awk '/E_TIM/ {CNT++}
{LN[CNT]; HD[$1]; MX[CNT,$1]=$2}
END { for (i in HD) printf "%10s", i; print "";
for (j in LN) {for (i in HD) printf "%10s", MX[j,i]; print ""}
}
' file
ETE E_TIM E_1 EOND E_2 E_3 E_4 EEC
15, 16, un, 26, un, un, 284, 81,
15, 17, un, 29, un, un, 249, 82,
A file content have
1 1:-0.289433 2:0.833778 3:0.314471 4:-0.289433 5:-0.81876 6:-0.456693 7:-0.17511 8:-0.644555 9:-0.00666341 10:-1.13603
I will like to have that column into row with numbers to be printed (red color) only after colon
output shud be like that
-0.289433... (1 Reply)
hi,
I have a requirement where in I read the values from a file using awk. The resulting data should be converted into row format from column format.
For ex: My log file login.lst contains the following
SERVER1 DB1
SERVER2 DB2
SERVER3 DB3
SERVER4 DB4
I use awk to grep only the server... (6 Replies)
Hi Friends
I have the following input data in 2 columns.
SNo 1
I1 Value
I2 Value
I3 Value
SNo 2
I4 Value
I5 Value
I6 Value
I7 Value
SNo 3
I8 Value
I9 Value
...............
................
SNo N (1 Reply)
Hello,
I have an input file like the following:
11_3_4
2_1_35
3_15__
_16989
Where '_' is a space. The data is in a table. Is there a way for the program to prompt the user for x1,y1 and x2,y2, where x1,y1 is the desired number (for example x=6 y=4 is a value of 4) and move to a desired spot... (2 Replies)
Hi,
I have a file like this
50 1 2 1374438
50 1 2 1682957
50 5 2 1453574
50 10 2 1985890
100 1 2 737307
100 5 2 1660204
100 10 2 2148483
and I want to convert this by... (1 Reply)
Hi,
i need to convert
SG_ERP1
SG_ERP2
SG_ERP3
in to:
SG_ERP1 SG_ERP2 SG_ERP3
It's possibile? (16 Replies)
Discussion started by: elilmal
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)