如何向现有词典中添加关键字?它没有.add()方法。


当前回答

dictionary[key] = value

其他回答

还有一个名字奇怪,行为怪异,但仍然很方便的dict.setdefault()。

This

value = my_dict.setdefault(key, default)

基本上就是这样:

try:
    value = my_dict[key]
except KeyError: # key not found
    value = my_dict[key] = default

例如。,

>>> mydict = {'a':1, 'b':2, 'c':3}
>>> mydict.setdefault('d', 4)
4 # returns new value at mydict['d']
>>> print(mydict)
{'a':1, 'b':2, 'c':3, 'd':4} # a new key/value pair was indeed added
# but see what happens when trying it on an existing key...
>>> mydict.setdefault('a', 111)
1 # old value was returned
>>> print(mydict)
{'a':1, 'b':2, 'c':3, 'd':4} # existing key was ignored

首先检查密钥是否已存在:

a={1:2,3:4}
a.get(1)
2
a.get(5)
None

然后可以添加新的键和值。

dictionary[key] = value

在不使用add的情况下向字典中添加关键字

        # Inserting/Updating single value
        # subscript notation method
        d['mynewkey'] = 'mynewvalue' # Updates if 'a' exists, else adds 'a'
        # OR
        d.update({'mynewkey': 'mynewvalue'})
        # OR
        d.update(dict('mynewkey'='mynewvalue'))
        # OR
        d.update('mynewkey'='mynewvalue')
        print(d)  # {'key': 'value', 'mynewkey': 'mynewvalue'}
        # To add/update multiple keys simultaneously, use d.update():
        x = {3:4, 5:6, 7:8}
        d.update(x)
        print(d) # {'key': 'value', 'mynewkey': 'mynewvalue', 3: 4, 5: 6, 7: 8}
        # update operator |= now works for dictionaries:
        d |= {'c':3,'d':4}
        # Assigning new key value pair using dictionary unpacking.
        data1 = {4:6, 9:10, 17:20}
        data2 = {20:30, 32:48, 90:100}
        data3 = { 38:"value", 99:"notvalid"}
        d = {**data1, **data2, **data3}
        # The merge operator | now works for dictionaries:
        data = data1 | {'c':3,'d':4}
        # Create a dictionary from two lists
        data = dict(zip(list_with_keys, list_with_values))

通过为字典中的键指定值,可以在字典中创建新的键/值对

d = {'key': 'value'}
print(d)  # {'key': 'value'}

d['mynewkey'] = 'mynewvalue'

print(d)  # {'key': 'value', 'mynewkey': 'mynewvalue'}

如果键不存在,则添加该键并指向该值。如果它存在,它所指向的当前值将被覆盖。