How to zero pad a sequence of integers in bash so that all have the same width?

Bash Problem Overview

I need to loop some values,

for i in $(seq $first $last)
do
    does something here
done

For $first and $last, i need it to be of fixed length 5. So if the input is 1, i need to add zeros in front such that it becomes 00001. It loops till 99999 for example, but the length has to be 5.

E.g.: 00002, 00042, 00212, 012312 and so forth.

Any idea on how i can do that?

Bash Solutions

Solution 1 - Bash

In your specific case though it's probably easiest to use the -f flag to seq to get it to format the numbers as it outputs the list. For example:

for i in $(seq -f "%05g" 10 15)
do
  echo $i
done

will produce the following output:

More generally, bash has printf as a built-in so you can pad output with zeroes as follows:

$ i=99
$ printf "%05d\n" $i
00099

You can use the -v flag to store the output in another variable:

$ i=99
$ printf -v j "%05d" $i
$ echo $j
00099

Notice that printf supports a slightly different format to seq so you need to use %05d instead of %05g.

Solution 2 - Bash

Easier still you can just do

for i in {00001..99999}; do
  echo $i
done

Solution 3 - Bash

If the end of sequence has maximal length of padding (for example, if you want 5 digits and command is "seq 1 10000"), than you can use "-w" flag for seq - it adds padding itself.

seq -w 1 10

produce

Solution 4 - Bash

use printf with "%05d" e.g.

printf "%05d" 1

Solution 5 - Bash

Very simple using printf

[jaypal:~/Temp] printf "%05d\n" 1
00001
[jaypal:~/Temp] printf "%05d\n" 2
00002

Solution 6 - Bash

Use awk like this:

awk -v start=1 -v end=10 'BEGIN{for (i=start; i<=end; i++) printf("%05d\n", i)}'

OUTPUT:

Update:

As pure bash alternative you can do this to get same output:

for i in {1..10}
do
   printf "%05d\n" $i
done

This way you can avoid using an external program seq which is NOT available on all the flavors of *nix.

Solution 7 - Bash

I pad output with more digits (zeros) than I need then use tail to only use the number of digits I am looking for. Notice that you have to use '6' in tail to get the last five digits :)

for i in $(seq 1 10)
do
RESULT=$(echo 00000$i | tail -c 6)
echo $RESULT
done

Solution 8 - Bash

If you want N digits, add 10^N and delete the first digit.

for (( num=100; num<=105; num++ ))
do
  echo ${num:1:3}
done

Output:

Solution 9 - Bash

Other way :

zeroos="000"
echo 

for num in {99..105};do
 echo ${zeroos:${#num}:${#zeroos}}${num}
done

So simple function to convert any number would be:

function leading_zero(){

    local num=$1
    local zeroos=00000
    echo ${zeroos:${#num}:${#zeroos}}${num} 

}

Solution 10 - Bash

One way without using external process forking is string manipulation, in a generic case it would look like this:

#start value
CNT=1

for [whatever iterative loop, seq, cat, find...];do
   # number of 0s is at least the amount of decimals needed, simple concatenation
   TEMP="000000$CNT"
   # for example 6 digits zero padded, get the last 6 character of the string
   echo ${TEMP:(-6)}
   # increment, if the for loop doesn't provide the number directly
   TEMP=$(( TEMP + 1 ))
done

This works quite well on WSL as well, where forking is a really heavy operation. I had a 110000 files list, using printf "%06d" $NUM took over 1 minute, the solution above ran in about 1 second.

Solution 11 - Bash

This will work also:

for i in {0..9}{0..9}{0..9}{0..9}
do
  echo "$i"
done

Solution 12 - Bash

If you're just after padding numbers with zeros to achieve fixed length, just add the nearest multiple of 10 eg. for 2 digits, add 10^2, then remove the first 1 before displaying output.

This solution works to pad/format single numbers of any length, or a whole sequence of numbers using a for loop.

# Padding 0s zeros:
# Pure bash without externals eg. awk, sed, seq, head, tail etc.
# works with echo, no need for printf

pad=100000		;# 5 digit fixed

for i in {0..99999}; do ((j=pad+i))
	echo ${j#?}
done

Tested on Mac OSX 10.6.8, Bash ver 3.2.48

Solution 13 - Bash

1.) Create a sequence of numbers 'seq' from 1 to 1000, and fix the width '-w' (width is determined by length of ending number, in this case 4 digits for 1000).

