Blasted data inputters :mad: they always have to screw my data up....My comma delimited file with three fields ( firstname,surname and address ) has been screwed up by people entering addresses like this (putting a comma in between the house number and the street name)
142,Stonewall Avenue
... (8 Replies)
Dear all,
I want below data to make it in column format.so i will see the data like this
cdrID teleServiceCode chargedPartyNumber ... ... ... ...
"egmailcom0w10ggzx00" 'sMS (5)' "716323770"
"m17ifi5z30w0z6o7200" 'sMS (5)' ... (7 Replies)
HI,
Need awk command to get date and time alone from
Input : "15:29:15 28.08.2010|SCHEDULE: Started program POSG1"
Output expected : "15:29:15 28.08.2010"
Please help. (9 Replies)
I have a file containing about 5 million rows, in the file there are some records which has extra delimiter at random position. (we dont know the positions), now we have to Count the delimeter from each row and if the count of delimeter is not matching then I want to delete those rows from the... (5 Replies)
Hi,
I have 2 files.
a.txt & b.txt
# a.txt contains the following text.
apple
grapes
# b.txt contains the following text.
banana
pine
My question is. (1 Reply)
I have a SQL query
SELECT
BLAH_ID,
BLAH_CODE,
DATEFORMAT(CALENDAR_DATE, 'MM-DD-YYYY') AS CALENDAR_DATE_STR,
HOURS,
'N',
FROM blah_tmp_tbl order by CALENDAR_DATE_STR,BLAH_ID,BLAH_CODE;
OUTPUT TO 'MyExport.CSV' quote '' FORMAT ASCII;
That gets me the below output;
... (2 Replies)
Hi Friends,
I have a file1.txt as below
29123973Ç2012-0529Ç35310124Ç000000000004762Ç00010Ç20Ç390ÇÇÇÇF
29123974Ç20120529Ç35310125Ç0000000000046770Ç00010Ç20Ç390ÇÇÇÇF
29123975Ç20120529Ç35310126Ç0000000000046804Ç00010Ç20Ç390ÇÇÇÇF
29123976Ç20120529Ç35310127Ç0000000000044820Ç00010Ç20Ç390ÇÇÇÇF
i have a file2.txt... (4 Replies)
Hi all I have 7 words ina file called "lookupfile"
CAD
CD
HT
RA
T1D
T2D
BD
in other file I have data like this in which columns are seaprated by comma but the names among above seven names are in one column menas comma between these seven words doesnt mean that they are separated by... (9 Replies)
Need awk solution. Please advise.
inputfile.txt
1,NY, 1111
2,MI, 222
3,NY,333
4,OH,444
5,OH,555
mapping.txt
NY NYNY
IL ILLINOIS
OH OHIO
Need to write a code which will compare 2nd column of inputfile.txt with mapping file and redirect output based on the... (2 Replies)
Discussion started by: vegasluxor
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)