从 dict 返回前 N 个键:值对 [英] Return first N key:value pairs from dict

问题描述

考虑以下字典，d:

d = {'a': 3, 'b': 2, 'c': 3, 'd': 4, 'e': 5}

我想从 d 返回前 N 个键:值对(在这种情况下 N <= 4).这样做的最有效方法是什么?

解决方案

没有前 n"个键这样的东西，因为 dict 不记得哪个键是最先插入的.

你可以得到任何n个键值对:

n_items = take(n, d.iteritems())

这使用了 中的 take 实现itertools 食谱:

from itertools import islicedef take(n, iterable):将可迭代对象的前 n 个项目作为列表返回"返回列表(islice(可迭代，n))

在线查看:ideone

Python 3.6 更新

n_items = take(n, d.items())

Consider the following dictionary, d:

d = {'a': 3, 'b': 2, 'c': 3, 'd': 4, 'e': 5}

I want to return the first N key:value pairs from d (N <= 4 in this case). What is the most efficient method of doing this?

解决方案

There's no such thing a the "first n" keys because a dict doesn't remember which keys were inserted first.

You can get any n key-value pairs though:

n_items = take(n, d.iteritems())

This uses the implementation of take from the itertools recipes:

from itertools import islice

def take(n, iterable):
    "Return first n items of the iterable as a list"
    return list(islice(iterable, n))

See it working online: ideone

Update for Python 3.6

n_items = take(n, d.items())

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