Hi,
I have the data like this
$1 $2
1 12
2 13
3 14
4 12
5 12
6 12
7 13
8 14
9 12
10 12
i want to compute average of $1 and $2 every 5th line (1-5 and 6-10)
Please help me with awk
Thank you (4 Replies)
Hi
I am looking for an awk script which can compute average of all the fields every 5th line. The file looks:
A B C D E F G H I J K L M
1 18 13 14 12 14 13 11 12 12 15 15 15
2 17 17 13 13 13 12 12 11 12 14 15 14
3 16 16 12 12 12 11 11 12 11 16 14 13
4 15 15 11 11 11 12 11 12 11... (6 Replies)
Hello,
Let's assume I have 100 files FILE_${m} (0<m<101). Each of them contains 100 lines and 10 columns.
I'd like to get in a file called "result" the average value of column 3, ONLY between lines 11 and 17, in order to plot that average as a function of the parameter m.
So far I can compute... (6 Replies)
I want to calculate the average line by line of some files with several lines on them, the files are identical, just want to average the 3rd columns of those files.:wall:
Example file:
File 1
001 0.046 0.667267
001 0.047 0.672028
001 0.048 0.656025
001 0.049 ... (2 Replies)
For the data
I would like to parse down and for each parsing
I want a cumulative averaging, stored in an array
that can be output.
I.e.
546/NR = 546
(546+344)/NR=(546+344)/2 = etc.
For N record input I want N values of the average (a block
averaging effectively)
Any... (3 Replies)
In the below awk I am trying to combine all matching $4 into a single $5 (up to the -), and count the lines in $6 and average all values in $7. The awk is close but it seems to only be using the last line in the file and skipping all others. The posted input is a sample of the file that is over... (3 Replies)
Hi forum members,
I'm trying to get an average of multiple columns in a csv file using awk. A small example of my input data is as follows:
cu,u3o8,au,ag
-9,20,-9,3.6
0.005,30,-9,-9
0.005,50,10,3.44
0.021,-9,8,3.35
The following code seems to do most of what I want
gawk -F","... (6 Replies)
Hello Team,
I am using the following awk script to calculate the SMA (Single Moving Average) for an specific period but now I would like to include the standard deviation output.
Could you please help me to modify this awk shell script
awk -F, -v points=5 ' { a = $2; ... (4 Replies)
Discussion started by: csierra
4 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)