Append value to out of range list

Question

I have a text file which contains a string. In this string there are some numbers which parse out by using regex. After the regex I get this as an output:

0.7.3.4

But lets assume that in another string i get this outout:

0.7.3

In this case I want to ad 00 in the end.

When I do:

lst = get_values.split('.')
print lst[0] -> 0
print lst[1] -> 7
print lst[2] -> 3
print lst[3] -> out of range obviously

So I wanna check if lst[3] is out range, then add "00" to it. Same goes for lst[2] and lst[1]. Here is "dummy pseudo code":

if not lst[3]:
    add "00"

The end output should look like this in the end:

070304

or maybe like this is a lst[x] is missing:

070300

And the first "0" is also removed just in case you wonder where it was.

Are you trying to compare version numbers? Because there are [tools for that](https://stackoverflow.com/q/11887762/1258041). — Lev Levitsky, Jun 19 '17 at 13:58
@LevLevitsky Nope, not comparing version number. More like checking if each item has a value. — user535081, Jun 19 '17 at 14:00
@user535081 should the output for `0.7` be `0700` or `070000`? — Zev Averbach, Jun 19 '17 at 14:31

score 1 · Accepted Answer · answered Jun 19 '17 at 14:22

This might work for you:

def format_string(str):
    parts = [0, 0, 0]
    for idx, num in enumerate(str.split('.')[1:]):
        parts[idx] = int(num)
    return '%02d%02d%02d' % tuple(parts)

print(format_string('0.7.3.4'))
print(format_string('0.7.3'))

Output:

070304
070300

If you want to make it more customizable:

def format_string(str, expected_parts=3, ignore_parts=1):
    input_parts = str.split('.')[ignore_parts:]
    output_parts = ['00'] * expected_parts

    for idx, num in enumerate(input_parts):
        output_parts[idx] = '%02d' % int(num)

    return ''.join(output_parts)

Append value to out of range list

1 Answers1