How to round decimals using bc in bash?

Question

A quick example of what I want using bash scripting:

#!/bin/bash
echo "Insert the price you want to calculate:"
read float
echo "This is the price without taxes:"
echo "scale=2; $float/1.18" |bc -l
read -p "Press any key to continue..."
bash scriptname.sh

Assuming that the price is: 48.86 The answer will be:41.406779661 (41.40 actually because I'm using scale=2;)

My Question is: How I round the second decimal to show the answer in this way?: 41.41

Solution 1

A bash round function:

round()
{
echo $(printf %.$2f $(echo "scale=$2;(((10^$2)*$1)+0.5)/(10^$2)" | bc))
};

Used in your code example:

#!/bin/bash
# the function "round()" was taken from 
# http://stempell.com/2009/08/rechnen-in-bash/

# the round function:
round()
{
echo $(printf %.$2f $(echo "scale=$2;(((10^$2)*$1)+0.5)/(10^$2)" | bc))
};

echo "Insert the price you want to calculate:"
read float
echo "This is the price without taxes:"
#echo "scale=2; $float/1.18" |bc -l
echo $(round $float/1.18 2);
read -p "Press any key to continue..."

Good luck :o)

Solution 2

Simplest solution:

printf %.2f $(echo "$float/1.18" | bc -l)

Solution 3

Bash/awk rounding:

echo "23.49" | awk '{printf("%d\n",$1 + 0.5)}'  

If you have python you can use something like this:

echo "4.678923" | python -c "print round(float(raw_input()))"

Solution 4

Here's a purely bc solution. Rounding rules: at +/- 0.5, round away from zero.

Put the scale you're looking for in $result_scale; your math should be where $MATH is located in the bc command list:

bc <<MATH
h=0
scale=0

/* the magnitude of the result scale */
t=(10 ^ $result_scale)

/* work with an extra digit */
scale=$result_scale + 1

/* your math into var: m */
m=($MATH)

/* rounding and output */
if (m < 0) h=-0.5
if (m > 0) h=0.5

a=(m * t + h)

scale=$result_scale
a / t
MATH

Solution 5

Here's an abbreviated version of your script, fixed to provide the output you want:

#!/bin/bash
float=48.86
echo "You asked for $float; This is the price without taxes:"
echo "scale=3; price=$float/1.18 +.005; scale=2; price/1 " | bc

Note that rounding up to nearest integer is equivalent to adding .5 and taking the floor, or rounding down (for positive numbers).

Also, the scale factor is applied at the time of operation; so (these are bc commands, you can paste them into your terminal):

float=48.86; rate=1.18; 
scale=2; p2=float/rate
scale=3; p3=float/rate
scale=4; p4=float/rate
print "Compare:  ",p2, " v ", p3, " v ", p4
Compare:  41.40 v 41.406 v 41.4067

# however, scale does not affect an entered value (nor addition)
scale=0
a=.005
9/10
0
9/10+a
.005

# let's try rounding
scale=2
p2+a
41.405
p3+a
41.411
(p2+a)/1
41.40
(p3+a)/1
41.41

View more solutions

Comments

[blackedx]

Thanks for the tip but it doesn't solve what I need. I have to do it just by using bash...

[Elle]

The python command is more readable and good for quick scripts. Also supports arbitrary digit rounding by adding e.g. ", 3" to the round function. Thanks

[Aquarius Power]

btw, if the number is negative, we have to use -0.5

[syntaxerror]

Couldn't get the example to work on the testing console! "Debugging" a little revealed that for e. g. $2 = 3 I had to use echo $(env printf %.3f $(echo "scale=3;((1000*$1)+0.5)/1000" | bc)). Mind the env before printf! This'll teach ya again that it's always important to understand what you're copy-pasting from elsewhere.

[syntaxerror]

@Aquarius Power thanks for the inspiration! I've now forked a version of the above script which will work with both negative and positive numbers.

[syntaxerror]

@muru about your recent edits: herestrings instead of echo .. |, shell arithmetic, what is the point of echo $()? This is easy to explain: your here strings will always require a writable /tmp directory! And my solution will also work on a read-only environment, e. g. an emergency root shell where / is not always writable by default. So there was a good reason why I coded it that way.

[muru]

Herestrings, I agree. But when is echo $() ever needed? That and indentation prompted my edits, the herestrings just happened.

[Adam Katz]

echo "$float" |awk '{printf "%.2f", $1/1.18}' will perform the question's requested math to the requested percision of hundredths. That's as much "using bash" as the bc call in the question.

[Adam Katz]

This is unnecessarily complicated and does all kinds of extra arithmetic to arrive at the simpler answers provided by migas and zuberuber.

[Six]

For Python you can just use something like python -c "print(round($num))" where num=4.678923. There's no need to muck about with stdin. You can also round to n digits like so: python -c "print(round($num, $n))".

[RobertG]

@AquariusPower +1 yes, it's different for negative numbers. To test for a negative (non-integer!) number, I tried a bit and settled for if [ echo "$1 / 1" | bc` -gt 0 ] ` - is there a more elegant way to check (except parsing the string for "-")?

[RobertG]

As noted also in a comment above (askubuntu.com/a/179949/512213), this would behave incorrectly for negative numbers, unfortunately.

[Jessica P]

@RobertG: Why? This solution doesn't use +0.5. Try printf "%.2f\n" "$(bc -l <<<"48.86/1.18")" "$(bc -l <<<"-48.86/1.18")" and you will get 41.41 and -41.41.

[dessert]

echo $(echo $(printf …)) is unnecessary, just do round(){ printf … ;} and call it with round …. If you want a following linebreak, change %.$2f to %.$2f\\n.

[dessert]

Also possible using a pipe: echo "$float/1.18" | bc -l | xargs printf %.2f

[Admin]

This looks neat, but bcr "5 / 2" 0 returns 2 instead of 3. Am I missing something?

[loic.jaouen]

it's quite funny to call awk or python to call printf, you'd better do just the same without them: bash echo "4.678923" | xargs printf "%.2f"

[Marc Tamsky]

I nominate this for "best answer".

[Torsten Bronger]

The solution in this answer gives me: bash: printf: 1.69491525423728813559: invalid number, depending on locale.

[Suuuehgi]

printf %.$2f already rounds and hence your round-construction redundant. Just take round() { printf %.$2f $1; }.

[x-yuri]

Good luck with paying taxes? :) A detailed explanation of your solution would be totally in order.

[x-yuri]

A detailed explanation would be in order. You could start with what exactly is scale.

[moonlight]

The awk version won't work with negative numbers, but printing a float with zero decimal places will: awk '{printf("%.0f", $1)}'