Python 如何一次迭代字典-n个键值对

Python 如何一次迭代字典-n个键值对,python,dictionary,Python,Dictionary,我有一本很大的字典,里面有成千上万的元素。我需要用这个字典作为参数执行一个函数。现在,我不想在一次执行中传递整个字典,而是希望成批执行函数——每次使用字典的x个键值对 我正在做以下工作: mydict = ##some large hash x = ##batch size def some_func(data): ##do something on data temp = {} for key,value in mydict.iteritems(): if len(te

我有一本很大的字典,里面有成千上万的元素。我需要用这个字典作为参数执行一个函数。现在,我不想在一次执行中传递整个字典,而是希望成批执行函数——每次使用字典的x个键值对

我正在做以下工作:

mydict = ##some large hash
x = ##batch size
def some_func(data):
    ##do something on data
temp = {}
for key,value in mydict.iteritems():
        if len(temp) != 0 and len(temp)%x == 0:
                some_func(temp)
                temp = {}
                temp[key] = value
        else:
                temp[key] = value
if temp != {}:
        some_func(temp)

对我来说,这看起来很粗糙。我想知道是否有一种优雅/更好的方法可以做到这一点。

我经常使用这个小工具:

import itertools

def chunked(it, size):
    it = iter(it)
    while True:
        p = tuple(itertools.islice(it, size))
        if not p:
            break
        yield p
对于您的用例:

for chunk in chunked(big_dict.iteritems(), batch_size):
    func(chunk)

以下是根据我先前的回答改编的两个解决方案

或者,您可以从字典中获取
项的列表
,并从该列表的片段中创建新的
dict
s。不过,这并不是最优的,因为它需要大量复制那本巨大的词典

def chunks(dictionary, size):
    items = dictionary.items()
    return (dict(items[i:i+size]) for i in range(0, len(items), size))
或者,您可以使用一些
itertools
模块的函数在循环时生成新的子字典。这与@georg的答案类似,只是使用了一个
for
循环

from itertools import chain, islice
def chunks(dictionary, size):
    iterator = dictionary.iteritems()
    for first in iterator:
        yield dict(chain([first], islice(iterator, size - 1)))
示例用法。对于这两种情况:

mydict = {i+1: chr(i+65) for i in range(26)}
for sub_d in chunks2(mydict, 10):
    some_func(sub_d)
发件人:


你可以试试,或者你好,乔治。谢谢你的回答。请解释一下
chunked
方法的性能。比我共享的解决方案更有效吗?@nish:我想应该是有效的
itertools
是用C编写的,比python快得多。在Python3中,它将是
big\u dict.items()而不是big\u dict.iteritems()
def chunked(iterable, n):
    """Break an iterable into lists of a given length::
        >>> list(chunked([1, 2, 3, 4, 5, 6, 7], 3))
        [[1, 2, 3], [4, 5, 6], [7]]
    If the length of ``iterable`` is not evenly divisible by ``n``, the last
    returned list will be shorter.
    This is useful for splitting up a computation on a large number of keys
    into batches, to be pickled and sent off to worker processes. One example
    is operations on rows in MySQL, which does not implement server-side
    cursors properly and would otherwise load the entire dataset into RAM on
    the client.
    """
    # Doesn't seem to run into any number-of-args limits.
    for group in (list(g) for g in izip_longest(*[iter(iterable)] * n,
                                                fillvalue=_marker)):
        if group[-1] is _marker:
            # If this is the last group, shuck off the padding:
            del group[group.index(_marker):]
        yield group