合并两个列表列表 - Python

This is a great primer but doesn't answer what I need: Combining two sorted lists in Python

这是一个很好的入门,但没有回答我的需要:在Python中组合两个排序列表

I have two Python lists, each is a list of datetime,value pairs:

我有两个Python列表,每个列表都包含日期时间,值对:

list_a = [['1241000884000', 3], ['1241004212000', 4], ['1241006473000', 11]]

And:

list_x = [['1241000884000', 16], ['1241000992000', 16], ['1241001121000', 17], ['1241001545000', 19], ['1241004212000', 20], ['1241006473000', 22]]

There are actually numerous list_a lists with different key/values.

实际上有许多list_a列表具有不同的键/值。

All list_a datetimes are in list_x.

所有list_a日期时间都在list_x中。

I want to make a list, list_c, corresponding to each list_a which has each datetime from list_x and value_a/value_x.

我想创建一个列表list_c,对应于每个list_a,其中包含list_x和value_a / value_x中的每个日期时间。

Bonus:

In my real program, list_a is actually a list within a dictionary like so. Taking the answer to the dictionary level would be:

在我的真实程序中,list_a实际上是字典中的列表。对词典级别的回答是:

dict = {object_a: [['1241000884000', 3], ['1241004212000', 4], ['1241006473000', 11]], object_b: [['1241004212000', 2]]}

I can figure that part out though.

我可以想出那个部分。

4 个解决方案

#1

Here's some code that does what you asked for. You can turn your list of pairs into a dictionary straightforwardly. Then keys that are shared can be found by intersecting the sets of keys. Finally, constructing the result dictionary is easy given the set of shared keys.

这里有一些代码可以满足您的要求。您可以直接将对子列表转换为字典。然后,通过交叉键组可以找到共享的键。最后,给定共享密钥集,构造结果字典很容易。

dict_a = dict(list_a)
dict_x = dict(list_x)

shared_keys = set(dict_a).intersection(set(dict_x))

result = dict((k, (dict_a[k], dict_x[k])) for k in shared_keys)

#2

"I want to make a list, list_c, corresponding to each list_a which has each datetime from list_x and value_a/value_x."

“我想创建一个列表list_c,对应于每个list_a,其中包含list_x和value_a / value_x中的每个日期时间。”

def merge_lists( list_a, list_x ):
    dict_x= dict(list_x)
    for k,v in list_a:
        if k in dict_x:
            yield k, (v, dict_x[k])

Something like that may work also.

这样的东西也可以起作用。

merged= list( merge_lists( someDict['object_a'], someDict['object_b'] )

This may be slightly quicker because it only makes one dictionary for lookups, and leaves the other list alone.

这可能会稍快一点,因为它只为查找创建一个字典,而另一个列表单独留下。

#3

Nothing beats a nice functional one-liner:

没有什么比一个很好的功能性单线:

reduce(lambda l1,l2: l1 + l2, list)

#4

Could try extend:

可以尝试扩展:

list_a.extend(list_b)

#1

dict_a = dict(list_a)
dict_x = dict(list_x)

shared_keys = set(dict_a).intersection(set(dict_x))

result = dict((k, (dict_a[k], dict_x[k])) for k in shared_keys)

#2