In bash how do I divide two variables and output the answer rounded upto 5 decimal digits?

Question

I took two variables as input and after dividing them, I want to get the output rounded up to 5 decimal digits. I have tried this way ->

sum=12
n=7
output=$("scale=5;sum/n"|bc)
echo $output

My code isn't showing any output. What can I do??

TestCase:

If sum=3345699 and n=1000000 then (sum/n)=3.345699, it should be changed into 3.34570.

`sum=12; n=7; output=$(bc<<<"scale=5;$sum/$n"); echo $output` — ymonad, Sep 06 '16 at 08:13

fedorqui · Accepted Answer · 2016-09-06T09:23:44.697

2

The problem here is that you missed the echo (or printf or any other thing) to provide the data to bc:

$ echo "scale=5; 12/7" | bc
1.71428

Also, as noted by cnicutar in comments, you need to use $ to refer to the variables. sum is a string, $sum is the value of the variable sum.
All together, your snippet should be like:

sum=12
n=7
output=$(echo "scale=5;$sum/$n" | bc)
echo "$output"

This returns 1.71428.

Otherwise, with "scale=5;sum/n"|bc you are just piping an assignment and makes bc fail:

$ "scale=5;sum/n"|bc
bash: scale=5;sum/n: No such file or directory

You then say that you want to have the result rounded, which does not happen right now:

$ sum=3345699
$ n=1000000
$ echo "scale=5;($sum/$n)" | bc
3.34569

This needs a different approach, since bc does not round. You can use printf together with %.Xf to round to X decimal numbers, which does:

$ printf "%.5f" "$(echo "scale=10;$sum/$n" | bc)"
3.34570

See I give it a big scale, so that then printf has decimals numbers enough to round properly.

edited Sep 06 '16 at 09:23

answered Sep 06 '16 at 08:21

fedorqui

275,237
103
548
598

1

Don't you have to say `$sum/$n` ? – cnicutar Sep 06 '16 at 08:27
@cnicutar absolutely. Thanks, corrected! – fedorqui Sep 06 '16 at 08:33
@jbsu32 what values do you give to `sum` and `n` to get `(sum/n)=3.345699`? – fedorqui Sep 06 '16 at 09:07
`sum=3345699` and `n=1000000` – jbsu32 Sep 06 '16 at 09:09
"you are just piping an assignment" I don't think that's exactly what happens : since the string is quoted, it is parsed by `bash` as a single token, and since it's the first token of the expression, `bash` tries to interpret it as a command, which explains the `No such file or directory` – Aaron Sep 06 '16 at 09:09
@jbsu32 `bc` does not perform rounding, it just truncates. You can use `printf` with `%.Xf` to round to X decimals. For example, `printf "%.5f" "$(echo "scale=10;$sum/$n" | bc)"`. – fedorqui Sep 06 '16 at 09:15
@fedorqui It works fine! Will you please add it to your answer? – jbsu32 Sep 06 '16 at 09:19
1

@jbsu32 sure, I just did it! – fedorqui Sep 06 '16 at 09:29

score 1 · Answer 2 · answered Sep 06 '16 at 08:33

1

sum and n, these are bash variables. you should add $ to get their values. So, the solution should be:

echo "scale=5;($sum/$n)"|bc
1.71428

answered Sep 06 '16 at 08:33

Ahsanul Haque

10,676
4
41
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P.... · Answer 3 · 2016-09-06T09:31:56.923

awk 'BEGIN{sum=12;n=7;printf "%0.5f\n", sum/n}'
1.71429

In this solution , awk uses printf to round up decimal to 5 places. If you wish to pass bash variables then use following :

 awk -v sum=12 -v n=7 'BEGIN{printf "%0.5f\n", sum/n}'
 1.71429

On side notes, awk seem to be good in arithmetic :

sh-4.1$  time echo "scale=5; 12/7" | bc ;  time echo "scale=5;($sum/$n)"|bc;time awk 'BEGIN{sum=12;n=7;printf "%0.5f\n", sum/n}'
1.71428

real    0m0.004s
user    0m0.001s
sys     0m0.002s
1.71428

real    0m0.004s
user    0m0.001s
sys     0m0.001s
1.71429

real    0m0.002s
user    0m0.001s
sys     0m0.000s
sh-4.1$

In bash how do I divide two variables and output the answer rounded upto 5 decimal digits?

3 Answers3

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