問題描述
我有一個(gè)大的迭代器，事實(shí)上，一個(gè)大的迭代器由以下給出:
I've got a large iterable, in fact, a large iterable given by:
itertools.permutations(range(10))
我想訪問第百萬個(gè)元素.我已經(jīng)用一些不同的方式解決了問題.
I would like to access to the millionth element. I alredy have problem solved in some different ways.
將可迭代對象轉(zhuǎn)換為列表并獲取第 1000000 個(gè)元素:
Casting iterable to list and getting 1000000th element:
return list(permutations(range(10)))[999999]
手動(dòng)跳過元素直到 999999:
Manually skiping elements till 999999:
p = permutations(range(10))
for i in xrange(999999): p.next()
return p.next()
手動(dòng)跳過元素 v2:
Manually skiping elements v2:
p = permutations(range(10))
for i, element in enumerate(p):
if i == 999999:
return element
使用 itertools 中的 islice:
Using islice from itertools:
return islice(permutations(range(10)), 999999, 1000000).next()
但我仍然不覺得它們都不是 python 的優(yōu)雅方式.第一個(gè)選項(xiàng)太昂貴了，它需要計(jì)算整個(gè)迭代來訪問單個(gè)元素.如果我沒記錯(cuò)的話，islice 在內(nèi)部進(jìn)行的計(jì)算與我在方法 2 中所做的相同，并且?guī)缀跖c第 3 次一樣，也許它有更多的冗余操作.
But I still don't feel like none of them is the python's elegant way to do that. First option is just too expensive, it needs to compute the whole iterable just to access a single element. If I'm not wrong, islice does internally the same computation I just did in method 2, and is almost exactly as 3rd, maybe it has even more redundant operations.
所以，我只是好奇，想知道在 python 中是否有其他方式可以訪問可迭代的具體元素，或者至少以更優(yōu)雅的方式跳過第一個(gè)元素，或者我是否只需要使用上述之一.
So, I'm just curious, wondering if there is in python some other way to access to a concrete element of an iterable, or at least to skip the first elements, in some more elegant way, or if I just need to use one of the aboves.
推薦答案
使用itertools 配方 consume 跳過 n 元素:
def consume(iterator, n):
"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)
注意那里的 islice() 調(diào)用；它使用 n, n，實(shí)際上不返回 anything，并且 next() 函數(shù)回退到默認(rèn)值.
Note the islice() call there; it uses n, n, effectively not returning anything, and the next() function falls back to the default.
簡化為您的示例，您希望跳過 999999 個(gè)元素，然后返回元素 1000000:
Simplified to your example, where you want to skip 999999 elements, then return element 1000000:
return next(islice(permutations(range(10)), 999999, 1000000))
islice() 在 C 中處理迭代器，這是 Python 循環(huán)無法比擬的.
islice() processes the iterator in C, something that Python loops cannot beat.
為了說明，以下是每種方法僅重復(fù) 10 次的時(shí)間:
To illustrate, here are the timings for just 10 repeats of each method:
>>> from itertools import islice, permutations
>>> from timeit import timeit
>>> def list_index():
... return list(permutations(range(10)))[999999]
...
>>> def for_loop():
... p = permutations(range(10))
... for i in xrange(999999): p.next()
... return p.next()
...
>>> def enumerate_loop():
... p = permutations(range(10))
... for i, element in enumerate(p):
... if i == 999999:
... return element
...
>>> def islice_next():
... return next(islice(permutations(range(10)), 999999, 1000000))
...
>>> timeit('f()', 'from __main__ import list_index as f', number=10)
5.550895929336548
>>> timeit('f()', 'from __main__ import for_loop as f', number=10)
1.6166789531707764
>>> timeit('f()', 'from __main__ import enumerate_loop as f', number=10)
1.2498459815979004
>>> timeit('f()', 'from __main__ import islice_next as f', number=10)
0.18969106674194336
islice() 方法比第二快的方法快近 7 倍.
The islice() method is nearly 7 times faster than the next fastest method.
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