Hi ,
I am using SUN OS Version 5.6.
I have a file that contains records of length 270. when I do 'set nu' in vi editor, I get the count as 86. whereas when I do "wc -l" on the command prompt, it shows the count as only 85. this is very strange. why would the 'wc' show 1 record less. The job... (3 Replies)
Greetings,
I need to search and count all the occurences of a word in all the files in a directory.
Any suggestions greatly appreciated.
Thanks (1 Reply)
Hi all
Can anybody suggest me, how to get the count of digits in a word
I tried
WORD=abcd1234
echo $WORD | grep -oE ] | wc -l
4
It works in bash command line, but not in scripts :mad: (12 Replies)
I want to count the number of occurences of say "200" in a file but that file also contains various stuff including dtaes like 2007 or smtg like 200.1 so count i am getting by doing grep -c "word" file is wrong
Please help!!!!! (8 Replies)
I am a newbie in UNIX shell script and seeking help on this UNIX function. Please give me a hand. Thanks.
I have a large file. Named as 'MyFile'. It was tab-delmited. I am told to write a shell function that counts the number of occurrences of the ord “mysring” in the file 'MyFile'. (1 Reply)
I'm trying to count the number of times each word in the file exist
for example if the file has:
today I have a lot to write, but I will not go for it. The main thing is that today I am looking for a way to get each word in this file with a word count after it specifying that this word has... (4 Replies)
I have the following string:
31-01-2012, 09:42:37;OK;94727132638;"Mozilla/5.0 (Linux; U; Android 2.2.1)";3G;WAP;I need a script which is counting the occurrence of semicolons ( ; ) but exclude the ones from the quotation marks.
In the string given as example there are 8 semicolons but the script... (3 Replies)
example:
i have the following text file...
i am very tired.
i am busy
i am hungry
i have to find the number of occurence of a particular word 'am' from the text file.. can any one give the shell script for it (34 Replies)
Hello,
I have a programming assignment to count number of occurrences of hours in particular file. Below is the code:
fname = raw_input("Enter file name: ")
if len(fname) < 1 : fname = "mbox-short.txt"
largest = None
fh = open(fname)
counts = dict()
test = list()
for line in fh:
... (2 Replies)
Discussion started by: infinitydon
2 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)