从迭代器中排除第一个元素（Python）

Question

I have a generator function A .

For example (in reality I have a more complex function A ),

def A():
    yield from [i**2 for i in range(20)]

Writing another generator function B , I want to enumerate all elements that A returns except the first element.

What are concise ways to implement this in Python 3?

Answer 1

使用itertools.islice ：

itertools.islice(generator,1,None)

Answer 2

Usually, you don't need this in an expression, so you just call next(it) , ignoring the results, to consume and discard the first element.

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However, if the iterator might be empty, you have to decide what you want to happen:

• Maybe you want to raise StopIteration , in which case next(it) is fine.
• Maybe you want to raise something else, in which case you next(it) inside an except StopIteration: raise SomethingElse() .
• Maybe you just want to leave the iterator empty, in which case you can call next(it, None) .

You can find examples of these in the stdlib and docs. For example, if you scan through the recipes in itertools :

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)

This is doing exactly what you want to do—skip the first element of b . And if iterable is empty, you don't want an error here; you just want to iterate nothing. So, next(b, None) .

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What if you need to do this in the middle of an expression?

Then you can write a function that skips the first element:

def skip_first(iterable):
    it = iter(iterable)
    next(it, None)
    return it

(Again, you have to decide what you want to happen for an empty iterable.)

This returns a first-skipped version of the iterator, so you can use it inline. (It also mutates the iterator you passed in, of course, but you normally only use on a temporary value that you're not keeping any references to, so that's not a problem.)

Or, if you need to return a generator instead of an arbitrary iterator (usually you don't):

def skip_first(iterable):
    it = iter(iterable)
    next(it, None)
    yield from it

Or you can use the more general version of the same idea, itertools.islice . The following have the same effect:

it = skip_first(it)
it = itertools.islice(it, 1, None)

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While we're on the itertools recipes, it's worth looking at consume :

def consume(iterator, n=None):
    "Advance the iterator n-steps ahead. If n is None, consume entirely."
    # Use functions that consume iterators at C speed.
    if n is None:
        # feed the entire iterator into a zero-length deque
        collections.deque(iterator, maxlen=0)
    else:
        # advance to the empty slice starting at position n
        next(islice(iterator, n, n), None)

Forget the None part; the interesting bit is that it skips n elements with an islice and a next . (Notice that it's mutating iterator in-place, not returning something.)

Answer 3

Disclaimer: See @abarnert and @Solaxun's answers above.

Just thought that the following should be mentioned

If you have eg original = iter((1,2,3,4,5))

Then

first, remaining = next(original), original

where remaining is the iterator without the first element.

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