Hello. I just found out about awk, and it appears that this could handle the problem I'm having right now.
I first stumbled on the thread How to extract first and last line of different record from a file, and that problem is almost similar to mine.
In my case, an ASCII file will contain the... (0 Replies)
hi there
I'm very new in programing and i've started with awk.
I'm processing 200 data files and I need to do some precessing on them.
The files have 3 columns with N-lines
for each line a have on the first and second value is the same for all the files and only the third is variable. like... (2 Replies)
Hi Guys,
I have a text file with ";" like separator
F1;F2;F3;F4;F5
444;100041;IT;GLOB;1800000000
444;100041;TM;GLOB;1000000000
444;10300264;IT;GLOB;2000000000
444;10300264;IT;GLOB;2500000000
I have to sum the cullums F5 for same F2 and F3 collums
The result must be:
... (7 Replies)
Hi everyone
I am very new at awk but think that that might be the best strategy for this. I have a matrix very similar to a correlation matrix and in practical terms I need to convert it into a list containing the values from the matrix (one value per line) with the first field of the line (row... (5 Replies)
My file is something like this :
03.097
03.094
03.093
03.095
03.091
04.089
06.093
07.225
08.196
06.097
06.094
05.096
04.086
I'd like to sum it from a given line to another one , e.g.: from line 10 until line 20
What s the awk way solving this ? (1 Reply)
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, every one. I have two files ,one is in matrix like this, one is a list with the same data as the matrix.
AB AE AC AD AA AF
SA 3 4 5 6 4 6
SC 5 7 2 8 4 3
SD 4 6 5 3 8 3
SE 45 ... (5 Replies)
Hi
I would like to know if it is possible to sum some specific fields.
I have this
x;x;x;x;x;x;x;x;467,390,611 Bytes;0.435291 GB;0.062247 GB;0.373045 GB;11,225;157
a;a;a;a;a;a;a;a;13,805,156,846 Bytes;12.857054 GB;1.838559 GB;11.018495 GB;151,063;18,933
b;b;b;b;b;b;b;b;232,797,478,723... (5 Replies)
I'm trying to sum a text file using AWK. Here is an example of the file:
600|3H68| 46
600|3H69| 46
600|3H6F| 290
600|3H6G| 24
600|3HDY| 1
600|3HDY| 3
600|3HE0| 1
600|3HE0| 3
I would like to sum the third field if the first... (7 Replies)
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)