Is There A Way Other Than 'try...except' And '.isdigit()' To Check User Input In Python 2?

I am currently trying to learn Python 2.7 via Learn Python The Hard Way, but have a question about Study Drill 5 of Exercise 35. The code I'm looking at is: choice = raw_input('>

Solution 1:

You can write your own is_digit function

def my_digit(input):
  digits = ['0','1','2','3','4','5','6','7','8','9']
  for i in list(input):
    if not i in digits:
       return False
  return True

Solution 2:

So, firstly read what jonrsharpe linked to in the comments and accept that try-except is the best way of doing this.

Then consider what it means to be an integer:

• Everything must be a digit (no decimal point, too)

So that's what you check for. You want everything to be a digit.

Thus, for a represents_integer(string) function:

for every letter in the string:
    check that it is one of "0", "1", "2", ..., "9"
    if it is not, this is not a number, so return false

if we are here, everything was satisfied, so return true

Note that the check that is is one of might require another loop, although there are faster ways.

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Finally, consider "", which won't work in this method (or GregS').

Solution 3:

Since you already learned sets, you can test that every character is a digit by something like

choice = choice.strip()
for d in choice:
    if d not in "0123456789":
        # harass user for their idiocy

how_much = int (choice)