ok this is my whole data:
there is a === before 8th line but neither nawk '/\===/{exit} NR>8{print $3}' file nor nawk 'NR>8{print $3} /\===/{exit}' file gave result
I have a H U G E file with over 1million entries in it.
Looks something like this:
USER0001|DEVICE001|VAR1
USER0001|DEVICE001|VAR2
USER0001|DEVICE001|VAR3
USER0001|DEVICE001|VAR4
USER0001|DEVICE001|VAR5
USER0001|DEVICE001|VAR6
USER0001|DEVICE002|VAR1
USER0001|DEVICE002|VAR2... (4 Replies)
How do I grep/check the on-hand value on the second line of show_prod script below? In this case it's a "3".
So if it's > 0, then run_this, otherwise, quit.
> ./show_prod
Product Status Onhand Price
shoe OK 3 1.1 (6 Replies)
I have an awk script to find the maximum value of the 2nd column of a 2 column datafile, but I need to find the top 5 maximum values of the 2nd column.
Here is the script that works for the maximum value.
awk 'BEGIN { subjectmax=$1 ; max=0} $2 >= max {subjectmax=$1 ; max=$2} END {print... (3 Replies)
Is there an awk script that can easily perform the following operation?
I have a data file that is in the format of
1944-12,5.6
1945-01,9.8
1945-02,6.7
1945-03,9.3
1945-04,5.9
1945-05,0.7
1945-06,0.0
1945-07,0.0
1945-08,0.0
1945-09,0.0
1945-10,0.2
1945-11,10.5
1945-12,22.3... (3 Replies)
Hi,
I need an awk script (or whatever shell-construct) that would take data like below and get the max value of 3 column, when grouping by the 1st column.
clientname,day-of-month,max-users
-----------------------------------
client1,20120610,5
client2,20120610,2
client3,20120610,7... (3 Replies)
I need to be able to search for a string in the first column and if that string exists than replace the nth column with "-9.99".
AW12000012012 2.38 1.51 3.01 1.66 0.90 0.91 1.22 0.82 0.57 1.67 2.31 3.63 0.00
AW12000012013 1.52 0.90 1.20 1.34 1.21 0.67 ... (14 Replies)
Hello Members,
Need your expert opinion how to tackle below.
I have an input file that looks like below:
USS|AWCC|AFGAW|93|70
USSAA|Roshan TDCA|AFGTD|93|72,79
ALB|Vodafone|ALBVF|355|69
ALGEE|Wataniya (Nedjma)|DZAWT|213|50,550
I like output file in below format:
... (7 Replies)
Hello All,
I am writing a shell script with following requirement:
1. I have one input file as below
CHE01,A,MSC,INO
CHE02,B,NST,INC
CHE03,C,STM,INP
2. In shell script I have predefined array as below:
Array1={A, B, C}
Array2= {U09, C04, A054} (6 Replies)
Discussion started by: angshuman
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)