05-14-2012
The input field separator in awk is called "FS".
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1. Shell Programming and Scripting
Can anybody help me out with this problem
" a shell program that takes one or any number of file names as input; sorts the lines of each file in ascending order and displays the non blank lines of each sorted file and merge them as one combined sorted file. The program generates an error... (1 Reply)
Discussion started by: arya
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2. Shell Programming and Scripting
Hi
I want to merge two or more files using perl in windows only(Just like Paste command in Unix script) . How can i do this.Is ther any single command to do this?
Thanks
Kunal (1 Reply)
Discussion started by: kunal_dixit
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3. Filesystems, Disks and Memory
why do inode indices starts from 1 unlike array indexes which starts from 0
its a question from "the design of unix operating system" of maurice j bach
id be glad if i get to know the answer quickly
:) (0 Replies)
Discussion started by: sairamdevotee
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4. Shell Programming and Scripting
Hello,
I have two txt files that look like this:
db.0.0.0.0:
Total number of NS records = 1
db.127.0.0.0:
Total number of NS records = 1
Total number of PTR records = 1
db.172.19.0.0:
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5. Shell Programming and Scripting
Hi,
I have a very basic knowledge of shell scripting & would like some help with a little problem I have. I sometimes use a program calle phronix & sometimes like to compare its results which are *.xml files. Which is easy enough but a friend wants to avoid typing the path to the files.... (2 Replies)
Discussion started by: ptrbee
2 Replies
6. Shell Programming and Scripting
Hi all,
I hope you can help me.
I got a file a and a file b
File a contains
a
b
c
d
e
f
g
h
File b contains
1
2
3 (8 Replies)
Discussion started by: stinkefisch
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7. Shell Programming and Scripting
Hi All,
I have a file (FileNames.txt) which contains the following data in it.
$ cat FileNames.txt
MYFILE17XXX208Sep191307.csv
MYFILE19XXX208Sep192124.csv
MYFILE20XXX208Sep192418.csv
MYFILE22XXX208Sep193234.csv
MYFILE21XXX208Sep193018.csv
MYFILE24XXX208Sep194053.csv... (5 Replies)
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8. Shell Programming and Scripting
I have two text files.
text file 1:
ID filePath col1 col2 col3
1 10584588.mol 269.126 190.958 23.237
2 10584549.mol 281.001 200.889 27.7414
3 10584511.mol 408.824 158.316 29.8561
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Discussion started by: LMHmedchem
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9. Programming
Hi friends,
I have implemented the merge sort algorith in c, before I put forward my question, you please have a look at my code.
// The array is sorted, as 1234
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int A = {4, 3, 2, 1};
void Merge_Sort(int , int, int);
void... (0 Replies)
Discussion started by: gabam
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10. Shell Programming and Scripting
Hello,
I have 40 data files where the first three columns are the same (in theory) and the 4th column is different. Here is an example of three files,
file 2: A_f0_r179_pred.txt
Id Group Name E0
1 V N(,)'1 0.2904
2 V N(,)'2 0.3180
3 V N(,)'3 0.3277
4 V N(,)'4 0.3675
5 V N(,)'5 0.3456
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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)