following output based on lowest to highest BE?
The following sort does not work.
$ sort -t. -k1,1n -k2,2n bfd.txt
BE31.116 0s 0s DOWN DAMP
BE31.116 0s 0s DOWN DAMP
BE31.117 0s 0s ... (7 Replies)
Hi All ,
I am having an input file like this
Input file
7 sks/jsjssj/ddjd/hjdjd/hdhd/Q 10 0.5 13
dkdkd/djdjd/djdjd/djd/QB 01 0.5
ldld/dkd/jdf/fjfjf/fjf/Q 0.5
10 sjs/jsdd/djdkd/dhd/Q 01 0.5 21
kdkd/djdd/djdd/jdd/djd/QB 01 0.5
dkdld/djdjd/djd/Q 01 0.5
... (9 Replies)
Discussion started by: kshitij
9 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)