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1. Shell Programming and Scripting
Hello,
i need help with awk.
I have this file:
cat number
DirB port 67 er_enc_out 0 er_bad_os 0
DirB port 71 er_enc_out 56 er_bad_os 0
DirB port 74 er_enc_out 0 er_bad_os 0
DirB port 75 ... (4 Replies)
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2. Shell Programming and Scripting
Oracle Enterprise Linux 6
This is my file. Two fields separated by space
$ cat testfile.txt
MARCH9 MARCH4
MARCH1 MARCH5
MARCH2 MARCH326
MARCH821 MARCH7
MARCH6 MARCH2
$
$
The following numeric sort, based on the first field's 6th character works as expected.
$
$ sort -n -k 1.6... (7 Replies)
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3. Shell Programming and Scripting
the below is sorted as it is. the fields that i'm interested in are the 4th and 5th field.
i want to sort the based on the 4th field.
my past attempt to do this was to do something like this:
awk '{print $4}'| awk '{print $1":"$2}' datafile | sort | uniq
however, if i do that, i lose... (2 Replies)
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4. Shell Programming and Scripting
I'm trying to update a text file via sed/awk, after a lot of searching I still can't find a code snippet that I can get to work.
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5. Programming
I have a vector of strings that contain a list of channels like this:
101,99,22HD,432,300HD
I have tried using the sort routine like this:
sort(mychans.begin(),mychans.end());
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6. Programming
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7. Shell Programming and Scripting
Hi,
I have a file that has data in it that says
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00:09:01.221 5678
00:12:23.321 93444
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8. Shell Programming and Scripting
Hi all,
Sorry the title is a mess, but did not find a better description at the time.
So here is my problem:
I have an input file:
8:Mass40s -- 0
48:Mass40s -- 0
67:Mass40s -- 0
86:Mass40s -- 0
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9. Programming
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10. Shell Programming and Scripting
Hello
I am trying to sort on the last field and it seems to have lost ideas on what to do. the file am sorting looks like this
Tan\da\1223
hey\1234
two\three\think\4579
i want to sort on the last fields (1223, 1234 and 4579).
thank you (2 Replies)
Discussion started by: ganiel24
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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)