Question

a = [
    {
        "id" : 15,
        "name" : "abc"
    },
    {
        "id" : 16,
        "name" : "xyz"
    },
    {
        "id" : 17,
        "name" : "pqr"
    }
]

b = [15,17]

我上面有两个列表，如果列表b中没有id，我想从列表a中删除该对象。任何帮助，该怎么做？

输出清单：

[
    {
        "id" : 15,
        "name" : "abc"
    },
    {
        "id" : 17,
        "name" : "pqr"
    }
]

Answer 1

迭代倒档，实现高效的就地删除。将b转换为set以进行恒定时间查找。

c = set(b)    
for i in reversed(range(len(a))):  # thanks to @juanpa.arrivillaga for this bit 
    if a[i]['id'] not in c:
        del a[i]

否则，使用列表推导并再次创建：

a = [i for i in a if i['id'] in c]

print(a)
[{'id': 15, 'name': 'abc'}, {'id': 17, 'name': 'pqr'}]

Answer 2

因此使用list comprehention：

[i for i in a if i['id']!=b[0] and i['name']!=b[1]]

Answer 3

您可以按如下方式使用列表推导：

a = [
     {
        "id" : 15,
        "name" : "abc"
    },
    {
        "id" : 16,
        "name" : "xyz"
    },
    {
        "id" : 17,
        "name" : "pqr"
    }]

b = [15, 17]

print [a[_] for _ in xrange(len(a)) if a[_]["id"] in b]

<强>输出：

[{'id': 15, 'name': 'abc'}, {'id': 17, 'name': 'pqr'}]

从python列表中删除对象

3 个答案: