Python：检查嵌套字典中是否存在键

Question

I am trying to check if a certain key exists in a nested dict.

eg:

x = [{
    '11': { 
        0: [
               {
                'bb_id': '122',
                'cc_id': '4343'
               }
           ],           
        
        1: [
               {
                'bb_id': '4334',
                'cc_id': '3443'
               },
               {
                'bb_id': '5345',
                'cc_id': '257'
               }
           ]
    }
}]

I need to check if the key '11' exists in x , and further if the key 0 exists in the value of the key '11' .

I've tried doing:

print(any(0 in d for d in x if '11' in d))

Answer 1

It seems this would achieve what you are trying to do:

any(['11' in d.keys() and 0 in d['11'].keys() for d in x])

Explanation:

• Iterate over each dictionary in the list named x ;
• Search each dictionary for a key of '11' , and search its value (which is expected to be a dictionary too) for a key of 0 ;
• If both conditions are met, return True ; else, return False .

---

In the question's comments, Sushanth has provided an even shorter and possibly more Pythonic way, using a generator and the dictionary's get() method with an empty dictionary as a fallback value:

any(d.get('11', {}).get(0) for d in x)

Answer 2

What you have here is a list of dicts with values as a dict of lists of dicts.

Try this one-liner list comprehension. x here is a list of dicts (in this case with a single dict). The code below returns True for every dict that is in x if '11' exists in its key AND if 0 exists in the key of value of '11' . Only if both conditions are met, you get a TRUE else FALSE -

1. Iterate over items of x
2. Iterate over the dicts inside items of x
3. Check if the items in x have a key = '11' AND if items in dicts inside items of x have key = 0
4. If both conditions are met, return True else False

#Items to detect
a = '11'
b = 0

#Iterate of the nested dictionaries and check conditions
result = [(k==a and b in v.keys()) for i in x for k,v in i.items()]
print(result)

[True]

Answer 3

A raw way to do that could just be an if condition:

if '11' in x[0]:
    print("11 in x")

if 0 in x[0]['11']:
    print("0 in 11")

You could also use a for loop:

for d in x:
    if '11' in d:
        print("11 in d")
        if any(d['11']) and 0 in d['11']:
            print("0 in 11")

Answer 4

has_key = lambda key, _dict: re.search("{%s" % str(key),   str(_dict).replace(" ", ""))

Answer 5

I came up with another way to do it.

(d.get('key1') or {}).get('key2')

Our example:

(x[0].get('11') or {}).get(0)

Explanation of Steps:

x[0] = Gets dictionary

(x[0].get('11') or {}) = Tries and gets key '11' key from dictionary x ; if the key doesn't exist, then just set a blank dictionary {} .

(x[0].get('11') or {}).get(0) = Remember .get() on a dictionary will return None if the 0 doesn't exist rather than throw an error.

get(key[, default])

Return the value for key if key is in the dictionary, else default. 
If default is not given, it defaults to None, so that this method never raises a KeyError.

See Python docs here for.get()