算法來比較兩個列表，並得到相同的元素在python

Question

我不得不名單，其中有他們一些共同的要素：

p = [('link1/d/b/c', 'target1/d/b/c'), ('link2/a/g/c', 'target2/a/g/c'), ..., ('linkn/b/b/f', 'targetn/b/b/f')] 

q = [['target1/d/b/c', 'target1', 123, 334], ['targetn/b/b/f', 'targetn', 23, 64], ... ,['targetx/f/f/f', 'targetx', 999, 888]] 

我試着對它們進行比較，找到共同的元素，然後做一些工作與結果：

do_job('target1/d/b/c', 'target1', 123, 334, 'link1/d/b/c') 

現在即時通訊使用簡單，很慢alghortihm：

for item in p: 
    link = item[0] 
    target = item[1] 
    for item2 in q: 
     target2 = item2[0] 
     if target2 == target: 
      do_some_job(...) 

我吼聲，那我需要比較這兩個列表，並獲得創建一個列表將包含所有的元素，如：

pq = [['target1/d/b/c', 'target1', 123, 334, 'link1/d/b/c'], ..., ['targetn/b/b/f', 'targetn', 23, 64, 'linkn/b/b/f']] 

，然後調用do_some_job(pq)與其說這是每次當我發現同一元素

的如何獲得它？

問候

Answer 1

使用chain()拼合兩個列表，然後用set()和intersection()得到共同的元素。

In [78]: from itertools import chain 

In [79]: p 
Out[79]: 
[('link1/d/b/c', 'target1/d/b/c'), 
('link2/a/g/c', 'target2/a/g/c'), 
('linkn/b/b/f', 'targetn/b/b/f')] 

In [80]: q 
Out[80]: 
[['target1/d/b/c', 'target1', 123, 334], 
['targetn/b/b/f', 'targetn', 23, 64], 
['targetx/f/f/f', 'targetx', 999, 888]] 

In [81]: set(chain(*p)).intersection(set(chain(*q))) 
Out[81]: set(['target1/d/b/c', 'targetn/b/b/f']) 

或使用列表理解與短路：

In [86]: [j for i in p for j in i if j in (z for y in q for z in y)] 
Out[86]: ['target1/d/b/c', 'targetn/b/b/f'] 

或使用any()：

In [87]: [j for i in p for j in i if any (j==z for y in q for z in y)] 
Out[87]: ['target1/d/b/c', 'targetn/b/b/f'] 

timeit：

In [93]: %timeit set(chain(*p)).intersection(set(chain(*q))) 
100000 loops, best of 3: 7.38 us per loop      ## winner 

In [94]: %timeit [j for i in p for j in i if j in (z for y in q for z in y)] 
10000 loops, best of 3: 24.9 us per loop 

In [95]: %timeit [j for i in p for j in i if any (j==z for y in q for z in y)] 
10000 loops, best of 3: 27.4 us per loop 

In [97]: %timeit [x for x in chain(*p) if x in chain(*q)] 
10000 loops, best of 3: 12.6 us per loop 

Answer 2

你或許應該使用的字典：

target_to_link = dict((v,k) for (k,v) in p) 
for item in q: 
    args = item + [target_to_link[item[0]] 
    do_some_job(*args) 

的target_to_link詞典讓你從你的目標的相應鏈接。只要確保你沒有幾個目標共享相同的鏈接...

在for循環，我們剛剛創建的，結合參數args的臨時列表您item（例如，['target1/d/b/c', 'target1', 123, 334]）與相應的鏈接，我們使用function(*args)語法...

---

如果您需要在p循環相反，你可以構建一個字典一樣

target_to_args = dict((k[0],k[1:]) for k in q) 

然後像做

for (link, target) in p: 
    args = [target] + target_to_args[target] + [link] 
    do_some_job(*args) 

Answer 3

與chain列表理解應該工作：

[x for x in chain(*p) if x in chain(*q)]