Question

我见过基于固定数字排序列表的解决方案： Sort a list by multiple attributes?

它有一个很好的排序解决方案： s = sorted（s，key = lambda x：（x [1]，x [2]））

以及itemgetter示例

但是我有不同数量的属性，一个例子可以有2个属性：

example_list = [{'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'x': 'd2_sort': 30},
    {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'x': 'd2_sort': 30},
    {'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'y': 'd2_sort': 35},
    {'d1_desc': 'c', 'd1_sort': 3, 'd2_desc': 'z': 'd2_sort': 38},
    etc.
]

但它也可以是1或3或更多。我不能使用像这样的lambda函数或itemgetter。但是我确实知道执行时的维数（即使它因情况而异）。所以我做了这个（参数设置为2昏暗的例子）：

example_list = [{'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'y', 'd2_sort': 35},
    {'d1_desc': 'c', 'd1_sort': 3, 'd2_desc': 'z', 'd2_sort': 38},
    {'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'z', 'd2_sort': 38},
    {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'c', 'd1_sort': 3, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'z', 'd2_sort': 38}
]

def order_get( a , nr):
    result = []
    for i in range(1, nr+1):
        result.append(a.get('d' + str(i) + '_sort'))
    return result

example_list.sort(key = lambda x: order_get(x, 2)) # for this example hard set to 2

In [82]: example_list
Out[82]: 
[{'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'x', 'd2_sort': 30},
 {'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'y', 'd2_sort': 35},
 {'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'z', 'd2_sort': 38},
 {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'x', 'd2_sort': 30},
 {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'x', 'd2_sort': 30},
 {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'z', 'd2_sort': 38},
 {'d1_desc': 'c', 'd1_sort': 3, 'd2_desc': 'x', 'd2_sort': 30},
 {'d1_desc': 'c', 'd1_sort': 3, 'd2_desc': 'z', 'd2_sort': 38}]

但这真的是最好的方法吗？我的意思是1）Pythonic和2）表现明智吗？这是一个常见问题吗？

Answer 1

我仍然会使用itemgetter因为它更快，你创建它一次并且每次都使用：

from operator import itemgetter

def make_getter(nr):
    keys = ('d%d_sort' % (n + 1) for n in xrange(nr))
    return itemgetter(*keys)

example_list.sort(key=make_getter(2))

创建itemgetter需要时间。如果您必须在多个列表中使用它，因为它始终相同，请将其存储为get_two = make_getter(2)并使用get_two作为key函数。

Answer 2

您可以支持任意数量的排序键，只要它们具有可预测的模式作为键。

假设您有d[X]_sort到d[Y]_sort，其中X和Y是整数，排序键全部以_sort结尾，其关键函数如下：

import re

def arb_kf(d): 
    li=filter(lambda s: s.endswith('_sort'), d) 
    rtr=[tuple(map(int, re.findall(r'([0-9]+)', k) + [d[k]])) for k in li]
    rtr.sort()            
    return rtr

使用你的dicts示例列表：

example_list = [{'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'y', 'd2_sort': 35},
    {'d1_desc': 'c', 'd1_sort': 3, 'd2_desc': 'z', 'd2_sort': 38},
    {'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'z', 'd2_sort': 38},
    {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'c', 'd1_sort': 3, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'z', 'd2_sort': 38}
]


>>> for d in sorted(example_list, key=arb_kf) :
...     print d  
{'d2_desc': 'x', 'd2_sort': 30, 'd1_sort': 1, 'd1_desc': 'a'}
{'d2_desc': 'y', 'd2_sort': 35, 'd1_sort': 1, 'd1_desc': 'a'}
{'d2_desc': 'z', 'd2_sort': 38, 'd1_sort': 1, 'd1_desc': 'a'}
{'d2_desc': 'x', 'd2_sort': 30, 'd1_sort': 2, 'd1_desc': 'b'}
{'d2_desc': 'x', 'd2_sort': 30, 'd1_sort': 2, 'd1_desc': 'b'}
{'d2_desc': 'z', 'd2_sort': 38, 'd1_sort': 2, 'd1_desc': 'b'}
{'d2_desc': 'x', 'd2_sort': 30, 'd1_sort': 3, 'd1_desc': 'c'}
{'d2_desc': 'z', 'd2_sort': 38, 'd1_sort': 3, 'd1_desc': 'c'}

假设d[X]_sort中的整数在某些词组中有所不同，并且您希望对较低的数字给予更多权重;即，d0_sort比不具有较低数字的字典具有更多的排序权重。

由于Python对元组元素进行排序，这是正确的：

>>> sorted([(1,99), (1,1,1), (0,50), (1,0,99)])
[(0, 50), (1, 0, 99), (1, 1, 1), (1, 99)]

由于key函数返回一个元组列表，在这种情况下也可以。

然后，如果您的示例列表中包含一个包含'd0_sort': 3的词典，那么它的排序将高于其中包含'd1_sort的任何内容：

example_list = [{'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'y', 'd2_sort': 35},
    {'d1_desc': 'c', 'd1_sort': 3, 'd2_desc': 'z', 'd2_sort': 38},
    {'d1_desc': 'a', 'd1_sort': 1, 'd2_desc': 'z', 'd2_sort': 38},
    {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'c', 'd1_sort': 3, 'd2_desc': 'x', 'd2_sort': 30},
    {'d1_desc': 'b', 'd1_sort': 2, 'd2_desc': 'z', 'd2_sort': 38},
    {'d1_desc': 'b', 'd0_sort': 3, 'd2_desc': 'z', 'd2_sort': 38}
]
>>> for d in sorted(example_list, key=arb_kf) :
...     print d  
{'d0_sort': 3, 'd2_desc': 'z', 'd2_sort': 38, 'd1_desc': 'b'}
{'d2_desc': 'x', 'd2_sort': 30, 'd1_sort': 1, 'd1_desc': 'a'}
{'d2_desc': 'y', 'd2_sort': 35, 'd1_sort': 1, 'd1_desc': 'a'}
{'d2_desc': 'z', 'd2_sort': 38, 'd1_sort': 1, 'd1_desc': 'a'}
{'d2_desc': 'x', 'd2_sort': 30, 'd1_sort': 2, 'd1_desc': 'b'}
{'d2_desc': 'x', 'd2_sort': 30, 'd1_sort': 2, 'd1_desc': 'b'}
{'d2_desc': 'z', 'd2_sort': 38, 'd1_sort': 2, 'd1_desc': 'b'}
{'d2_desc': 'x', 'd2_sort': 30, 'd1_sort': 3, 'd1_desc': 'c'}
{'d2_desc': 'z', 'd2_sort': 38, 'd1_sort': 3, 'd1_desc': 'c'}

对不同数量的属性进行动态排序

2 个答案: