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# The pop() method can accept either one or two parameters:
# The name of the key you want to remove (mandatory).
# The value that should be returned if a key cannot be found (optional).
dictionary.pop(key_to_remove, not_found)
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# Basic syntax:
del dictionary['key']
# Example usage:
dictionary = {'a': 3, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
del dictionary['c'] # Remove the 'c' key:value pair from dictionary
dictionary
--> {'a': 3, 'b': 2, 'd': 4, 'e': 5}
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dict.pop('key')
#optionally you can give value to return if key doesn't exist (default is None)
dict.pop('key', 'key not found')
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dict = {'an':30, 'example':18}
#1 Del
del dict['an']
#2 Pop (returns the value deleted, but can also be used alone)
#You can optionally set a default return value in case key is not found
dict.pop('example') #deletes example and returns 18
dict.pop('test', 'Key not found') #returns 'Key not found'
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>>> # initialise a dictionary with the keys “city”, “name”, “food”
>>> person1_information = {'city': 'San Francisco', 'name': 'Sam', "food": "shrimps"}
>>> # delete the key, value pair with the key “food”
>>> del person1_information["food"]
>>> # print the present personal1_information. Note that the key, value pair “food”: “shrimps” is not there anymore.
>>> print(person1_information)
{'city': 'San Francisco', 'name': 'Sam'}
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a_dictionary = {"one": 1, "two" : 2, "three": 3}
desired_value = 2
for key, value in a_dictionary.items():
if value == desired_value:
del a_dictionary[key]
break
print(a_dictionary)
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OUTPUT
{'one': 1, 'three': 3}
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# empty dictionary
dictionary = {}
# lists
list_1 = [1, 2, 3, 4, 5]
list_2 = ["a", "b", "c", "d", "e"]
# populate a dictionary.
for key, value in zip(list_1, list_2):
dictionary[key] = value
# output
print(dictionary)
# Item to delete
item_1 = "c"
# Delete item from dictionary - based on values specified
for key, value in dictionary.items():
if value == item_1:
dictionary.pop(key)
break
print(dictionary)