Bash Rounding to 2 decimal places


 
Thread Tools Search this Thread
Top Forums Shell Programming and Scripting Bash Rounding to 2 decimal places
Prev   Next
# 1  
Old 03-10-2013
Bash Rounding to 2 decimal places

I have a number in a bash variable n, and want to round it to 2 decimal places. How can I do that?

Code:
n=0.0867268

Need to have
Code:
num=0.09

 
Login or Register to Ask a Question

Previous Thread | Next Thread

9 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

Sum the fields with 6 decimal places - getting only 2 decimal places as output

I used the below script to Sum up a field in a file based on some unique values. But the problem is when it is summing up the units, it is truncating to 2 decimals and not 6 decimals as in the input file (Input file has the units with up to 6 Decimals – Sample data below, when the units in the 2... (4 Replies)
Discussion started by: brlsubbu
4 Replies

2. UNIX for Dummies Questions & Answers

Rounding off a decimal

How to round off a decimal number to higher whole number using ceil command in unix? Eg. 4.41 or 4.11 or 4.51 should be rounded off to 5. (11 Replies)
Discussion started by: SanjayKumar28
11 Replies

3. Shell Programming and Scripting

Rounding decimal values in a column

Hi, I wanted to round all the values in a column to nearest integer. I have multiple files with only two columns and I want to round values in column 2. e.g input_file A1 23.971578 A2 34.624976 A3 46.403446 A4 375 A5 1 A6 3 A7 ... (3 Replies)
Discussion started by: ashu0001
3 Replies

4. Shell Programming and Scripting

Rounding off decimal values

Hi Friends, This is my last post for today. My input file is chr1 100 200 chr1 123 300 chr1 300 400 chr1 420 520 chr10 132344343 132348674 When I try using this command awk '{v=($3+$2)/2; print $0"\t"v}' 1 This is my output chr1 100 200 150 chr1 123 300 211.5 (2 Replies)
Discussion started by: jacobs.smith
2 Replies

5. Shell Programming and Scripting

Arithmetic but keep 2 decimal places

I am trying to perform arithmetric, for example, to increment the value of variable $a (say 3) by 0.05 but when I tried the following expression let a=a+0.05 or a=$((a+0.05)) both returned 3.0499999999999998 I want to keep 2 decimal places so it returns 3.05 instead. (6 Replies)
Discussion started by: piynik
6 Replies

6. Shell Programming and Scripting

Decimal places

i need to multiplay a number with 1.00.. so that the output should contain two decimal places at end.. for example... 236 * 1.00 = 236.00 245.8 * 1.00 = 245.80 but when i perform multiplication it shows output as. 236 245.8 can anyone help me to get the actual output of... (11 Replies)
Discussion started by: arunmanas
11 Replies

7. Shell Programming and Scripting

Four decimal places with awk

i have a script in which awk prints "($2-1700)/10000" and the answer is -0.07,but i want the answer in 4 decimal places. that is -0.0700. How can i sue awk to get my results in four decimal places (4 Replies)
Discussion started by: tomjones
4 Replies

8. UNIX for Dummies Questions & Answers

Rounding a decimal

Hi, I am currently using tcsh I am trying to round a decimal number to the ten-thousandths place For instance: 1.23456 is rounded up towards 1.2346 I am not looking for truncation, but for rounding. Anyone know how to do this with awk or expr? Thanks (2 Replies)
Discussion started by: miniwheats
2 Replies

9. Shell Programming and Scripting

trimm up the decimal places in output

I have a perl script that reports the avg time of a application call and the total number of calls. This works fine, however I would like to trim the number of decimal places reported from 12 to like 3 and I don't know how. Any suggestions? Here is what I use to get the avg time... for $eRef (... (2 Replies)
Discussion started by: theninja
2 Replies
Login or Register to Ask a Question
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)