Get ceiling integer from number in linux (BASH)

问题:

How would I do something like:

ceiling(N/500)

N representing a number.

But in a linux Bash script

回答1:

Call out to a scripting language with a ceil function. Given $NUMBER:

python -c "from math import ceil; print ceil($NUMBER/500.0)"

or

perl -w -e "use POSIX; print ceil($NUMBER/500.0), qq{\n}"

回答2:

Why use external script languages? You get floor by default. To get ceil, do

$ divide=8; by=3; let result=($divide+$by-1)/$by; echo $result
3
$ divide=9; by=3; let result=($divide+$by-1)/$by; echo $result
3
$ divide=10; by=3; let result=($divide+$by-1)/$by; echo $result
4
$ divide=11; by=3; let result=($divide+$by-1)/$by; echo $result
4
$ divide=12; by=3; let result=($divide+$by-1)/$by; echo $result
4
$ divide=13; by=3; let result=($divide+$by-1)/$by; echo $result
5
....

To take negative numbers into account you can beef it up a bit. Probably cleaner ways out there but for starters

$ divide=-10; by=10; neg=; if [ $divide -lt 0 ]; then let divide=-$divide; neg=1; fi; let result=($divide+$by-1)/$by; if [ $neg ]; then let result=-$result; fi; echo $result
-1

$ divide=10; by=10; neg=; if [ $divide -lt 0 ]; then let divide=-$divide; neg=1; fi; let result=($divide+$by-1)/$by; if [ $neg ]; then let result=-$result; fi; echo $result
1

回答3:

Here's a solution using bc (which should be installed just about everywhere):

ceiling_divide() {
  ceiling_result=`echo "($1 + $2 - 1)/$2" | bc`
}

Here's another purely in bash:

# Call it with two numbers.
# It has no error checking.
# It places the result in a global since return() will sometimes truncate at 255.

# Short form from comments (thanks: Jonathan Leffler)
ceiling_divide() {
  ceiling_result=$((($1+$2-1)/$2))
}

# Long drawn out form.
ceiling_divide() {
  # Normal integer divide.
  ceiling_result=$(($1/$2))
  # If there is any remainder...
  if [ $(($1%$2)) -gt 0 ]; then
    # rount up to the next integer
    ceiling_result=$((ceiling_result + 1))
  fi
  # debugging
  # echo $ceiling_result
}

回答4:

You can use awk

#!/bin/bash
number="$1"
divisor="$2"
ceiling() {
  awk -vnumber="$number" -vdiv="$divisor" '
  function ceiling(x){return x%1 ? int(x)+1 : x}
  BEGIN{ print ceiling(number/div) }'
}
ceiling

output

$ ./shell.sh 1.234 500
1

Or if there's a choice, you can use a better shell that does floating point, eg Zsh

integer ceiling_result
ceiling_divide() {
  ceiling_result=$(($1/$2))
  echo $((ceiling_result+1))
}

ceiling_divide 1.234 500

回答5:

Mathematically, the function of ceiling can be define with floor, ceiling(x) = -floor(-x). And, floor is the default when converting a positive float to integer.

if [ $N -gt 0 ]; then expr 1 - $(expr $(expr 1 - $N) / 500); else expr $N / 500; fi

Ref. https://en.wikipedia.org/wiki/Floor_and_ceiling_functions

回答6:

Floor () {
  DIVIDEND=${1}
  DIVISOR=${2}
  RESULT=$(( ( ${DIVIDEND} - ( ${DIVIDEND} % ${DIVISOR}) )/${DIVISOR} ))
  echo ${RESULT}
}
R=$( Floor 8 3 )
echo ${R}

Ceiling () {
  DIVIDEND=${1}
  DIVISOR=${2}
  $(( ( ( ${DIVIDEND} - ( ${DIVIDEND} % ${DIVISOR}) )/${DIVISOR} ) + 1 ))
  echo ${RESULT}
}
R=$( Ceiling 8 3 )
echo ${R}

