CODEWITHSUNDEEP
pythonpython-3.xdictionary
Ming Chow
Ming Chow

Reputation: 11

Python: check if a key exists in a nested dict

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

e.g:

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))

Upvotes: 0

Views: 5246

Answers (5)

Freddie von Stange
Freddie von Stange

Reputation: 331

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()

Upvotes: 1

pitamer
pitamer

Reputation: 1064

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)

Upvotes: 3

Umang shrestha
Umang shrestha

Reputation: 1

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

Upvotes: 0

Akshay Sehgal
Akshay Sehgal

Reputation: 19322

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]

Upvotes: 1

Alessandro Polverino
Alessandro Polverino

Reputation: 33

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")

Upvotes: 0

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