我有一个Python脚本，它把一个整数列表作为输入，我需要一次处理四个整数。不幸的是，我无法控制输入，否则我将它作为一个四元素元组列表传入。目前，我以这种方式迭代它:
for i in range(0, len(ints), 4):
# dummy op for example code
foo += ints[i] * ints[i + 1] + ints[i + 2] * ints[i + 3]
不过，它看起来很像“C-think”，这让我怀疑有一种更python的方式来处理这种情况。该列表在迭代后被丢弃，因此不需要保留。也许这样会更好?
while ints:
foo += ints[0] * ints[1] + ints[2] * ints[3]
ints[0:4] = []
不过，感觉还是不太对。: - /
相关问题:在Python中如何将列表分割成大小均匀的块?
当前回答
在你的第二种方法中，我将通过这样做进入下一组4人:
ints = ints[4:]
然而，我还没有做过任何绩效评估，所以我不知道哪种方法更有效。
话虽如此，我通常会选择第一种方法。这并不漂亮，但这通常是与外部世界接触的结果。
其他回答
为什么不使用列表理解
l = [1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
n = 4
filler = 0
fills = len(l) % n
chunks = ((l + [filler] * fills)[x * n:x * n + n] for x in range(int((len(l) + n - 1)/n)))
print(chunks)
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 0]]
在我的特殊情况下，我需要填充项目，以重复最后一个元素，直到它达到大小，所以我改变了这个答案，以适应我的需要。
大小为4的输入输出示例:
Input = [1,2,3,4,5,6,7,8]
Output= [[1,2,3,4], [5,6,7,8]]
Input = [[1,2,3,4,5,6,7]]
Output= [[1,2,3,4], [5,6,7,7]]
Input = [1,2,3,4,5]
Output= [[1,2,3,4], [5,5,5,5]]
def chunker(seq, size):
res = []
for el in seq:
res.append(el)
if len(res) == size:
yield res
res = []
if res:
res = res + (size - len(res)) * [res[-1]]
yield res
def chunker(iterable, n):
"""Yield iterable in chunk sizes.
>>> chunks = chunker('ABCDEF', n=4)
>>> chunks.next()
['A', 'B', 'C', 'D']
>>> chunks.next()
['E', 'F']
"""
it = iter(iterable)
while True:
chunk = []
for i in range(n):
try:
chunk.append(next(it))
except StopIteration:
yield chunk
raise StopIteration
yield chunk
if __name__ == '__main__':
import doctest
doctest.testmod()
还有另一个答案，它的优点是:
1)容易理解 2)适用于任何可迭代对象，而不仅仅是序列(上面的一些答案会阻塞文件句柄) 3)不立即将数据块加载到内存 4)不会在内存中生成对同一迭代器的块长的引用列表 5)在列表的末尾没有填充填充值
话虽如此，我还没有计算它的时间，所以它可能比一些更聪明的方法慢，而且考虑到用例，一些优势可能是无关紧要的。
def chunkiter(iterable, size):
def inneriter(first, iterator, size):
yield first
for _ in xrange(size - 1):
yield iterator.next()
it = iter(iterable)
while True:
yield inneriter(it.next(), it, size)
In [2]: i = chunkiter('abcdefgh', 3)
In [3]: for ii in i:
for c in ii:
print c,
print ''
...:
a b c
d e f
g h
Update: A couple of drawbacks due to the fact the inner and outer loops are pulling values from the same iterator: 1) continue doesn't work as expected in the outer loop - it just continues on to the next item rather than skipping a chunk. However, this doesn't seem like a problem as there's nothing to test in the outer loop. 2) break doesn't work as expected in the inner loop - control will wind up in the inner loop again with the next item in the iterator. To skip whole chunks, either wrap the inner iterator (ii above) in a tuple, e.g. for c in tuple(ii), or set a flag and exhaust the iterator.
如果列表很大，执行效率最高的方法是使用生成器:
def get_chunk(iterable, chunk_size):
result = []
for item in iterable:
result.append(item)
if len(result) == chunk_size:
yield tuple(result)
result = []
if len(result) > 0:
yield tuple(result)
for x in get_chunk([1,2,3,4,5,6,7,8,9,10], 3):
print x
(1, 2, 3)
(4, 5, 6)
(7, 8, 9)
(10,)
