用Python过滤列表

词典源列表

[
    {'main_id': 0, '_id': ObjectId('111111111111111'), 'key1': 'val1'}, 
    {'main_id': 1, '_id': ObjectId('222222222222222'), 'key1': 'val2'}, 
    {'main_id': 1, '_id': ObjectId('333333333333333'), 'key1': 'val3'}, 
    {'main_id': 4, '_id': ObjectId('444444444444444'), 'key1': 'val4'},
    {'main_id': 2, '_id': ObjectId('555555555555555'), 'key1': 'val5'},
]

[
    {'main_id': 0, '_id': ObjectId('111111111111111'), 'key1': 'val1'}, 
    {'main_id': 1, '_id': ObjectId('222222222222222'), 'key1': 'val2'}, 
    {'main_id': 2, '_id': ObjectId('555555555555555'), 'key1': 'val3'},
]

已筛选词典列表

[
    {'main_id': 0, '_id': ObjectId('111111111111111'), 'key1': 'val1'}, 
    {'main_id': 1, '_id': ObjectId('222222222222222'), 'key1': 'val2'}, 
    {'main_id': 1, '_id': ObjectId('333333333333333'), 'key1': 'val3'}, 
    {'main_id': 4, '_id': ObjectId('444444444444444'), 'key1': 'val4'},
    {'main_id': 2, '_id': ObjectId('555555555555555'), 'key1': 'val5'},
]

[
    {'main_id': 0, '_id': ObjectId('111111111111111'), 'key1': 'val1'}, 
    {'main_id': 1, '_id': ObjectId('222222222222222'), 'key1': 'val2'}, 
    {'main_id': 2, '_id': ObjectId('555555555555555'), 'key1': 'val3'},
]

我想获得一个新的词典列表：

[
    {'main_id': 1, '_id': ObjectId('333333333333333'), 'key1': 'val3'}, 
    {'main_id': 4, '_id': ObjectId('444444444444444'), 'key1': 'val4'},
]

换句话说，我希望一个新列表包含筛选列表中不存在的值。 你的想法？

一行：

full = [
    {'main_id': 0, '_id': ObjectId('111111111111111'), 'key1': 'val1'}, 
    {'main_id': 1, '_id': ObjectId('222222222222222'), 'key1': 'val2'}, 
    {'main_id': 1, '_id': ObjectId('333333333333333'), 'key1': 'val3'}, 
    {'main_id': 4, '_id': ObjectId('444444444444444'), 'key1': 'val4'},
    {'main_id': 2, '_id': ObjectId('555555555555555'), 'key1': 'val5'},
]
filtered = [
    {'main_id': 0, '_id': ObjectId('111111111111111'), 'key1': 'val6'}, 
    {'main_id': 1, '_id': ObjectId('222222222222222'), 'key1': 'val7'}, 
    {'main_id': 2, '_id': ObjectId('555555555555555'), 'key1': 'val8'},
]

diff = [x for x in full if x not in filtered]

result = [s for s in original if s not in filtered]

---

或使用：

使用原始数据：

# dummy!
ObjectId = str


original = [
    {'main_id': 0, '_id': ObjectId('111111111111111'), 'key1': 'val1'}, 
    {'main_id': 1, '_id': ObjectId('222222222222222'), 'key1': 'val2'}, 
    {'main_id': 1, '_id': ObjectId('333333333333333'), 'key1': 'val3'}, 
    {'main_id': 4, '_id': ObjectId('444444444444444'), 'key1': 'val4'},
    {'main_id': 2, '_id': ObjectId('555555555555555'), 'key1': 'val5'},
]

filtered = [
    {'main_id': 0, '_id': ObjectId('111111111111111'), 'key1': 'val1'}, 
    {'main_id': 1, '_id': ObjectId('222222222222222'), 'key1': 'val2'}, 
    {'main_id': 2, '_id': ObjectId('555555555555555'), 'key1': 'val3'},
]

new_list = [d for d in original if d not in filtered]

print(new_list)

---

在您的示例中，筛选列表不是第一个列表的子集。为什么？我认为

val3

应该是过滤列表中的

val5

？可能最好先将

filtered

复制到一个集合中，因为成员资格检查会大大加快。（显然，这取决于输入的大小以及性能/内存使用的需求）。啊，CPU与RAM。经典的决斗。但它不允许你复制到设置-将抛出dict obj是不可破坏的。