How to get the n next values of a generator in a list (python)

问题

I have made a generator to read a file word by word and it works nicely.

def word_reader(file):
    for line in open(file):
        for p in line.split():
            yield p

reader = word_reader('txtfile')
next(reader)

What is the easiest way of getting the n next values in a list?

回答1:

Use itertools.islice:

list(itertools.islice(it, n))

回答2:

EDIT: Use itertools.islice. The pattern below that I originally proposed is bad idea — it crashes when it yields less than n values, and this behaviour depends on subtle issues, so people reading such code are unlikely to understand it's precise semantics.

There is also
[next(it) for _ in range(n)]
which might(?) be clearer to people not familiar with itertools; but if you deal with iterators a lot, itertools is a worthy addition to your toolset.

What happens if `next(it)` was exhausted and raises `StopIteration`?

(i.e. when it had less than n values to yield)

When I wrote the above line a couple years ago, I probably thought a StopIteration will have the clever side effect of cleanly terminating the list comprehension. But no, the whole comprehension will crash passing the StopIteration upwards. (It'd exit cleanly only if the exception originated from the range(n) iterator.)

Which is probably not the behavior you want.

But it gets worse. The following is supposed to be equivalent to the list comprehension (especially on Python 3):

list(next(it) for _ in range(n))

It isn't. The inner part is shorthand for a generator function; list() knows it's done when it raises StopIteration anywhere.
=> This version copes safely when there aren't n values and returns a shorter list. (Like itertools.islice().)

[Executions on: 2.7, 3.4]

But that's too going to change! The fact a generator silently exits when any code inside it raises StopIteration is a known wart, addressed by PEP 479. From Python 3.7 (or 3.5 with a future import) that's going to cause a RuntimeError instead of cleanly finishing the generator. I.e. it'll become similar to the list comprehension's behaviour. (Tested on a recent HEAD build)

回答3:

for word, i in zip(word_reader(file), xrange(n)):
    ...

回答4:

To get the first n values of a generator, you can use more_itertools.take.

If you plan to iterate over the words in chunks (eg. 100 at a time), you can use more_itertools.chunked (https://more-itertools.readthedocs.io/en/latest/api.html):

import more_itertools
for words in more_itertools.chunked(reader, n=100):
    # process 100 words

回答5:

Use cytoolz.take.

>>> from cytoolz import take
>>> list(take(2, [10, 20, 30, 40, 50]))
[10, 20]

来源：https://stackoverflow.com/questions/4152376/how-to-get-the-n-next-values-of-a-generator-in-a-list-python

标签

python

list

generator