Python3: Find word in list using while-loop

Question

A simple question, that I've actually had a hard time finding a solution to, as most suggest using the much easier to work with, for loop.

I have an assignment that tells me explicitly to use the while loop to return either true or false, based on if a string occurs in a list.

My current solution (which works) is as follows:

def word_in_list(word, words):
x=0
length = int(len(words)) -1

while words[x] != word and x < length:
    x = x+1

    if words[x] == word:
        print (True)

    elif x == length and words[x] != word:
        print (False)

My biggest issue has been limiting the function to the length of the list, otherwise I get an error stating "list index out of range".

However, as this is a rather clunky way of doing something so simple, I'm looking for suggestions on how to streamline it a little - if there's a way that is. And as stated, the for-loop, that you'd normally use, is not an option.

parkerduckworth · Accepted Answer · 2018-01-27 18:47:07Z

2

here is something that runs a bit more efficiently. Each iteration checks two indices, rather than one.

def word_in_list(word, words):
    l, r = 0, len(words) - 1
    while l <= r:
        if words[l] == word or words[r] == word:
            return True
        l += 1
        r -= 1
    return False

Hope that helps!

answered Jan 27, 2018 at 18:47

parkerduckworth

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Comments

jpp · Accepted Answer · 2018-01-27 20:07:42Z

1

Fixed a small error and cleaned up a little, but as a simple exercise requiring the use of while the algorithm fundamentally looks fine.

def word_in_list(word, words):

    x = 0
    length = len(words) - 1

    while x < length:

        if words[x] == word:
            return True

        elif x == length:
            return False

        x += 1

word_in_list('hello', ['now', 'this', 'hello', 'world'])   # True
word_in_list('hello2', ['now', 'this', 'hello', 'world'])  # False

edited Jan 27, 2018 at 20:07

answered Jan 27, 2018 at 18:37

jpp

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2 Comments

Keyur Potdar Over a year ago

Using words[x] != word in elif after if words[x] == word: seems redundant.

Holmberg Over a year ago

Thank you! This do indeed look much better, however it's a small mistake in your example, as you add -1 in two places, which equals -2, missed it as well initially, but now it works great! :)

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