如何从 python 字典中删除 key: pairs over a certain number？

Question

给定一个数字和一个字典，“remove_numbers_larger_than”删除任何值大于给定数字的键。返回修改后的字典。

def remove_numbers_larger_than(dictionary, number):

    for k, v in dictionary.items():

        if v > number:

            del d[k]

    return dictionary

Answer 1

如果您需要dict就地修改给定，请先复制键，以便从字典中删除键不会干扰迭代。def remove_numbers_larger_than(dictionary, number):&nbsp; &nbsp; keys = list(dictionary)&nbsp; &nbsp; for k in keys:&nbsp; &nbsp; &nbsp; &nbsp; if dictionary[k] > number:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; del dictionary[k]&nbsp; &nbsp; return dictionary如果您有空创建一个新的dict，def remove_numbers_larger_than(dictionary, number):&nbsp; &nbsp; return dict(kv for kv in dictionary.items() if kv[1] <= number)

Answer 2

就地方法可以稍微简化一下。整个字典只迭代一次，辅助列表只包含有问题的键，而不是所有键。for&nbsp;k&nbsp;in&nbsp;[k&nbsp;for&nbsp;k,&nbsp;v&nbsp;in&nbsp;dictionary.items()&nbsp;if&nbsp;v&nbsp;>&nbsp;number]: &nbsp;&nbsp;&nbsp;&nbsp;del&nbsp;dictionary[k]

Answer 3

您可以使用此代码段，它将遍历字典，选择小于或等于给定数字的数字，并使用所选数字创建一个新字典。def remove_numbers_larger_than(dictionary, number):&nbsp; &nbsp; return { key: dictionary[key] for key in dictionary if dictionary[key] <= number }