Hi Friends
How to do sum on a column?
I have a file like:
FRED 500.01 TX
SMITH 50.10 NY
HARRY 5.00 CA
555.11
Sum on second column.
I am trying using nawk like
nawk 'BEGIN {FS="|"}; {printf $1"+"}'
Thanks a lot for your help
S :) (2 Replies)
I have a text file with two columns
the first column is an integer and the second column is date
how do i sum up the first column according to the date
example
123 jan1
232 jan1
473 jan2
467 jan2
356 jan3
376 jan3
my result should be
355 jan1
940 jan2
732 jan3
how do i... (2 Replies)
Dear all,
I have a rather large file of numbers which i would like to read into a script and then do some maths on a specific column( e.g column).
so far i have been using the following awk command
awk '{print $4}' infile.txt > out.tmp to strip out the desired column within the in perl... (3 Replies)
I have a file which contains data as below:
-----------------------------------------------------------------------------------------------
GSPWeb Statistics for the period of last 20 days... (3 Replies)
Hi,
I am writing a comparator script, which comapre two txt files(column by column)
below are the precondition of this comparator
1)columns of file are not seperated
Ex.
file1.txt
8888812341181892
1243548895685687
8945896789897789
1111111111111111
file2.txt
9578956789567897... (2 Replies)
Dear all,
I have one file like
LABEL A B C D E F G H I J K L M N
G02100 64651.3 25630.7 8225.21 51238 267324 268005 234001 52410.9 18598.2 10611 10754.7 122535 267170 36631.4
G02100 12030.3 8260.15 8569.91 ... (4 Replies)
a,b,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z,aa,bb,cc,dd,ee,ff,gg,hh,ii
a thru ii are digits and strings....
The awk needed....if coloumn 9 == i (coloumn 9 is string ), output the sum of x's(coloumn 22 ) in all records and sum of y's (coloumn 23 ) in all records in a file (records.txt).... (6 Replies)
hi All,
i have a file in which only one column is there.,
test.txt
======
-900.01
-900.02
-900.03
-900.04
-900.05
-900.06
-900.07
-900.08
-900.09
900.01
900.02
900.03
900.04
900.05 (4 Replies)
Hello,
I am facing issue in summing up a column in unix.I am displaying a column sum up to 4 decimal places and below is the code snippet
sed '1d' abc.csv | cut -d',' -f7 | awk '{s+=$1}END{ printf("%.4f\n",s)}'
-170552450514.8603
example of data values in the column(not... (3 Replies)
Discussion started by: karthik adiga
3 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)