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蟒蛇。 检查索引是否存在于可能的字典列表中

[英]Python. Check if index exists in a list of possible dictionaries

I have this: 我有这个:

list_name = [0, 1, 2, 3]
list_name[0] = {}
list_name[0]['test'] = 'any value'

I want to know if a key of the list exists or not. 我想知道列表中的键是否存在。 Usually I use: 通常我使用:

if 3 not in list_name:
    print("this doesn't exist")
else:
    print("exists")

And testing this for example for number 3 is working. 例如,对3号进行测试就可以了。 It says "exists". 它说“存在”。 If I check for number 999 is working, it says "this doesn't exist". 如果我检查数字999是否正常工作,则显示“此地址不存在”。

The problem is that it is not working for 0. As you can see, 0 value in the list has a dictionary. 问题在于它不适用于0。如您所见,列表中的0值具有字典。 And I need to check if 0 exists or not in the list (doesn't matter if it has a dictionary or not). 而且我需要检查列表中是否存在0(无论是否具有字典都没有关系)。 How to achieve this? 如何实现呢? Using python3, thank you. 使用python3,谢谢。

If element 0 exists in the list, then the length of the list must be greater than zero. 如果列表中存在元素0 ,则列表的长度必须大于零。 So you could use: 因此,您可以使用:

if len(list_name) > 0:
    print("0 exists")
else:
    print("0 does not exist")

As a side note, {} is a dictionary, not an array. 另外, {}是字典,而不是数组。

Use try except for check index is exists or not 使用try except检查索引是否存在

try:
    if list_name[6]:
        print("exists")

except IndexError:
    print("this doesn't exist")

Output 输出量

this doesn't exist 这个不存在

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