[SOLVED] sperating a number by threes

[Garrett85]

Below is my script, my question is in three comment lines in the function? If you know how I can pull this off please let me know. Thanks.

#! /bin/bash

# Program to take user input number and return it as a sentence.


# Get user input

read -p "Enter a number no higher than the billions range" NUM

NUMBEROFDIGITS=${#NUM}


###################################
### FUNCTION-FindNumberOfGroups ###
###################################

FindNumberOfGroups()
{
    if [[ `expr ${NUMBEROFDIGITS} % 3` -eq 0 ]]; then # line is good for determining if there is an odd number of digits when dividing by 3
        NUMBEROFGROUPS=$( expr ${NUMBEROFDIGITS} / 3 )
        echo ${NUMBEROFGROUPS}
    else
        echo "remainder"
        NUMBEROFGROUPS=$( expr ${NUMBEROFDIGITS} / 3 )
        let NUMBEROFGROUPS=NUMBEROFGROUPS+1
        echo ${NUMBEROFGROUPS}
        # here I need to seperate NUMBEROFDIGITS divided by 3 from it's remainder
        # if the number is 4500 I need to seperate the number into two groups and add two leading zeros to the left number, e.g. 004 500
        # if the number is 61382 I need to seperate the number into two groups and add one leading zero to the left number, e.g. 061 382
    fi
}

####################
### END FUNCTION ###
####################

FindNumberOfGroups

[Irithori]

printf has the capability to insert a thousand separator.
Caution: The char depends on LC_LOCALE, so it is not always a dot. (But one can set LC_LOCALE before calling printf)

The idea is:
printf outputs a separated version of a decimal number, which can then be further processed to e.g. replace the dot with a white space.

Now, the requirement of padding the leading group with zeros is a bit evil.
Because one needs to know the length of the printed decimal beforehand, and the dots (if any) also take a place.
Hence the case statement.

#!/usr/bin/env bash

NUM=12345678901234
let MOD=${#NUM}%3
let GRPS=${#NUM}/3

case $MOD in
1)
  let LENGTH=${#NUM}+2+$GRPS;;
2)
  let LENGTH=${#NUM}+1+$GRPS;;
*)
  let LENGTH=${#NUM}+$GRPS-1;;
esac

# zero padded, dot separated number assigned to a variable
a=$(printf "%'0*d\n" $LENGTH $NUM)
echo $a

# If there wouldnt be a requirement for leading zeros, it is easier.
# No need for $LENGTH
printf "%'d\n" $NUM

[drl]

Hi.

Not tested, thought experiment: perhaps look at reminder (%) of division of length by 3, print that number of leading zeros ("...0*d ...", ... remainder) ... cheers, drl

[watael]

let's make it look a bit more "bashy"

#!/bin/bash

number=12345678901234
(( modulo=${#number}%3 , periods=${#number}/3 ))

(( modulo == 1 ? (length=${#number}+periods+2) : 
   modulo == 2 ? (length=${#number}+periods+1) :
                 (length=${#number}+periods-1) ))

printf "%'0*d\n" $length $number

printf "%'d\n" $number

i've been tedious

var=$1
v=${#var}

for ((i=v; i>=0; i--))
do
   t=${var:$i}
   ((${#t}>=3)) && { var=${var%$t*}; ar=( $t ${ar[@]} ); t="";}
done

printf -vval '%.3d ' $t ${ar[@]}

echo $val

[Irithori]

Ok, you know bash

- The code iterates backwards and number by number over $var.
- $t is assigned the part between the current position and the end of $var
- Once the length of $t is three, $t is cut from $var and also an array is build.
- So after the loop, there is a nice array of 3-digit numbers and (possibly) a one or two digit number in $t
- the printf assigns $val with a formated $t and all elements of ${ar[@]}

I didnt know about the -v VAL option. Thanks.

[alf55]

Below is my script, my question is in three comment lines in the function? If you know how I can pull this off please let me know. Thanks.

#! /bin/bash

# Program to take user input number and return it as a sentence.


# Get user input

read -p "Enter a number no higher than the billions range" NUM

NUMBEROFDIGITS=${#NUM}


###################################
### FUNCTION-FindNumberOfGroups ###
###################################

FindNumberOfGroups()
{
    if [[ `expr ${NUMBEROFDIGITS} % 3` -eq 0 ]]; then # line is good for determining if there is an odd number of digits when dividing by 3
        NUMBEROFGROUPS=$( expr ${NUMBEROFDIGITS} / 3 )
        echo ${NUMBEROFGROUPS}
    else
        echo "remainder"
        NUMBEROFGROUPS=$( expr ${NUMBEROFDIGITS} / 3 )
        let NUMBEROFGROUPS=NUMBEROFGROUPS+1
        echo ${NUMBEROFGROUPS}
        # here I need to seperate NUMBEROFDIGITS divided by 3 from it's remainder
        # if the number is 4500 I need to seperate the number into two groups and add two leading zeros to the left number, e.g. 004 500
        # if the number is 61382 I need to seperate the number into two groups and add one leading zero to the left number, e.g. 061 382
    fi
}

####################
### END FUNCTION ###
####################

FindNumberOfGroups

Assuming this is not an assignement, here is an example in "bash" (rather than sh or ksh).

FindNumberOfGroups()
{
    local index
    local padding_string="000"
    local num="${1}"
    local len_mod_3=$((${#num} % 3))
    if [ "${len_mod_3}" != "0" ] ; then
        num="${padding_string:0:$((3 - ${len_mod_3}))}${num}"
        len_mod_3=$((${#num} % 3))
    fi
    group_cnt=$((${#num} / 3))
    #for index in $(seq 0 3 ${len_of_num}); do
     for index in $(seq 0 3 ${#num}); do
       [ $index -ne 0 ] && printf " "
       printf "${num:${index}:3}"
    done
    printf "\n"
}

Which produces an output of:

bash$ FindNumberOfGroups 123456789
123 456 789
bash$ echo

bash$ FindNumberOfGroups 12345678
012 345 678
bash$ echo

bash$ FindNumberOfGroups 1234567
001 234 567
bash$ echo

bash$ FindNumberOfGroups 123
123  
bash$ echo

bash$ FindNumberOfGroups 12
012  
bash$ echo

bash$ FindNumberOfGroups 1
001  
bash$

[watael]

you forgot to define `$len_of_num` to `seq', it should be ${#num}