On Tue, Apr 23, 2019 at 10:28:29AM -0700, Brett Cannon wrote:
> Given "abcdefabcdefabcdef", what is the last result of "abc"?
> x.rindex("abc") will tell you.
>
> Given [1, 2, 3, 4, 5, 1, 2, 3, 4, 5] where is the last result of 3?
> reversed(x).index(3) will tell you (or x[::-1]).
That first version doesn't work, as list_reverseiterator objects don't
have an index method. You're not the only person to make that error, I
too often forget that reverse() returns an iterator, not a list.
The second is easy to get wrong, because it returns the wrong index:
# Get the item following the last instance of spam.
index = x[::-1].index(spam)
print(x[index+1])
In your example, the correct index is 7 but the returned value is 2.
> Notice how with lists you can easily reverse them and still get at the
> value since you are searching per index.
"Easily" hides a lot of copying behind the scenes. If the list is a
non-trivial size, that can be very wasteful, especially if you're doing
it in a loop, or hidden in a function. Don't think about the case of a
ten element list, think of a ten-billion element list.
Personally, I don't think I've every used list.index, let alone needed
rindex. But I think we underestimate the difficulty and cost of faking
an rindex method from index for those who need it (if anyone does).
> But with strings, you searching by
> a subslice that can be greater than 1 in which case you can't use a similar
> approach.
Of course you can: you "just" need to reverse the substring as well.
The conversions will be even more fiddly and error-prone:
py> s = "abc spam def spam ghi"
py> s.rindex('spam') == len(s) - s[::-1].index('spam'[::-1]) - len('spam')
True
but it can be done.
--
Steven
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