Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

python check if word is in certain elements of a list

Tags:

python

list

word

I was wondering if there was a better way to put:

if word==wordList[0] or word==wordList[2] or word==wordList[3] or word==worldList[4]
like image 283
Sam Avatar asked Dec 09 '22 02:12

Sam


2 Answers

word in wordList

Or, if you want to check the 4 first,

word in wordList[:4]
like image 68
Vincent Savard Avatar answered Dec 30 '22 11:12

Vincent Savard


Very simple task, and so many ways to deal with it. Exciting! Here is what I think:

If you know for sure that wordList is small (else it might be too inefficient), then I recommend using this one:

b = word in (wordList[:1] + wordList[2:])

Otherwise I would probably go for this (still, it depends!):

b = word in (w for i, w in enumerate(wordList) if i != 1)

For example, if you want to ignore several indexes:

ignore = frozenset([5, 17])
b = word in (w for i, w in enumerate(wordList) if i not in ignore)

This is pythonic and it scales.


However, there are noteworthy alternatives:

### Constructing a tuple ad-hoc. Easy to read/understand, but doesn't scale.
# Note lack of index 1.
b = word in (wordList[0], wordList[2], wordList[3], wordList[4])

### Playing around with iterators. Scales, but rather hard to understand.
from itertools import chain, islice
b = word in chain(islice(wordList, None, 1), islice(wordList, 2, None))

### More efficient, if condition is to be evaluated many times in a loop.
from itertools import chain
words = frozenset(chain(wordList[:1], wordList[2:]))
b = word in words
like image 34
robert Avatar answered Dec 30 '22 11:12

robert