python check if a string end with a number in a range valid

Question

In my test code, I want to assert that a string ends with a number. Say the number is between [0,3):

assert_equals('/api_vod_asset/v0/assets/0', '/api_vod_asset/v0/assets/number') #valid

assert_equals('/api_vod_asset/v0/assets/1', '/api_vod_asset/v0/assets/number') #valid

assert_equals('/api_vod_asset/v0/assets/5', '/api_vod_asset/v0/assets/number') #invalid

How to use regular expression or some other technique for number ?

Answer 1

You would probably want to use assertRegex:

test_case.assertRegex('/api_vod_asset/v0/assets/0', '/api_vod_asset/v0/assets/[012]')

The one above works in the case of [0,3) range. If you do not want that restriction, you would likely want to have:

test_case.assertRegex('/api_vod_asset/v0/assets/0', '/api_vod_asset/v0/assets/[\d]')

All the code above works after the following lines have been added to your snippet:

import unittest as ut
test_case = ut.TestCase()

Answer 2

If it will always be at the same place in the string, you can use string.split

something like

def check_range(to_test, valid_range):
    return int(to_test.split('/')[-1]) in range(valid_range[0], valid_range[1] + 1)

then you can do

check_range('/api_vod_asset/v0/assets/0', [0,3]) #True
check_range('/api_vod_asset/v0/assets/1', [0,3]) #True
check_range('/api_vod_asset/v0/assets/5', [0,3]) #False

Answer 3

First, use a regex like \d+$ to get any number (\d+) at the end of the string ($)...

>>> m = re.search(r"\d+$", s)

... then check whether you have a match and whether the number is in the required range:

>>> m is not None and 0 <= int(m.group()) < 3

Or use a range, if you prefer this notation (assuming upper bound of [0,3) is exclusive):

>>> m is not None and int(m.group()) in range(0, 3)

Answer 4

I want to assert that a string ends with a number

if int(myString[-1]) in [0,1,2]:
     do something...