问题是嵌套。你必须一直工作到元素。 例如:
for item in list1:
for list2_item in list2:
if item[0] in list2_item:
print(item)
输出:
['i2_instance_floating_ip_association']
['i1_v1_instance_volume_attach']
另一种方法是首先拼合两份名单,并将其转换成集:
flat1 = set(x[0] for x in list1)
flat2 = set(y for x in list2 for y in x)
print(flat1 & flat2)
输出:
{'i1_v1_instance_volume_attach', 'i2_instance_floating_ip_association'}
加一些打印的调试和学习会发生什么:
for item in list1:
print('item', item)
for list2_item in list2:
print('list2_item', list2_item)
if item[0] in list2_item:
print(' found', item[0], 'in', list2_item)
else:
print(' did not find', item[0], 'in', list2_item)
输出:
item ['i2_instance_floating_ip_association']
list2_item ['i2_instance_floating_ip_association', 'i2_v1_instance_volume_attach']
found i2_instance_floating_ip_association in ['i2_instance_floating_ip_association', 'i2_v1_instance_volume_attach']
list2_item ['i1_instance_floating_ip_association', 'i1_v1_instance_volume_attach']
did not find i2_instance_floating_ip_association in ['i1_instance_floating_ip_association', 'i1_v1_instance_volume_attach']
item ['i1_v1_instance_volume_attach']
list2_item ['i2_instance_floating_ip_association', 'i2_v1_instance_volume_attach']
did not find i1_v1_instance_volume_attach in ['i2_instance_floating_ip_association', 'i2_v1_instance_volume_attach']
list2_item ['i1_instance_floating_ip_association', 'i1_v1_instance_volume_attach']
found i1_v1_instance_volume_attach in ['i1_instance_floating_ip_association', 'i1_v1_instance_volume_attach']
我正在使用你的方法输出。我认为遍历像[我为我在list1中,如果我在列表2]应该已经通过了所有元素。 –
只需在loop版本中添加一些打印件即可查看您获得的物品实际是什么样子。 –