I have a data file like below, where Time is in the second column
There are about 20 or so columns. The data file contains data every 5 seconds (as shown above), I need to find a way to average the data every 30 minutes. Is there any way to do this?
Hello,
I have three columns of data of the format below:
<name> <volume> <size>
a 2 1.2
a 2 1.1
b 3 1.7
c 0.7 1.9
c 0.7 1.9
c 0.7 1.8
What I... (3 Replies)
Hello. Im just starting to learn awk so hang in there with me...I have a large text file formatted as such everything is in a single column
ID001
value 1
value 2
value....n
ID002
value 1
value 2
value... n
I want to be able to calculate the average for values for each ID from the... (18 Replies)
Hello all,
I'm trying to perform an averaging procedure which selects a selection of rows, average the corresponding value, selects the next set of rows and average the corresponding values etc.
The data below illustrates what I want to do. Given two columns (day and value),
I want to... (2 Replies)
Dear all,
I have the data in the following format. I want to do average of each NR= 5 (rows) for all the 3 ($1,$2, $3) columns and want to print average result in another file in the same format. I dont know how to write code for this in 'awk', can some one help me to write a code for this in... (1 Reply)
Hi, I only have a very limited understanding and experience with writing code and I was hoping I could get some help.
I have a dataset of two columns (txt format, numbers in each row separated by a tab)
Eg.
1 5
2 5
3 6
4 7
5 6
6 6
7 ... (5 Replies)
A happy Monday to you all,
I have a .csv file which contains data taken every 5 seconds. I want to average these 5 second data points into 30 minute averages!
date co2
25/06/2011 08:04 8.31
25/06/2011 08:04 8.32
25/06/2011 08:04 8.33... (18 Replies)
Hey all,
I have a set of 5-second data as shown below. I need to find an hourly average of this data.
date co2
25/06/2011 08:04:00 8.30
25/06/2011 08:04:05 8.31
25/06/2011 08:04:10 8.32
25/06/2011 08:04:15 8.33
25/06/2011 08:04:20 ... (5 Replies)
Hi,
I am trying to average the values from 3 files with the same format. They are very large files so I will describe the file and show some it of. Basically the file has 83 columns (with nearly 7000 rows). The first three columns are the same for each file while the remaining 80 are values... (3 Replies)
Hello all, I need to compute a row-wise average of files with a single column based on the pattern of the filenames. I really appreciate any help on this. it would just be very difficult to do them manually as the rows are mounting to 100,000 lines. the filenames are as below with convention as... (2 Replies)
I have a file of the form.
16:00:26,83.33 16:05:26,83.33 16:10:26,83.33 16:15:26,83.33 16:20:26,90.26 16:25:26,83.33 16:30:26,83.33 17:00:26,83.33 17:05:26,83.33 17:10:26,83.33 17:15:26,83.33 17:20:26,90.26 17:25:26,83.33 17:30:26,83.33
For the timestamp 16:00:00 to 16:55:00, I need to... (5 Replies)
Discussion started by: Saidul
5 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)