CODEWITHSUNDEEP
bashshellawk
user244333
user244333

Reputation:

How to compare two decimal numbers in bash/awk?

I am trying to compare two decimal values but I am getting errors. I used 

if [ "$(echo $result1 '>' $result2 | bc -l)" -eq 1 ];then

as suggested by the other Stack Overflow thread.

I am getting errors.

What is the correct way to go about this?

Upvotes: 34

Views: 91464

Answers (8)

Marek Manduch
Marek Manduch

Reputation: 2491

For shell script I couldn't use double brackets (()). So, what helped me was to split it in two rows and do the comparison in the classic way.

low_limit=4.2
value=3.9
        
result=$(echo "${value}<${low_limit}" | bc)
    
if [ $result = 1 ]; then
  echo too low; 
else 
  echo not too low; 
fi

Upvotes: 1

Timor Kodal
Timor Kodal

Reputation: 364

if [[ `echo "$result1 $result2" | awk '{print ($1 > $2)}'` == 1 ]]; then
  echo "$result1 is greater than $result2"
fi

Upvotes: 9

Samir
Samir

Reputation: 723

Following up on Dennis's reply:

Although his reply is correct for decimal points, bash throws (standard_in) 1: syntax error with floating point arithmetic.

result1=12
result2=1.27554e-05


if (( $(echo "$result1 > $result2" | bc -l) )); then
    echo "r1 > r2"
else
    echo "r1 < r2"
fi

This returns incorrect output with a warning although with an exit code of 0.

(standard_in) 1: syntax error
r1 < r2

While there is no clear solution to this (discussion thread 1 and thread 2), I used following partial fix by rounding off floating point results using awk followed by use of bc command as in Dennis's reply and this thread

Round off to a desired decimal place: Following will get recursive directory space in TB with rounding off at the second decimal place.

result2=$(du -s "/home/foo/videos" | tail -n1 | awk '{$1=$1/(1024^3); printf "%.2f", $1;}')

You can then use bash arithmetic as above or using [[ ]] enclosure as in following thread.

if (( $(echo "$result1 > $result2" | bc -l) )); then
    echo "r1 > r2"
else
    echo "r1 < r2"
fi

or using -eq operator where bc output of 1 is true and 0 is false

if [[ $(bc <<< "$result1 < $result2") -eq 1 ]]; then
    echo "r1 < r2"
else
    echo "r1 > r2"
fi

Upvotes: 5

Zombo
Zombo

Reputation: 1

if awk 'BEGIN{exit ARGV[1]>ARGV[2]}' "$z" "$y"
then
  echo z not greater than y
else
  echo z greater than y
fi

Upvotes: 11

Bernie Bored
Bernie Bored

Reputation: 11

Can't bash force type conversion? For example:

($result1 + 0) < ($result2 + 0)

Upvotes: -1

Pierre Prot
Pierre Prot

Reputation: 11

Why use bc ?

for i in $(seq -3 0.5 4) ; do echo $i ; if [[ (( "$i" < 2 )) ]] ; then echo "... is < 2";fi; done

The only problem : the comparison "<" doesn't work with negative numbers : they are taken as their absolute value.

Upvotes: -2

progz
progz

Reputation: 327

You can also echo an if...else statement to bc.

- echo $result1 '>' $result2
+ echo "if (${result1} > ${result2}) 1 else 0"

(
#export IFS=2  # example why quoting is important
result1="2.3" 
result2="1.7" 
if [ "$(echo $result1 '>' $result2 | bc -l)" -eq 1 ]; then echo yes; else echo no;fi
if [ "$(echo "if (${result1} > ${result2}) 1 else 0" | bc -l)" -eq 1 ];then echo yes; else echo no; fi
if echo $result1 $result2 | awk '{exit !( $1 > $2)}'; then echo yes; else echo no; fi
)

Upvotes: 0

Dennis Williamson
Dennis Williamson

Reputation: 360325

You can do it using Bash's numeric context:

if (( $(echo "$result1 > $result2" | bc -l) )); then

bc will output 0 or 1 and the (( )) will interpret them as false or true respectively.

The same thing using AWK:

if (( $(echo "$result1 $result2" | awk '{print ($1 > $2)}') )); then

Upvotes: 51

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