Comparing a string with multiple substrings of another string

Each Answer to this Q is separated by one/two green lines.

Is there another simpler way to write code that basically checks every character of the string 'abcde'

if input == 'a' or input == 'ab' or input == 'abc' or input == 'abcd' or input == 'abcde':
    return True

This should do the same thing as what you put.

return 'abcde'.startswith(input)

Don’t name variables input, since it will shadow the builtin function input(). Its considered bad practice to do this, and easy enough to just choose another variable name.

You could use a set to check if the input matches any of the substrings:

lookups = {'a', 'ab', 'abc', 'abcd', 'abcde'}

my_input = input()

if my_input in lookups:
    return True

We could also generate this set using a set comprehension:

characters="abcde"

lookups = {characters[:i] for i in range(1, len(characters) + 1)}

my_input = input()

if my_input in lookups:
    return True

For large sets of combinations, the benefit of using a set over a list is that you get constant time O(1) lookups for searching. This is much better than using a list, which will give you linear O(N) lookups.

There are multiple cute ways to do it.
startwith is probably the most efficient one, but these should work too:

using lstrip:

return 'abcde'.lstrip(input)!='abcde'

using list comprehension:

return any(['abcde'[:i+1] == input for i in range(len('abcde'))])

using regex:

   pattern = re.compile('^'+input)
   return bool(pattern.match('abcde'))

or just:

  return 'abcde'[:len(input)]==input

You could probably try something like this:

def your_function():
    # Add as much chars that you want here
    chars = "abcde"

    # Assuming you are taking the user input from terminal
    user_input = input()

    # Loop over every substring from chars
    for i in range(len(chars) + 1):
        if chars[:i] == user_input:
            return True

    return False

Let me know if this helps!

You can try this:

If input in ['a', 'ab', 'abc', 'abcd', 'abcde']:
    return True
else:
   return False


The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .