"Python：在多個列表中查找相同的項目"

Question

我有一個任意數量的列表的列表，例如：

[[1,2,3], [3,4,5], [5,6,7], [7,8,9]]

Answer 1

與您手動執行相同的方式：

seen = set()
repeated = set()
for l in list_of_lists:
  for i in set(l):
    if i in seen:
      repeated.add(i)
    else:
      seen.add(i)

順便說一句，這是一些人正在尋找的一個班輪（不包括進口）（應該比另一種方法效率低）

from itertools import *
reduce(set.union, (starmap(set.intersection, combinations(map(set, ll), 2))))

Answer 2

最干凈的方法可能是使用reduce：

def findCommon(L):
    def R(a, b, seen=set()):
        a.update(b & seen)
        seen.update(b)
        return a
    return reduce(R, map(set, L), set())

result = findCommon([[1,2,3], [3,4,5], [5,6,7], [7,8,9]])

結果是一個集合，但如果您確實需要一個列表，只需執行list(result) 。

Answer 3

這只找到所有列表共有的元素（即交集）：

 set.intersection(*[set(list) for list in list_of_lists])

Answer 4

參考： http : //docs.python.org/library/stdtypes.html#set

#!/usr/bin/python

ll = [[1,2,3], [3,4,5], [5,6,7], [7,8,9]]
ls = [set(l) for l in ll]

su = ls[0]  #union
ssd = ls[0] #symmetric_difference
for s in ls[1:]:
  su = su.union(s)
  ssd = ssd.symmetric_difference(s)

result = su.difference(ssd)
print list(result)

=>

[3, 5, 7]

修改和采用FP，

ll = [[1,2,3], [3,4,5], [5,6,7], [7,8,9]]

u = reduce(set.union, map(set, ll))
sd = reduce(set.symmetric_difference, map(set, ll))
print u - sd

=>

[3, 5, 7]

Answer 5

嘗試這個：

data = [[1,2,3], [3,4,5], [5,6,7], [7,8,9], [1,2,3]]

res = set()

for i in data:
    for j in data:
        if i is not j:
            res |= set(i) & set(j)

print res

Answer 6

您可以使用字典來獲取每個

from collections import defaultdict

init_list = [[1,2,3], [3,4,5], [5,6,7], [7,8,9]]
#defaultdict, every new key will have a int(0) as default value
d = defaultdict(int)
for values in init_list:
  #Transform each list in a set to avoid false positives like [[1,1],[2,2]]
  for v in set(values):
    d[v] += 1

#Get only the ones that are more than once
final_list = [ value for value,number in d.items() if number > 1 ]

Answer 7

>>> sets = [[1,2,3], [3,4,5], [5,6,7], [7,8,9]]
>>> seen = set()
>>> duplicates = set()
>>> 
>>> for subset in map(set, sets) :
...     duplicates |= (subset & seen)
...     seen |= subset
... 
>>> print(duplicates)
set([3, 5, 7])
>>> 

我嘗試使用 map/reduce 進行單行答案，但還不能完全理解。

Answer 8

我還在學習，但想分享我的答案。

num_list1 = ['3', '6', '5', '8', '33', '12', '7', '4', '72', '2', '42', '13']
num_list2 = ['3', '6', '13', '5', '7', '89', '12', '3', '33', '34', '1', '344', '42']

result = [int(num) for num in num_list1 if num in num_list2)
print(result)

Answer 9

l=[[1,2,3], [3,4,5], [5,6,7], [7,8,9]]
d={}
for x in l:
    for y in x:
        if not d.has_key(y):
            d[y]=0
        d[y]+=1
[x for x,y in d.iteritems() if y>1]

Answer 10

這是我的去處：

seen = set()
result = set()
for s in map(set, [[1,2,3], [3,4,5], [5,6,7], [7,8,9]]):
    result.update(s & seen)
    seen.update(s)
print result

這打印：

set([3, 5, 7])

Answer 11

您可以使用集合，請參閱http://docs.python.org/library/stdtypes.html#set

Answer 12

展平，排序，1 for 循環比較之前和之后的數字