Hello.
I face this (2 side) problem.
Some lines with this structure.
...........
12345678 4
12345989 13
12346356 205
12346644 74
12346819 22
.........
The first field (timestamp) is growing (or at least equal).
1)Sum the second fields if the first_field/500 are... (8 Replies)
Hi friends,
This is sed & awk type question.
I have a text file which has numbers spread all over the file. I want to sum the series of numbers whenever i find it and produce an output file with the sum. For example
###start of input text file ####
abc
def
ghi
1
2
3
4
kjld
random... (3 Replies)
Hi friends,
This is sed & awk type question. It is slightly different from my previous question.
I have a text file which has numbers spread all over the file. I want to sum the series of numbers (but no more than 10 numbers in series) whenever i find it and produce an output file with the... (4 Replies)
Hi,
I am using the following code to find the sum of the values of column 286 in a file. It will have the Decimal values with the scale of 2. Delimiter is '|^'
cut -d'|^' -f286 filename|cut -c3-| awk '{ x += $1 } END { printf("%.2f\n", x) }'
There are around 50k records in this file... (2 Replies)
Hello,
I am new to Linux environment , I working on Linux script which should send auto email based on the specific condition from log file. Below is the sample log file
Name m/c usage
abc xxx 10
abc xxx 20
abc xxx 5
xyz ... (6 Replies)
I need to sum values in text file in case duplicate row are present with same name and different value below is example of data in file i have and format i need.
Data in text file
20170308
PM,U,2
PM,U,113
PM,I,123
DA,U,135
DA,I,113
DA,I,1
20170309
PM,U,2
PM,U,1
PM,I,123
PM,I,1... (3 Replies)
In the awk below I am trying to add a penalty to a score to each matching $1 in file2 based on the sum of $3+$4 (variable TL) from file1. Then the $4 value in file1 is divided by TL and multiplied by 100 (this valvue is variable S). Finally, $2 in file2 - S gives the updated $2 result in file2.... (2 Replies)
I have two files, file1.table is the count table, and the other is the range condition file2.range.
file1.table
chr start end count
N1 0 48 1
N1 48 181 2
N1 181 193 0
N1 193 326 2
N1 326 457 0
N1 457 471 1
N1 471 590 2
N1 590 604 1
N1 604 752 1
N1 752 875 1
file2.range... (12 Replies)
Discussion started by: yifangt
12 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)