2.) Also, select which numbers you want using 'sed -n' (in this case, we select numbers 1-100).

3.) 'echo' out each number. Numbers are stored in the variable 'i', accessed using the '$'.

Pros: This code is pretty clean.

Cons: 'seq' isn't native to all Linux systems (as I understand)

for i in `seq -w 1 1000 | sed -n '1,100p'`; 
do 
    echo $i; 
done

Solution 14 - Bash

TL;DR

$ seq 1 10 | awk '{printf("%05d\n", $1)}'

Input(Pattern 1. Slow):

$ seq 1 10 | xargs -n 1 printf "%05d\n"

Input(Pattern 2. Fast):

$ seq 1 10 | awk '{printf("%05d\n", $1)}'

Output(same result in each case):

Explanation

I'd like to suggest the above patterns. These implementations can be used as a command so that we can use them again with ease. All you have to care about in these commands is the length of the numbers after being converted.(like changing the number %05d into %09d.) Plus, it's also applicable to other solutions such as the following. The example is too dependent on my environment, so your output might be different, but I think you can tell the usefulness easily.

$ wc -l * | awk '{printf("%05d\n", $1)}'
00007
00001
00001
00001
00013
00017
00001
00001
00001
00043

And like this:

$ wc -l * | awk '{printf("%05d\n", $1)}' | sort | uniq
00001
00007
00013
00017
00043

Moreover, if you write in this manner, we can also execute the commands asynchronously. (I found a nice article: https://www.dataart.com/en/blog/linux-pipes-tips-tricks)

disclaimer: I'm not sure of this, and I am not a *nix expert.

Performance test:

Super Slow:

$ time seq 1 1000 | xargs -n 1 printf "%09d\n" > test
seq 1 1000  0.00s user 0.00s system 48% cpu 0.008 total
xargs -n 1 printf "%09d\n" > test  1.14s user 2.17s system 84% cpu 3.929 total

Relatively Fast:

for i in {1..1000}
do
   printf "%09d\n" $i
done
$ time sh k.sh > test
sh k.sh > test  0.01s user 0.01s system 74% cpu 0.021 total


for i in {1..1000000}
do
   printf "%09d\n" $i
done
$ time sh k.sh > test
sh k.sh > test  7.10s user 1.52s system 99% cpu 8.669 total

Fast:

$ time seq 1 1000 | awk '{printf("%09d\n", $1)}' > test
seq 1 1000  0.00s user 0.00s system 47% cpu 0.008 total
awk '{printf("%09d\n", $1)}' > test  0.00s user 0.00s system 52% cpu 0.009 total


$ time seq 1 1000000 | awk '{printf("%09d\n", $1)}' > test
seq 1 1000000  0.27s user 0.00s system 28% cpu 0.927 total
awk '{printf("%09d\n", $1)}' > test  0.92s user 0.01s system 99% cpu 0.937 total

If you have to implement the higher performance solution, probably it may require other techniques, not using the shell script.

Solution 15 - Bash

you don't need awk for that — either seq or jot alone suffices :

% seq -f '%05.f' 6     # bsd-seq
00001
00002
00003
00004
00005
00006

% gseq -f '%05.f' 6    # gnu-seq
00001
00002
00003
00004
00005
00006

% jot -w '%05.f' 6
00001
00002
00003
00004
00005
00006

…… unless you're going into bigint territory :

% gawk -Mbe '

  function __(_,___) {
      return +_<+___?___:_
  }
  BEGIN {
        _+=_^=_<_                 
      ____="%0*.f\n"   
  } {                      
       ___=__($--_, !+$++_)                
     _____=__(++_+--_, length(______=+$NF)) 
     do {                     
        printf(____,_____,___)
     }  while (___++<______) 
                                                       
  }' <<< '999999999999999999996 1000000000000000000003'

0999999999999999999996
0999999999999999999997
0999999999999999999998
0999999999999999999999
1000000000000000000000
1000000000000000000001
1000000000000000000002
1000000000000000000003

——————————————————————————————————————————————————

If you need to print out a HUGE range of numbers, then this approach maybe a bit faster -

printing out every integer from 1 to 1 million, left-zero-padded to 9-digits wide, in 0.049s
*caveat : I didn't have the spare time to make it cover all input ranges :: it's just a proof of concept accepting increments of powers of 10