python - 在 python 中的两个列表列表之间查找公共(public)元素的最快方法

Question

我有一个列表如下。

mylist = 
[  
   [  
      [  
         "chocolate_pudding",
         920.8000000000001
      ],
      [  
         "caramel_pudding",
         345.59999999999997
      ],
      [  
         "pudding",
         248.0
      ],
      [  
         "banana_pudding",
         27.599999999999998
      ]
   ],
   [  
      [  
         "biscuits",
         190.8
      ],
      [  
         "chocolates",
         33.599999999999994
      ],
      [  
         "chocolate_pudding",
         920.8000000000001
      ]
   ],
   [  
      [  
         "tiramusu",
         145.8
      ]
   ],
   [  
      [  
         "cakes",
         139.29999999999998
      ]
   ],
   [  
      [  
         "butter_cakes",
         133.0
      ]
   ],
   [  
      [  
         "chocolate_pudding",
         920.8000000000001
      ]
   ]
]

我想找到在列表中出现多次的元素(例如，["chocolate_pudding", 920.8000000000001])并想删除重复的元素，同时保留第一个条目。

所以，我的输出应该如下所示。

mylist = 
[  
   [  
      [  
         "chocolate_pudding",
         920.8000000000001
      ],
      [  
         "caramel_pudding",
         345.59999999999997
      ],
      [  
         "pudding",
         248.0
      ],
      [  
         "banana_pudding",
         27.599999999999998
      ]
   ],
   [  
      [  
         "biscuits",
         190.8
      ],
      [  
         "chocolates",
         33.599999999999994
      ]
   ],
   [  
      [  
         "tiramusu",
         145.8
      ]
   ],
   [  
      [  
         "cakes",
         139.29999999999998
      ]
   ],
   [  
      [  
         "butter_cakes",
         133.0
      ]
   ]
]

我一直在尝试的代码如下。

mylist_copy = mylist

for item in mylist:
    myindex = mylist.index(item)
    #print(item)

    for single_item in item:
        #print(single_item)
        for item_copy in mylist_copy:
            if mylist_copy.index(item_copy) != myindex:
                if single_item in item_copy:
                    print(single_item)

因为它有很多 for 循环，所以我想要一种有效的方法来完成它。注:我也试过了；

mylist_copy = mylist

for item in mylist:
    myindex = mylist.index(item)
    for item_copy in mylist_copy:
          if mylist_copy.index(item_copy) != myindex:
                print(set(item).intersection(item_copy))

但是，交集不支持列表。

在 python 中有没有简单快捷的方法？

Answer 1

使用 set() 对象并保留子列表的顺序:

mylist = [[["chocolate_pudding", 920.8000000000001], ["caramel_pudding", 345.59999999999997], 
          ["pudding", 248.0], ["banana_pudding", 27.599999999999998]], [["biscuits", 190.8], 
          ["chocolates", 33.599999999999994], ["chocolate_pudding", 920.8000000000001]], 
          [["tiramusu", 145.8]], [["cakes", 139.29999999999998]], [["butter_cakes", 133.0]], 
          [["chocolate_pudding", 920.8000000000001]]]

result, foods = [], set()
for sub_l in mylist:
    new_sublist = []
    for i in sub_l:
        if i[0] not in foods:     # on the 1st occurrence of `foodstuff` name
            new_sublist.append(i)
            foods.add(i[0])       # add `foodstuff` into set of unique foods
    if new_sublist: result.append(new_sublist)

print(result)

输出:

[[['chocolate_pudding', 920.8000000000001], ['caramel_pudding', 345.59999999999997], ['pudding', 248.0], ['banana_pudding', 27.599999999999998]], [['biscuits', 190.8], ['chocolates', 33.599999999999994]], [['tiramusu', 145.8]], [['cakes', 139.29999999999998]], [['butter_cakes', 133.0]]]