To calculate average by taking command line argument

Question

The task is that if input numbers:

Output=(sum of all except first_number_in_series)/first_number in series Then output average should be 1+2+9+8/(first_number_in series)=20/4=5

I have tried the following code, however not able to achieve the task. I will be grateful if anyone can point out the mistake.

#!/bin/bash
sum=0
count=1
for x in $*
do
    if [ $count -eq 1 ]
    then
        p=$x
    else
        sum=$(($sum + $x))
    fi
    ((count++))
done
echo "scale=3;$sum/$p" | bc

For me, it looks like you include first number in your average, and it's the main problem. I'm going to check if I'm right. — GingerPlusPlus, Feb 07 '16 at 18:59
I think the problem may be elsewhere; your script does what you want, but maybe you're calling it differently? It's expecting the numbers to be passed to it while your answer below specifically reads from stdin and ignores any numbers passed to it. Show us the error you're getting. — Jeff Schaller, Feb 07 '16 at 19:51
Feels like it would be easier to `shift` p off before starting to do the sum. — azzid, Feb 07 '16 at 19:57

score 2 · Answer 1 · answered Feb 07 '16 at 21:05

Since you need floating point calculation, you will end up using bc, or awk, anyways. Why not use Awk to solve the whole problem? Here is a Awk only solution, I used n for numerator and d for denominator:

$ printf "4\n1\n2\n9\n8\n" | awk '{if (NR == 1) {d = $0}; if (NR != 1) {n += $0}} END{printf "%.03f\n", n/(d*1.0)}'
5.000

score 1 · Answer 2 · answered Feb 26 '20 at 07:18

1

A simple POSIX solution:

average() { printf '%s\n' 'scale=2' "($*)/$#" | tr ' ' + | bc;}

Then run it like so:

average 4 1 2 9 8

Output: 4.80

answered Feb 26 '20 at 07:18

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eduardogr · Answer 3 · 2016-02-07T20:01:19.923

0

You were including the first number in the sum and you wrote a bad condition:

if [ count -eq "1" ]

instead of

if [ $count -eq 1 ]

$ operator let you access to a variable and you were using 1 as a String instead of a integer.

#!/bin/bash
read n
p=$n
sum=0
count=1
while [ $count -le $p ]
do
    read n
    x=$n
    count=$(($count + 1))
    sum=$(($sum + $x))
done
result=`echo $sum $p | awk '{printf "%.3f", $1/$2 }'`
echo $result

edited Feb 07 '16 at 20:01

answered Feb 07 '16 at 19:00

eduardogr

215
1
7

sir output with this code is syntax error in standard input – frp farhan Feb 07 '16 at 19:03
Compiler Message Wrong Answer Error (stderr) (standard_in) 2: syntax error – frp farhan Feb 07 '16 at 19:04
I'm executing it in my shell right now, it's working... I just added an else branch.. – eduardogr Feb 07 '16 at 19:05
pls check the same code by putting on the challenge page https://www.hackerrank.com/challenges/bash-tutorials---compute-the-average it gives the error @GingerPlusPlus u also check sir – frp farhan Feb 07 '16 at 19:07
The problem is with the hackerank interpreter... something happens with bc... – eduardogr Feb 07 '16 at 19:26
I'm doing it with awk right now but hackerrank can't do it... i don't know why, but it's working on my computer... – eduardogr Feb 07 '16 at 19:37
Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/35415/discussion-between-edugr-and-farhan-patel). – eduardogr Feb 07 '16 at 19:43
It's done... it's working on hackerrank i got the challenge :) – eduardogr Feb 07 '16 at 20:02
using which solution its working in hackerrank interpreter???@EduGR – frp farhan Feb 08 '16 at 03:06
The solution of my comment...i updated it and i got the challenge in Hackerrank – eduardogr Feb 08 '16 at 11:12
yes gud it is amazing....thnx for the concept.... – frp farhan Feb 08 '16 at 11:38

score 0 · Accepted Answer · 2016-02-08T06:00:47.600

0

You failed to say that the numbers are given to the script in the stdin.

For that, this code will work:

#!/bin/bash
readarray -t x
count="${x[0]}" ; unset x[0]
for y in ${x[@]}; do (( sum+=y )); done
a="$(echo "scale=8; $sum/$count" | bc)"
LC_ALL=C printf '%0.3f\n' "$a"

Test it as this:

$ printf '%s\n' 4 1 2 9 8 | ./script
5.000
$ printf '%s\n' 6 1 2 9 8 13 25 | ./script
9.667

edited Feb 08 '16 at 06:00

answered Feb 07 '16 at 23:40

What is the difference between stdin and normal inputs?? I am new to bash, it will be greatful if u can clearify this for me...thnx – frp farhan Feb 08 '16 at 03:37
Also, in the above code if the output is 9.6666, then answer is expected to be 9.667,then how to achieve this task in ur testcase 2 wherein the answer is 9.6666 how to display rounded off value such as 9.667? – frp farhan Feb 08 '16 at 03:40
The difference is like the difference between this two lines: `./script 1 2 3 4` and `echo 1 2 3 4 | ./script`. The first supply the arguments in the array of "positional arguments" (similar to c ARGV[]), the second supply the values in the stdin input (no ARGV is set). – Feb 08 '16 at 04:42
Rounding is a very slippery slope. You can get what you ask for with printf: `LC_ALL=C; a=9.666666; printf '%0.3f' "$a"` (output 9.667), but even that is bound to fail in some other corner cases. – Feb 08 '16 at 04:48
In which cases will it result into failure??? and in our code, where shall i include printf '%0.3f'?? – frp farhan Feb 08 '16 at 05:32
[Read this](https://unix.stackexchange.com/questions/240112/weird-float-rounding-behavior-with-printf) (as one example) and code edited in the answer. Something else still? – Feb 08 '16 at 06:03
thnx dear....for further any queries will surely discuss with u...thnx a lot...wish u a vry happy and prosperous future ahead!!! also for sed and awk utilities pls guide me any good material or reference from which i can prepare them...thnx once again – frp farhan Feb 08 '16 at 06:22

To calculate average by taking command line argument

4 Answers4