In python, how to compare two numerical strings without casting them to int()?

Question

For example, to check if they are greater than, lesser than or equal to each other without using int() and def.

num1 = "67"
num2 = "1954"

• chepner: Why wouldn't you use int?
• desperate: the question specifies not to use it
• MisterMiyagi: Why would the question specify not to use it?
• user2390182: Would you need this to work for negative numbers?

Answer 1

Left pad with zero and then compare the strings lexicographically:

num1 = "67"
num2 = "1954"
if num1.zfill(10) < num2.zfill(10):
 print("67 is less than 1954")

Note here that the left padding trick leaves the 2 numbers with the same string length. So we are doing something like comparing 0067 to 1954, in which case the dictionary order is in agreement with the numerical order.

• user2390182: What about negative numbers?
• desperate: what if the user inputs a number and you don't know what the len() would be?
• Tim Biegeleisen: @schwobaseggl Good edge case, I hadn't thought of this. The padding logic would get more complex.
• user2390182: Yeah, also then the lexicographic order must be reversed. That's quite awkward
• Tim Biegeleisen: @desperate Assuming your above comment be directed at me, you may just choose a large enough padding. There probably isn't any real upper limit to the size of the padding.

Answer 2

Easiest without padding to an unknown length:

if (len(num1), num1) < (len(num2), num2):
 print(num1, "<", num2)

• desperate: it is not needed for negative numbers
• desperate: can you explain the purpose of putting the variable after the comma in the brackets, for eg (len(num1), num1)? I am new to this
• user2390182: Instead of only the strings, you compare tuples (pairs) of length and string, in your example: (2, "67") < (4, "1954"). Tuple comparison works lexicographically, as well, in that the second elements are only compared if the first elements are the same. Hence, shorter numbers will always be considered smaller. Numbers of equal length will be compared alphabetically.
• ekhumoro: This method fails for e.g. num1 = '4'; num2 = '03'.