Hi,
I am new to awk/unix and am trying to put together an awk script to perform an action similar to vlookup between the two csv files.
Here are the contents of the two files:
File 1:
Date,ParentID,Number,Area,Volume,Dimensions
2014-01-01,ABC,247,83430.33,857.84,8110.76... (9 Replies)
Hi,
Is there possible to do vlookup in Mysql one table from another table based on one column values and placed the data in same table?
if it is possible in mysql itself pls share links for reference.
Here is the ex: i need to vlookup the cus.id in table to and place the cus.name in 4th... (3 Replies)
Hi folks,
awk 'NR==FNR {m=$0; next} $1 in m{$0=m} {print}' file2 file1
Works a charm for a vlookup type query, sourced from https://www.unix.com/shell-programming-and-scripting/215998-vlookup-using-awk.html
However my column content has white spaces and numbers. Example
file1
The Man... (6 Replies)
Code used to find the server from cloum 3 and update needtotakesnap
Output came from above command
awk 'NR==FNR{A;next}$3 in A{$3 = "needtotakesnap " $3}1' /home/Others/active-server.txt /home/Others/all-server |grep server1
879 dummy server1_217_silver dummy 00870 TDEV 2071575
831 Tier1... (3 Replies)
I need to vlookup and check the server not found.
Source file 1
server1
server2
server3
server4
server5_root
server6_silver
server7
server7-test
server7-temp
Source file 2
server1_bronze (6 Replies)
Hello Folks,
What I wish to do is:
If first column matches in main and new file, then
paste $COL2 into output file. Something like vlookup. Please see also bold text in expected output.
mainfile
11
22
33
44
55
66
77
88
99
100
101
102 (4 Replies)
Hello, i am trying to print group name column(etc/group) on script (etc/passwd) since group name is not listed on etc/passwd columns. Im trying to do a vlookup. but i cant figure out how i can insert the vlookup command FNR==NR inside the print out command or the output. I also tried exporting... (2 Replies)
Hi I just want again to ask for help on what command to use to vlookup f1 group name in "/etc/group" matching f3 of it to "/etc/passwd" f4. I do need to display group name in the output of /etc/passwd without using awk or NR==FNR command. thank you
while
IFS=: read -r f1 f2 f3 f4 f5 f6 f7... (4 Replies)
Discussion started by: joonisio
4 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)