python - 带条件的嵌套列表的列表理解

Question

我有一个 list

list1 = [ [2,3,4] ,[5,6,7] ,[1,5,8] ,[2,3,{'key1':12}] ]

从list1中我需要删除具有字典元素的列表并生成list2，它应该是[ [2,3,4] ,[5,6,7] ,[1,5,8] ]

使用 for 循环我可以得到 list2 就像

for e1 in list1:
    condition = 0
    for e2 in e1:
        if type(e2) is dict:
            condition = 1
    if not condition:
        list2.append(e1)

如何通过列表理解获得 list2 的结果？

Answer 1

尝试:

list1 = [[2, 3, 4], [5, 6, 7], [1, 5, 8], [2, 3, {"key1": 12}]]

list2 = [l for l in list1 if dict not in map(type, l)]
print(list2)

打印:

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

---

或者使用any()(感谢@chepner):

list2 = [l for l in list1 if not any(isinstance(x, dict) for x in l)]
print(list2)