使用 Python 从列表中删除特定重复项

Question

Let's say I have a list of objects in which I want to remove duplicates that have the same id and then keep the one that have public: true

recipes = [
       {'id': 1, 'public': True}, 
       {'id': 1, 'public': False}, 
       {'id': 2, 'public': False}, 
       {'id': 3, 'public': True},
       {'id': 3, 'public': False}
    ]

list_ = []
[list_.append(r) for r in recipes if r.id not in list_]

print(list_)

What would be the second condition to add?

Wanted result:

    recipes = [
       {'id': 1, public: True},  
       {'id': 2, public: False}, 
       {'id': 3, public: True},
    ]

Answer 1

You can use count to check if the entry is a duplicate.

recipes = [
   {'id': 1, 'public': True}, 
   {'id': 1, 'public': False}, 
   {'id': 2, 'public': False}, 
   {'id': 3, 'public': True},
   {'id': 3, 'public': False}
]

list = []
[list.append(r) for r in recipes if [i['id'] for i in recipes].count (r['id']) == 1 or r['public']]

return list

Answer 2

This answer has O(n) time, hope it's enough:

ids={}  # Use a set to check for dups
l=[]    # Answer list
for i in recipes: # O(n) time
    if i['id'] in ids:
        if i['public']:
              ids[i['id']]=i
    else:
        ids[i['id']]=len(l)  # renumber the index for later O(1) use
                             # len(l) is a O(1) function, so no need to keep a counter
        l.append(i)

Sorry it's not a one-liner,but I think this is good practice :)

Answer 3

You could first sort the list according to the 'public' key, in reverse order, so that all the 'public': True entries are on top. Then do a list comprehension to filter out unique 'id' values sequentially:

recipes = [{'id': 1, 'public': True}, 
           {'id': 1, 'public': False}, 
           {'id': 2, 'public': False}, 
           {'id': 3, 'public': True},
           {'id': 3, 'public': False}]

recipes.sort(key = lambda r: r['public'], reverse=True)

recipes_unique = []
[recipes_unique.append(r) for r in recipes if 
    r['id'] not in [s['id'] for s in recipes_unique]]

display(recipes_unique)

This delivers the desired result (if necessary, you could of course sort by 'id' again):

[{'id': 1, 'public': True},
 {'id': 3, 'public': True},
 {'id': 2, 'public': False}]