将元组与不按顺序排列的元素进行比较

I have a two dicts that I'm converting bot to tuples . Both dict contain the same elements, but they are not generated using the same logic .

我有两个dicts，我正在将bot转换为元组。两个dict都包含相同的元素，但它们不是使用相同的逻辑生成的。

For example let's say I have a tuple like this :

例如，假设我有一个这样的元组：

(('a',5),('n',4),('c',8))

And the 2nd tuple is like that :

而第二个元组是这样的：

(('c',8),('n',4),('a',5))

Their original dicts is like that (probably, I can't know how will the elements be ordered in the dict however they are generated from two dicts that contains the same elements (no more, no less) :

他们原来的说法是这样的（可能，我不知道如何在dict中对元素进行排序，但是它们是由包含相同元素的两个dicts生成的（不多也不少）：

{'a':5,'c': 8,'n':4}

For a human being both tuples are similar, but for a computer they are not .

对于一个人而言，两个元组都是相似的，但对于一个计算机则不然。

How can I go with checking if two tuples are similar ?

我怎样才能检查两个元组是否相似？

5 个解决方案

#1

Sort both the tuple and compare them. For example:

对元组进行排序并进行比较。例如：

>>> tuple_1 = (('a',5),('n',4),('c',8))
>>> tuple_2 = (('c',8),('n',4),('a',5))

# Non-sorted --> unequal; issue you are currently facing
>>> tuple_1 == tuple_2
False

# comparing sorted tuples -- equal 
>>> sorted(tuple_1) == sorted(tuple_2)
True

If the elements of both the tuples are unique, you may also compare them via using set as:

如果两个元组的元素都是唯一的，您也可以通过使用set来比较它们：

>>> set(tuple_1) == set(tuple_2)
True

As a side note, you do not have to convert the dict to tuple in order to compare the content of dictionaries. You may directly compare the dict objects as:

作为旁注，您不必将字典转换为元组以便比较字典的内容。您可以直接将dict对象比较为：

>>> {1: 2, 3: 4} == {3: 4, 1: 2}
True

#2

You could also convert the tuples to a dict and then compare them.

您还可以将元组转换为dict然后进行比较。

Not sure if this is faster/slower/the same as sorting them.

不确定这是否更快/更慢/与排序相同。

dict(t1) == dict(t2)

#3

def similarTuple(a,b):
    h = {}
    for e in a:
        h[e] = True
    for e in b:
        if e not in h:
            return False
    return True

#4

You could also make a set of both tuples and compare those:

您还可以创建一组两个元组并进行比较：

a = (('a', 0), ('b', 1), ('c', 2))
b = (('b', 1), ('a', 0), ('c', 2))
set(a) == set(b)

#5

trying for a one liner, exploring getting a bool result from list comprehensions

尝试一个班轮，探索从列表推导得到一个bool结果

not efficient, possibly amusing

没有效率，可能很有趣

def SameElementsP(a,b):
    return not ( [1 for e in a if e not in b] or [1 for e in b if e not in a] )

#1