Grab the first word in a list that is found in a s

Question

So, I have a list of words like so:

activationWords = ['cactus', 'cacti', 'rofl']

And I want to find any of those words and return the first word of any of those words appearing in a random string. I'll use this string as an example:

str = "Wow, rofl I found a cactus in a cacti pile."

As you can see with the above example string, the first instance of a word in the list is "rofl". I want to be able to detect that and return the word into a string that I can use to my discretion. How would I do this?

Keep in mind that that string is just an example. Each time I run this it will be using a different string.

Answer 1

You can use str.find() here:

>>> activationWords = ['cactus', 'cacti', 'rofl']
>>> s = "Wow, rofl I found a cactus in a cacti pile."
>>> temp = [(s.find(i), i) for i in activationWords if i in s]
>>> print temp
[(20, 'cactus'), (32, 'cacti'), (5, 'rofl')]
>>> print min(temp)[1]
rofl

str.find() finds where the string is in the other string. It returns the index of the first letter.

[(s.find(i), i) for i in activationWords] is a list comprehension that returns a list of tuples, the first item being the result of str.find(), the second being the word.

Then we use min() here to get the lowest number, and [1] to get the word.

Note that you should not name a string str. It overrides the built-in type.