Simple math statements in bash in a for loop

问题:

I'm quite new to bash scripting and usually avoid it all costs but I need to write a bash script to execute some simple things on a remote cluster. I'm having problems with a for loop that does the following:

for i in {1..20}
do
    for j in {1..20}
    do
        echo (i*i + j*j ) **.5  <--- Pseudo code!
    done
done

Can you help me with this simple math? I've thrown $'s everywhere and can't write it properly. If you could help me understand how variables are named/assigned in bash for loops and the limitations of bash math interpretation (how do you do the square root?) I'd be very grateful. Thanks!

回答1:

Here's a decent solution:

for i in {1..20}
do
   for j in {1..20}
   do
       echo "scale = 3; sqrt($i*$i + $j*$j)" | bc
   done
done

Output will be:

1.414
2.236
3.162
2.236
[...etc...]

回答2:

Arithmetic expansion needs $((...)) notation, so something like:

echo $((i*i + j*j))

However, bash only uses integers so you may need to use an external tool such as dc.

E.g.

dc -e "18k $i $i * $j $j * + v p"

回答3:

Shell math can be done in several ways.

echo $(( i*i + j*j ))
echo $[ i*i + j*j ]
expr "$i" '*' "$i" '+' "$j" '*' "$j"

However, this can only handle integer arithmetic. Instead, you can use bc:

echo "scale = 5; sqrt( $i*$i + $j*$j)" | bc

Change scale to the number of decimal places desired.

回答4:

#!/bin/bash
for i in {1..20}; do
    for j in {1..20}; do
        echo 5k$i $i\* $j $j\*+vp | dc
    done
done

回答5:

Use double paren to evaluate a variable.

variableA=$((variableB*variableC))

Only for ints though.

回答6:

does your remote cluster only have bash? if not, try and see if you have awk

awk 'BEGIN{
  for(i=1;i<=20;i++){
    for(j=1;j<=20;j++){
       print ( i*i + j*j ) ** 0.5
    }
  }
}'

回答7:

Typically you would use $((1*3)), but your case won't work as bash doesn't support floating point numbers. You'll have to use an external tool like awk, bc or dc: http://mywiki.wooledge.org/BashFAQ/022

回答8:

The code

echo $[(($i * $i) + ($j * $j)) ** $X]

will work if $X is an integer. You're trying to take the square root, and I'm not sure if bash's built-in arithmetic will do that. You'll probably be better off using a more powerful calculator tool (like bc, et al.) for this.

回答9:

Bash doesn't offer mathematical functions. However, you almost certainly have the korn shell installed. This should work:

#!/bin/ksh

for i in {1..20}
do
    for j in {1..20}
    do
        x=$((sqrt(i*i + j*j)))
        echo "sqrt($i^2 + $j^2) = $x"
    done
done

The beginning of the output is

sqrt(1^2 + 1^2) = 1.41421356237309505
sqrt(1^2 + 2^2) = 2.2360679774997897
sqrt(1^2 + 3^2) = 3.16227766016837933
sqrt(1^2 + 4^2) = 4.12310562561766055
sqrt(1^2 + 5^2) = 5.09901951359278483
sqrt(1^2 + 6^2) = 6.08276253029821969
sqrt(1^2 + 7^2) = 7.07106781186547524
sqrt(1^2 + 8^2) = 8.06225774829854965
sqrt(1^2 + 9^2) = 9.05538513813741663
sqrt(1^2 + 10^2) = 10.0498756211208903

回答10:

with zsh, this will work

for i in {1..20};do
  for j in {1..20};do
     echo $((($i*$i + $j*$j)**.5))
  done
done

回答11:

Another form of integer math expressions in Bash puts the double parentheses on the outside of the entire expression for assignment operations:

(( var = i ** 2 ))
(( i++ ))
(( position += delta ))

As you can see, dollar signs are not needed here (nor inside $(())). Also, spaces are permitted around the equal sign.

Also, this form can be used in conditionals:

sixpacks=8             # note spaces not allowed here
(( beerprice = 8 ))    # but you can use spaces this way
budget=50

# you can do assignments inside a conditional just like C
until (( (( beertotal = sixpacks * beerprice )) <= budget ))
do
    (( sixpacks-- ))
done
echo "Buy ${sixpacks} six-packs at \$${beerprice} each for a total of \$${beertotal}."

Or you can replace all that with this, of course:

beerprice=8
budget=50

# integer division
(( sixpacks = budget / beerprice ))
(( beertotal = sixpacks * beerprice ))
echo "Buy ${sixpacks} six-packs at \$${beerprice} each for a total of \$${beertotal}."

Bash also has a let statement:

let a=2**16
let 'a>>=1'    # bitwise shift - some operations need to be quoted or escaped
(( a>>=1 ))    # but not inside (())