get(), takes keys and values( optional ) argument ( inputs ) .
Returns : the value of the input key , None if not found and optional parameter is not there.
The original dictionary remain same without any change.
get() with key having value
my_dict={'a':'One','b':'Two','c':'Three'}
print(my_dict.get('b'))
Output is here
Two
get() with key not present
my_dict={'a':'One','b':'Two','c':'Three'}
print(my_dict.get('d'))
print(my_dict)
Output ( there is no change in dictionary )
None
{'a': 'One', 'b': 'Two', 'c': 'Three'}
get() with optional value for not present key
If the key is not present then we can get a fixed return value.
my_dict={'a':'One','b':'Two','c':'Three'}
print(my_dict.get('d','not found'))
Output
not found
Example 1: Accessing Nested Dictionary Safely
my_dict = {'person': {'name': 'John', 'age': 30}}
print(my_dict.get('person', {}).get('age', 'Not found')) # Output: 30
Example 2: Using Functions for Default Values
def default_value():
return "Default value"
my_dict = {'a': 1}
print(my_dict.get('b', default_value())) # Output: "Default value"
Question using get()
Find out the frequency of occurrence of chars in a string?
my_str='Welcome to Python'
my_dict={}
for i in my_str:
my_dict[i]=my_dict.get(i,0)+1
print(my_dict)
Output
{'W': 1, 'e': 2, 'l': 1, 'c': 1, 'o': 3, 'm': 1, ' ': 2,
't': 2, 'P': 1, 'y': 1, 'h': 1, 'n': 1}
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