Hi
I have a file like
a,1
b,2
d,3
a,2
b,3
c,7
Result Desired:
a,3
b,5
d,3
c,7
i.e on the bases of 1st field the addition is done of the 2nd field and result printed out. (3 Replies)
Hi
I need to do some thing like "find and insert before that " in a file which contains many records. This will be clear with the following example.
The original data record should be some thing like this
60119827 RTMS_LOCATION_CDR INSTANT_POSITION_QUERY 1236574686123083rtmssrv7 ... (8 Replies)
Hi, my requirement is to sum values in a row.
eg:
input is: sum,value1,value2,value3,.....,value N
Required Output: sum,<summation of N values>
Please help me... (5 Replies)
Hi everyone,
I'm just wondering how could I using awk language merge two files by comparison of one their row.
I mean, I have one file like this:
file#1:
21/07/2009 11:45:00 100.0000000 27.2727280
21/07/2009 11:50:00 75.9856644 25.2492676
21/07/2009 11:55:00 51.9713287 23.2258072... (4 Replies)
HI,
I need to arrange values of a colum in row.
e.g.
input file :
Alpha<>123
AAAA<>6754
Beta<>456
BBBB<>63784
CCC<>783
Gama<>789
Alpha<>555
AAAA<>6754
BBBB<>63784
Beta<>666
CCC<>783
Gama<>888 (9 Replies)
hi
i have an input file in which there are diffrent values for xxxx,yyyyyy,zzzzzzz how can i arrange the dynamic values of x,y&z in a row.
input file:
xxxxx 1
yyyyyy 4
yyyyyy 5
zzzzzzzz 7
yyyyyy 13
zzzzzzzz 7
zzzzzzzz 6
yyyyyy 14
yyyyyy 12
zzzzzzzz 4
yyyyyy 4
yyyyyy 5
yyyyyy 6... (6 Replies)
Hi all,
I want to compute for the average of a file with null values (NaN) for each row. any help on how to do it. the sample file looks like this.
1.4 1.2 1.5 NaN 1.6
1.3 1.1 NaN 1.3 NaN
2.4 1.3 1.5 NaN 1.5
NaN 1.2 NaN 1.4 NaN
I need to do a row-wise averaging such that it will sum only... (14 Replies)
Hello guys.
Please can you help me with this.. Thanks in advance:b:
Input file
2134 6371 N
2150 6371 M
2166 6371 S
2138 6417 N
2154 6417 M
2170 6417 S
2157 6603 N
2173 6603 M
2189 6603 S
desired uotput
6371 2134N 2150M 2166S
6417 2138N 2154M 2170S... (2 Replies)
Hello out there,
file.txt:
comp51820_c1_seq1 42 N 0:0:0:0:0:0 1:0:0:0:0:0 0:0:0:0:0:0 3:0:0:0:0:0 0:0:0:0:0:0
comp51820_c1_seq1 43 N 0:0:0:0:0:0 0:1:0:0:0:0 0:0:0:0:0:0 0:3:0:0:0:0 0:0:0:0:0:0
comp51820_c1_seq1 44 N 0:0:4:0:3:1 0:0:1:9:0:0 10:0:0:0:0:0 0:3:3:2:2:6 2:2:2:5:60:3... (16 Replies)
Discussion started by: pathunkathunk
16 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)