问题描述
如果我有这两个列表:
la = [1, 2, 3] lb = [4, 5, 6]
我可以如下迭代它们:
for i in range(min(len(la), len(lb))): print la[i], lb[i]
或者更多的蟒蛇
for a, b in zip(la, lb): print a, b
<小时>
如果我有两本字典怎么办?
What if I have two dictionaries?
da = {'a': 1, 'b': 2, 'c': 3} db = {'a': 4, 'b': 5, 'c': 6}
同样,我可以手动迭代:
Again, I can iterate manually:
for key in set(da.keys()) & set(db.keys()): print key, da[key], db[key]
是否有一些内置方法可以让我进行如下迭代?
Is there some builtin method that allows me to iterate as follows?
for key, value_a, value_b in common_entries(da, db): print key, value_a, value_b
推荐答案
没有内置函数或方法可以做到这一点.但是,您可以轻松定义自己的.
There is no built-in function or method that can do this. However, you could easily define your own.
def common_entries(*dcts): if not dcts: return for i in set(dcts[0]).intersection(*dcts[1:]): yield (i,) + tuple(d[i] for d in dcts)
这建立在手动方法"的基础上.您提供,但像 zip 一样,可用于任意数量的字典.
This builds on the "manual method" you provide, but, like zip, can be used for any number of dictionaries.
>>> da = {'a': 1, 'b': 2, 'c': 3} >>> db = {'a': 4, 'b': 5, 'c': 6} >>> list(common_entries(da, db)) [('c', 3, 6), ('b', 2, 5), ('a', 1, 4)]
当只提供一个字典作为参数时,它本质上返回 dct.items().
When only one dictionary is provided as an argument, it essentially returns dct.items().
>>> list(common_entries(da)) [('c', 3), ('b', 2), ('a', 1)]
没有字典,它返回一个空的生成器(就像 zip())
With no dictionaries, it returns an empty generator (just like zip())
>>> list(common_entries()) []