Use Franklin52's soln if input file is exactly like
In case you input differs a little like
Then use soln in post #4 in above link which is:
It increments cnt value when a new Cell id value is found (Every Cell id value is stored in array once found and if a Cell id value is not found in array, means 1st occurance, then cnt is incremented).
Hi,
I have a file like this:
Some_String_Here 123 123 123 321 321 321 3432 3221 557 886 321 321
I would like to find only the unique values in the files and get the following output:
Some_String_Here 123 321 3432 3221 557 886
I am trying to get this done using awk. Can someone please... (5 Replies)
Hello Everyone
I need your help in fixing this issue., I have a log file which has data of users logging in to an application.
I want to search for a particular pattern in the log
ISSessionValidated=N
If this key word is found , the above 8 lines will contain the name of the user who's... (12 Replies)
Hello Guys
I have a flat file with '|~|' delimited
When I use to record count using below command
awk -FS"+" ' {print $colno}' filename | wc -l
the count is fine
But when I am trying to find the unique number of record the o/p is always 1
awk -FS"+" ' {print $colno}'... (11 Replies)
I need to take the second column of a .csv file and count the number of instances of each unique value in that same second column. I'd like the output to be value,count sorted by most instances. Thanks for any guidance!
Data example:
317476,317756,0
816063,318861,0
313123,319091,0... (4 Replies)
Hello,
I need some sort of way to extract every date contained in a file, and count how many of those dates there are.
Here are the specifics:
The date format I'm looking for is mm/dd/yyyy
I only need to look after line 45 in the file (that's where the data begins)
The columns of... (2 Replies)
Hello,
I`m a complete newbie to coding, please help with this problem.
I have multiple files in a directory, I have to loop through the contents of each file and extract number of unique isoforms in that file. Each file is tab delimited and only the line with the first parent (column 3)... (1 Reply)
Hi, I have tab-deliminated data similar to the following:
dot is-big 2
dot is-round 3
dot is-gray 4
cat is-big 3
hot in-summer 5
I want to count the frequency of each individual "unique" value in the 1st column. Thus, the desired output would be as follows:
dot 3
cat 1
hot 1
is... (5 Replies)
Looking for a little help here.
I have 1000's of text files within a multiple folders.
YYYY/
/MM
/1000's Files
Eg.
2014/01/1000 files
2014/02/1237 files
2014/03/1400 files
There are folders for each year and each month, and within each monthly folder there are... (4 Replies)
Hello experts,
I am converting a number into its binary output as :
read n
echo "obase=2;$n" | bc
I wish to count the maximum continuous occurrences of the digit 1.
Example :
1. The binary equivalent of 5 = 101. Hence the output must be 1.
2. The binary... (3 Replies)
What is an efficient way of counting the number of unique values in a 400 column by 1000 row array and outputting the counts per column, assuming the unique values in the array are:
A, B, C, D
In other words the output should look like: Value COL1 COL2 COL3
A 50 51 52... (16 Replies)
Discussion started by: Geneanalyst
16 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)