问题描述
我可以想到一个更简单的解决方案,itertools.zip_longest()
如果产生的元组中存在用于填充较短的可迭代对象的哨兵值,请使用并引发异常:
from itertools import zip_longest
def zip_equal(*iterables):
sentinel = object()
for combo in zip_longest(*iterables, fillvalue=sentinel):
if sentinel in combo:
raise ValueError('Iterables have different lengths')
yield combo
不幸的是,我们不能使用zip()
withyield
from
来避免每次迭代都进行测试的Python代码循环。一旦最短的迭代器用完,zip()
将推进所有先前的迭代器,从而吞噬证据,如果其中仅包含一个额外的项目。
解决方法
我正在寻找一种对zip
多个可迭代对象引发异常的好方法,如果可迭代对象的长度不相等。
如果可迭代对象是列表或具有某种len
方法,则此解决方案简单易行:
def zip_equal(it1,it2):
if len(it1) != len(it2):
raise ValueError("Lengths of iterables are different")
return zip(it1,it2)
但是,如果it1
和it2
是生成器,则前一个函数将失败,因为未定义长度TypeError: object of type 'generator'
has no len()
。
我认为该itertools
模块提供了一种实现该目标的简单方法,但是到目前为止,我还没有找到它。我想出了这个自制的解决方案:
def zip_equal(it1,it2):
exhausted = False
while True:
try:
el1 = next(it1)
if exhausted: # in a previous iteration it2 was exhausted but it1 still has elements
raise ValueError("it1 and it2 have different lengths")
except StopIteration:
exhausted = True
# it2 must be exhausted too.
try:
el2 = next(it2)
# here it2 is not exhausted.
if exhausted: # it1 was exhausted => raise
raise ValueError("it1 and it2 have different lengths")
except StopIteration:
# here it2 is exhausted
if not exhausted:
# but it1 was not exhausted => raise
raise ValueError("it1 and it2 have different lengths")
exhausted = True
if not exhausted:
yield (el1,el2)
else:
return
可以使用以下代码测试该解决方案:
it1 = (x for x in ['a','b','c']) # it1 has length 3
it2 = (x for x in [0,1,2,3]) # it2 has length 4
list(zip_equal(it1,it2)) # len(it1) < len(it2) => raise
it1 = (x for x in ['a',3]) # it2 has length 4
list(zip_equal(it2,it1)) # len(it2) > len(it1) => raise
it1 = (x for x in ['a','c','d']) # it1 has length 4
it2 = (x for x in [0,3]) # it2 has length 4
list(zip_equal(it1,it2)) # like zip (or izip in python2)
我是否在忽略任何其他解决方案?我的zip_equal
函数有更简单的实现吗?