Hi
I have a input file in the format
ABC,111,2008Q2, 49K
ABC,111,2008Q3, 0K
ABC,111,2008Q4, 0K
ABC,222,2008Q2, 49K
ABC,222,2008Q3, 0K
ABC,222,2008Q4, 0K
XYZ,111,2008Q2, 49K
XYZ,111,2008Q3, 0K
XYZ,111,2008Q4, 0K
XYZ,222,2008Q2, 49K
XYZ,222,2008Q3, 0K
XYZ,222,2008Q4, 0K
The output file... (3 Replies)
Hi guys
Could anyone advise me how to convert my rows into columns from a file
My file would be similar to this:
A11 A12 A13 A14 A15 ... A1n
A21 A22 A23
A31
A41
A51
...
Am1 Am2 Am3 Am4 Am5 ... Amn
The number of rows is not the same to the number of columns
Thanks in advance (2 Replies)
hi,
Apologies if this has been covered.
I have requirement where i have to convert a single column into multiple column.
My data will be like this -
2
3
4
5
6
Output required -
2 3 4 5 6 (1 Reply)
Hi Gurus,
How to convert rows in to columns using linux shell scripting
Input is like (sample.txt)
ABC
DEF
GHI
JKL
MNO
PQR
STU
VWX
YZA
BCD
output should be (sampleoutput.csv)
ABC,DEF,GHI,JKL,MNO
PQR,STU,VWX,YZA,BCD (2 Replies)
Hi Everyone,
Could someone shed some lights on how to convert the records in rows form into column basis.
172.29.59.12
IBM,8255-E8B
102691P
8
65536 MB
6100-04-11-1140
172.29.59.15
IBM,8255-E8B
102698P
4
45056 MB
6100-04-11-1140
IP SYS MODEL ... (6 Replies)
I am looking to print the data in columns and after every 3 words it should be a new row.
cat example.out | awk 'END { for (i = 0; ++i < m;) print _;print _ }{ _ = _ x ? _ OFS $1 : $1}' m=1| grep -i INNER
I am looking to print in a new line after every 3 words.
... (2 Replies)
hi folks,
I have a sample data like what is shown below:
1,ID=1000
1,Org=CedarparkHospital
1,cn=john
1,sn=doe
1,uid=User001
2,uid=User002
2,ID=2000
2,cn=steve
2,sn=jobs
2,Org=Providence
I would like to convert it into the below format:
1,1000,CedarparkHospital,john,doe,User001... (11 Replies)
Discussion started by: vskr72
11 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)