python 列表取交集

Question

{代码...} 我想取 a 和 b 的交集，结果应该是 [3,4],[3,5] ,就是第一元素要一样我的笨办法是先取 a 和 b 第一个元素的列表，取它们的交集，然后再分别从 a 和 b 中找请问有没有更好的办法？ 补充：对不起大家我没...

怪我咯 · Answer

Because I don’t know the details you want yet, let me give you a tentative answer first:

from collections import defaultdict

a = [[1,2],[3,4],[3,6]]
b = [[1,2],[2,3],[3,5]]

dic_a = defaultdict(list)
dic_b = defaultdict(list)

for item in a:
    dic_a[item[0]].append(item)

for item in b:
    dic_b[item[0]].append(item)

common_keys = set(dic_a.keys()) & set(dic_b.keys())

for key in common_keys:
    print 'key:', key, 'a:', dic_a[key], 'b:',dic_b[key]

Questions I answered: Python-QA

天蓬老师 · Answer

My idea is to create a dictionary and then search it directly

a = [[1,2],[3,4]]
b = [[2,3],[3,5]]
dic_a = {}
dic_b = {}
for item in a:
    if item[0] in dic_a:
        dic_a[item[0]].append(item)
    else:
        dic_a[item[0]] = item
for item in b:
    if item[0] in dic_b:
        dic_b[item[0]].append(item)
    else:
        dic_b[item[0]] = item
for key in dic_a:
    if key in dic_b:
        print dic_a[key],dic_b[key]

PHP中文网 · Answer

Isn’t it supposed to be like this?

list(set(a).intersection(set(b)))

Modification:

Is this okay?

a = [[1, 2], [3, 4], [2, 6]]
b = [[2, 3], [3, 5],[1,5]]
for i in itertools.product(a, b):
    if i[0][0] == i[1][0]:
        print(i)

伊谢尔伦 · Answer

I am not here to answer the question. The poster I saw is using line_profiler. I would like to ask. I just downloaded line_profiler. I don’t know if it is installed. I downloaded the source code and then ran it directly python setup.py install But when I use it, I get an error: NameError: name 'profile' is not defined. I don't know the reason. How to solve it?