Hi,
I have a space delimited text file that looks like the following:
Aa 100 200
Bb 300 100
Cc X 500
Dd 600 X
Basically, I want to take the average of columns 2 and 3 and print it in column 4. However if there is an X in either column 2 or 3, I want to print the non-X value. Therefore... (11 Replies)
Hi, I need help with the awk command.
I have a folder with aprox 500 files each one with two columns and I want to print in a new file, the average of column 1 and average of column 2 and the name of each file.
Input files are:
File-1:
100 99
20 99
50 99
50 99
File-2:
200 85... (3 Replies)
I have a dataset with 120 columns. I would like to write a script, that takes the average of every two columns, starting from columns 2 and 3, and moving consecutively in frames of 3 columns, all the way until the last column.
The first column in the output file would be the averages of columns... (1 Reply)
Hi experts,
I want to group by average, for multiple columns starting column $7 until NF,
group by ($1-$5), please help
For just 7th column, I can do
awk '
NR>1{
arr += $7
count += 1
}
END{
for (a in arr) {
print a, arr/count
... (10 Replies)
I have the need to match the first two columns and when they match, calculate the percent of average for the third columns. The following awk script does not give me the expected results.
awk 'NR==FNR {T=$3; next} $1,$2 in T {P=T/$3*100; printf "%s %s %.0f\n", $1, $2, (P>=0)?P:-P}' diff.file... (1 Reply)
I have the following format of input from multiple files
File 1
24.01 -81.01 1.0
24.02 -81.02 5.0
24.03 -81.03 0.0
File 2
24.01 -81.01 2.0
24.02 -81.02 -5.0
24.03 -81.03 10.0
I need to scan through the files and when the first 2 columns match I... (18 Replies)
Hi Friends,
I have files with columns like this. This sample input below is partial.
Please check below for main file link. Each file will have only two rows.
... (8 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)
I have files that have the following columns
chr pos ref alt sample 1 sample 2 sample 3
chr2 179644035 G A 1,107 0,1 58,67
chr7 151945167 G T 142,101 100,200 500,700
chr13 31789169 CTT CT,C 6,37,8 0,0,0 15,46,89
chr22 ... (3 Replies)
Discussion started by: nans
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)