問題描述

如果我有這兩個列表:

la = [1, 2, 3]
lb = [4, 5, 6]

我可以如下迭代它們:

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

<小時>

如果我有兩本字典怎么辦?

What if I have two dictionaries?

da = {'a': 1, 'b': 2, 'c': 3}
db = {'a': 4, 'b': 5, 'c': 6}

同樣，我可以手動迭代:

Again, I can iterate manually:

for key in set(da.keys()) & set(db.keys()):
    print key, da[key], db[key]

是否有一些內(nèi)置方法可以讓我進(jìn)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 

推薦答案

沒有內(nèi)置函數(shù)或方法可以做到這一點(diǎn).但是，您可以輕松定義自己的.

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)

這建立在手動方法"的基礎(chǔ)上.您提供，但像 zip 一樣，可用于任意數(shù)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)]

當(dāng)只提供一個字典作為參數(shù)時，它本質(zhì)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)]

沒有字典，它返回一個空的生成器(就像 zip())

With no dictionaries, it returns an empty generator (just like zip())

>>> list(common_entries())
[]

這篇關(guān)于Python 相當(dāng)于 zip 的字典的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！