回答7:

Expanding a bit on Kalle's great answer, here's the algorithm nicely packed in a function:

ceildiv() {
    local num=$1
    local div=$2
    echo $(( (num + div - 1) / div ))
}

or as a one-liner:

ceildiv(){ echo $((($1+$2-1)/$2)); }

If you want to get fancy, you could use a more robust version validates input to check if they're numerical, also handles negative numbers:

ceildiv() {
    local num=${1:-0}
    local div=${2:-1}
    if ! ((div)); then
        return 1
    fi
    if ((num >= 0)); then
        echo $(( (num + div - 1) / div ))
    else
        echo $(( -(-num + div - 1) / div ))
    fi
}

This uses a "fake" ceil for negative numbers, to the highest absolute integer, ie, -10 / 3 = -4 and not -3 as it should, as -3 > -4. If you want a "true" ceil, use $(( num / div )) instead after the else

And then use it like:

$ ceildiv 10 3
4
$ ceildiv 501 500
2
$ ceildiv 0 3
0
$ ceildiv -10 1
-10
$ ceildiv -10 3
-4

回答8:

Using the gorgeous 'printf' 1 will round up to the next integer

printf %.0f $float
or
printf %.0f `your calculation formula`
or
printf %.0f $(your calculation formula)

ref: how to remove decimal from a variable?

回答9:

This is a simple solution using Awk:

If you want the ceil of ($a/$b) use

echo "$a $b" | awk '{print int( ($1/$2) + 1 )}'

and the floor use

echo "$a $b" | awk '{print int($1/$2)}'

Note that I just echo the dividend '$a' as the first field of the line to awk and the divisor '$b' as the second.

回答10:

This function wont't add 1, if the division returns a non-floating number.

function ceiling {
    DIVIDEND=${1}
    DIVISOR=${2}
    if [ $(( DIVIDEND % DIVISOR )) -gt 0 ]; then
            RESULT=$(( ( ( $DIVIDEND - ( $DIVIDEND % $DIVISOR ) ) / $DIVISOR ) + 1 ))
    else
            RESULT=$(( $DIVIDEND / $DIVISOR ))
    fi
    echo $RESULT
}

Use it like this:

echo $( ceiling 100 33 )
> 4

回答11:

Some more concise Awk logic

awk '
function ceil(ip) {
  print ip%1 ? int(ip)+1 : ip
}  
BEGIN {
  ceil(1000/500)
  ceil(1001/500)
}
'

Result

2
3

回答12:

Without specifying any function, we can use the following awk script:

echo x y | awk '{ r=$1 % $2; q=$1/y; if (r != 0) q=int(q+1); print q}'

Not sure this one get any logical error or not. Please correct.

回答13:

If you are already familiar with the Python library, then rather than learn bc, you might want to define this bash function:

pc () { pyexpr="from math import *; print($@)"; python -c "$pyexpr"; }

Then:

pc "ceil(3/4)"
1

but also any valid python expression works:

pc pi / 4
0.7853981633974483

pc "'\n'.join(['Pythagoras said that %3.2f^2 + %3.2f^2 is always %3.2f'
    % (sin(ai), cos(ai), sin(ai)**2 + cos(ai)**2)
    for ai in [pi / 4 * k for k in range(8)]])"
Pythagoras said that 0.00^2 + 1.00^2 is always 1.00
Pythagoras said that 0.71^2 + 0.71^2 is always 1.00
Pythagoras said that 1.00^2 + 0.00^2 is always 1.00
Pythagoras said that 0.71^2 + -0.71^2 is always 1.00
Pythagoras said that 0.00^2 + -1.00^2 is always 1.00
Pythagoras said that -0.71^2 + -0.71^2 is always 1.00
Pythagoras said that -1.00^2 + -0.00^2 is always 1.00
Pythagoras said that -0.71^2 + 0.71^2 is always 1.00