Question

我的代码是：

a = {"key1": 5 , "key2": "8 hours ago", "key3": 2}
b = {"key1": 7 , "key2": "42 hours ago", "key3": 9}
c = {"key1": 6 , "key2": "1 hour ago", "key3": 1}
undecorated = [a, b, c]

undecorated.sort(key=operator.itemgetter('key2'))

结果是：

[
{"key1": 6 , "key2": "1 hour ago", "key3": 1}
{"key1": 7 , "key2": "42 hours ago", "key3": 9}
{"key1": 5 , "key2": "8 hours ago", "key3": 2}
]

但结果需要：

[
{"key1": 6 , "key2": "1 hour ago", "key3": 1}
{"key1": 5 , "key2": "8 hours ago", "key3": 2}
{"key1": 7 , "key2": "42 hours ago", "key3": 9}
]

Answer 1

如果key2的值与此处显示的一样正常，您可以执行以下操作：

undecorated.sort(key=lambda d: int(d['key2'].split()[0]))

Answer 2

正如评论中已经解释的那样，由于您的比较键是字符串，因此它们按字典顺序进行比较。您可以使用dateparser模块（pip install dateparser）将字符串解析为datetime正确比较的对象：

>>> from dateparser import parse
>>> undecorated.sort(key=lambda x: parse(x["key2"]), reverse=True)
>>> undecorated
[{'key1': 6, 'key2': '1 hour ago', 'key3': 1}, {'key1': 5, 'key2': '8 hours ago', 'key3': 2}, {'key1': 7, 'key2': '42 hours ago', 'key3': 9}]

Answer 3

由于比较键有点复杂，我更喜欢将它设为函数而不是lambda，它可以包含try-except，并且更具可读性：

a = {"key1": 5, "key2": "8 hours ago", "key3": 2}
b = {"key1": 7, "key2": "42 hours ago", "key3": 9}
c = {"key1": 6, "key2": "1 hour ago", "key3": 1}
undecorated = [a, b, c]


def compare_key(my_dict):
    value = my_dict.get('key2')
    try:
        result = int(value.split()[0])
    except:
        result = 0
    return result


undecorated.sort(key=compare_key)

print(undecorated)

如何按相对日期值对字典列表进行排序？

3 个答案: