CODEWITHSUNDEEP
pythonstringpython-3.x
2Xchampion
2Xchampion

Reputation: 656

How to specify a fill character to string and split it by chunk?

I am attempting to split a string of integers by every n character and then convert them to unicode. Unfortunately, I came across a problem... Since the leading 0s has been dropped from the string, it makes me confused about how to handle them. Here is an example：

print(num2txt(97097114103104))

Here the leading 97 is a valid number in the ASCII table and I can simple add a "0" before the string so that this whole string will be correctly split into something like this:

['097', '097', '114', '103', '104']

However, what if the first three digits are actually valid in the ASCII table? Like this:

100097114103

In this case I want it to be split into 

['100', '097', '114']

But how do I make sure this does not need a "0" anymore if something happens in my function?

My current code is below:

def num2txt(num, k=3):

    line = str(num) 
    line = "0" + line
    line = [line[i:i+k] for i in range(0, len(line), k)]

    return line

Upvotes: 1

Views: 421

Answers (2)

Sede
Sede

Reputation: 61225

I would do this using math.ceil to round up the length of my string to the multiple of 3 then format my string. From there to split my string what other better way than using a generator function?

import math
def num2txt(num, k):
    num_str = "{:0>{width}}".format(num, width=math.ceil(len(str(num))/k)*k)
    for n in range(0, len(num_str), k):
        yield num_str[n:n+k]

Demo:

>>> list(num2txt(97097114103104, 3))
['097', '097', '114', '103', '104']

Upvotes: 1

TigerhawkT3
TigerhawkT3

Reputation: 49318

Just add leading zeros when necessary.

def num2txt(num, k=3):

    line = str(num)
    padding = len(line)%k
    if padding:
        line = '0'*(k-padding) + line
    line = [line[i:i+k] for i in range(0, len(line), k)]

    return line

Upvotes: 2

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