Python比较标记化列表 [英] Python Compare Tokenized Lists

问题描述

我需要针对该问题的最快解决方案，因为它将应用于巨大的数据集:

I need the fastest-possible solution to this problem as it will be applied to a huge data set:

给出此主列表:

m=['abc','bcd','cde','def']

...以及列表的此参考列表:

...and this reference list of lists:

r=[['abc','def'],['bcd','cde'],['abc','def','bcd']]

我想将r中的每个列表与主列表(m)进行比较，并生成一个新的列表.此新对象的匹配度为1(基于m的顺序)，0为不匹配项.因此，新对象(列表列表)将始终具有与m相同长度的列表. 这是我根据上面的m和r期望的结果:

I'd like to compare each list within r to the master list (m) and generate a new list of lists. This new object will have a 1 for matches based on the order in m and 0 for non-matches. So the new object (list of lists) will always have the lists of the same length as m. Here's what I would expect based on m and r above:

[[1,0,0,1],[0,1,1,0],[1,1,0,1]]

因为r的第一个元素是['abc','def']并且具有匹配项 加上第m个元素的第1个和第4个元素，则结果为[1,0,0,1].

Because the first element of r is ['abc','def'] and has a match with the 1st and 4th elements of m, the result is then [1,0,0,1].

到目前为止，这是我的方法(可能太慢了，缺少零):

Here's my approach so far (probably way too slow and is missing zeros):

output=[]
for i in r:
    output.append([1 for x in m if x in i])

导致:

[[1, 1], [1, 1], [1, 1, 1]]

提前谢谢！

推荐答案

您可以使用嵌套列表推导，如下所示:

You can use a nested list comprehension like this:

>>> m = ['abc','bcd','cde','def']
>>> r = [['abc','def'],['bcd','cde'],['abc','def','bcd']]
>>> [[1 if mx in rx else 0 for mx in m] for rx in r]
[[1, 0, 0, 1], [0, 1, 1, 0], [1, 1, 0, 1]]

此外，您可以使用int(...)缩短1 if ... else 0，并且可以将r的子列表转换为set，这样单个mx in rx的查找会更快.

Also, you could shorten the 1 if ... else 0 using int(...), and you can convert the sublists of r to set, so that the individual mx in rx lookups are faster.

>>> [[int(mx in rx) for mx in m] for rx in r]
[[1, 0, 0, 1], [0, 1, 1, 0], [1, 1, 0, 1]]
>>> [[int(mx in rx) for mx in m] for rx in map(set, r)]
[[1, 0, 0, 1], [0, 1, 1, 0], [1, 1, 0, 1]]

虽然int(...)比1 if ... else 0短，但它似乎也较慢，因此您可能不应该使用它.在重复查找之前将r的子列表转换为set应该可以加快较长列表的速度，但是对于您的示例列表很短，实际上这比幼稚的方法要慢.

While int(...) is a bit shorter than 1 if ... else 0, it also seems to be slower, so you probably should not use that. Converting the sublists of r to set prior to the repeated lookup should speed things up for longer lists, but for you very short example lists, it's in fact slower than the naive approach.

>>> %timeit [[1 if mx in rx else 0 for mx in m] for rx in r]
100000 loops, best of 3: 4.74 µs per loop
>>> %timeit [[int(mx in rx) for mx in m] for rx in r]
100000 loops, best of 3: 8.07 µs per loop
>>> %timeit [[1 if mx in rx else 0 for mx in m] for rx in map(set, r)]
100000 loops, best of 3: 5.82 µs per loop

对于更长的列表，如预期的那样，使用set会变得更快:

For longer lists, using set becomes faster, as would be expected:

>>> m = [random.randint(1, 100) for _ in range(50)]
>>> r = [[random.randint(1,100) for _ in range(10)] for _ in range(20)]
>>> %timeit [[1 if mx in rx else 0 for mx in m] for rx in r]
1000 loops, best of 3: 412 µs per loop
>>> %timeit [[1 if mx in rx else 0 for mx in m] for rx in map(set, r)]
10000 loops, best of 3: 208 µs per loop

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