快速迭代元组列表

I wonder whether there's a quicker and less time consuming way to iterate over a list of tuples, finding the right match. What I do is:

我想知道是否有一种更快,更省时的方法来迭代元组列表,找到正确的匹配。我所做的是:

# this is a very long list.
my_list = [ (old1, new1), (old2, new2), (old3, new3), ... (oldN, newN)]

# go through entire list and look for match
for j in my_list:
    if j[0] == VALUE:
        PAIR_FOUND = True
        MATCHING_VALUE = j[1]
        break

this code can take quite some time to execute, depending on the number of items in the list. I'm sure there's a better way of doing this.

此代码可能需要相当长的时间才能执行,具体取决于列表中的项目数。我确信有更好的方法可以做到这一点。

5 个解决方案

#1

Assuming a bit more memory usage is not a problem and if the first item of your tuple is hashable, you can create a dict out of your list of tuples and then looking up the value is as simple as looking up a key from the dict. Something like:

假设更多的内存使用不是问题,如果你的元组的第一项是可清除的,你可以从元组列表中创建一个字典,然后查找值就像从字典中查找键一样简单。就像是:

dct = dict(tuples)
val = dct.get(key) # None if item not found else the corresponding value

EDIT: To create a reverse mapping, use something like:

编辑:要创建反向映射,请使用以下内容:

revDct = dict((val, key) for (key, val) in tuples)

#2

I think that you can use

我认为你可以使用

for j,k in my_list:
  [ ... stuff ... ]

#3

The code can be cleaned up, but if you are using a list to store your tuples, any such lookup will be O(N).

可以清除代码,但是如果使用列表来存储元组,则任何此类查找都将为O(N)。

If lookup speed is important, you should use a dict to store your tuples. The key should be the 0th element of your tuples, since that's what you're searching on. You can easily create a dict from your list:

如果查找速度很重要,则应使用dict存储元组。键应该是元组的第0个元素,因为那是你正在搜索的内容。您可以从列表中轻松创建一个dict:

my_dict = dict(my_list)

Then, (VALUE, my_dict[VALUE]) will give you your matching tuple (assuming VALUE exists).

然后,(VALUE,my_dict [VALUE])将为您提供匹配的元组(假设存在VALUE)。

#4

The question is dead but still knowing one more way doesn't hurt:

这个问题已经死了但仍然知道另一种方式并没有伤害:

my_list = [ (old1, new1), (old2, new2), (old3, new3), ... (oldN, newN)]

for first,*args in my_list:
    if first == Value:
        PAIR_FOUND = True
        MATCHING_VALUE = args
        break

#5

I wonder whether the below method is what you want.

我想知道下面的方法是否是你想要的。

You can use defaultdict.

您可以使用defaultdict。

>>> from collections import defaultdict
>>> s = [('red',1), ('blue',2), ('red',3), ('blue',4), ('red',1), ('blue',4)]
>>> d = defaultdict(list)
>>> for k, v in s:
       d[k].append(v)    
>>> sorted(d.items())
[('blue', [2, 4, 4]), ('red', [1, 3, 1])]

#1