我重写了公认的答案，并为新手提供了方便。

#break it down and understand it intuitively.
new = {}
for (key, value) in data:
    if key not in new:
        new[key] = [] # this is core of setdefault equals to new.setdefault(key, [])
        new[key].append(value)
    else:
        new[key].append(value)


# easy with setdefault
new = {}
for (key, value) in data:
    group = new.setdefault(key, []) # it is new[key] = []
    group.append(value)



# even simpler with defaultdict
new = defaultdict(list)
for (key, value) in data:
    new[key].append(value) # all keys have a default value of empty list []

另外，我将这些方法分类为参考:

dict_methods_11 = {
            'views':['keys', 'values', 'items'],
            'add':['update','setdefault'],
            'remove':['pop', 'popitem','clear'],
            'retrieve':['get',],
            'copy':['copy','fromkeys'],}

下面是一些setdefault的例子来展示它的有用性:

"""
d = {}
# To add a key->value pair, do the following:
d.setdefault(key, []).append(value)

# To retrieve a list of the values for a key
list_of_values = d[key]

# To remove a key->value pair is still easy, if
# you don't mind leaving empty lists behind when
# the last value for a given key is removed:
d[key].remove(value)

# Despite the empty lists, it's still possible to 
# test for the existance of values easily:
if d.has_key(key) and d[key]:
    pass # d has some values for key

# Note: Each value can exist multiple times!
"""
e = {}
print e
e.setdefault('Cars', []).append('Toyota')
print e
e.setdefault('Motorcycles', []).append('Yamaha')
print e
e.setdefault('Airplanes', []).append('Boeing')
print e
e.setdefault('Cars', []).append('Honda')
print e
e.setdefault('Cars', []).append('BMW')
print e
e.setdefault('Cars', []).append('Toyota')
print e

# NOTE: now e['Cars'] == ['Toyota', 'Honda', 'BMW', 'Toyota']
e['Cars'].remove('Toyota')
print e
# NOTE: it's still true that ('Toyota' in e['Cars'])

我重写了公认的答案，并为新手提供了方便。

#break it down and understand it intuitively.
new = {}
for (key, value) in data:
    if key not in new:
        new[key] = [] # this is core of setdefault equals to new.setdefault(key, [])
        new[key].append(value)
    else:
        new[key].append(value)


# easy with setdefault
new = {}
for (key, value) in data:
    group = new.setdefault(key, []) # it is new[key] = []
    group.append(value)



# even simpler with defaultdict
new = defaultdict(list)
for (key, value) in data:
    new[key].append(value) # all keys have a default value of empty list []

另外，我将这些方法分类为参考:

dict_methods_11 = {
            'views':['keys', 'values', 'items'],
            'add':['update','setdefault'],
            'remove':['pop', 'popitem','clear'],
            'retrieve':['get',],
            'copy':['copy','fromkeys'],}

我经常使用setdefault，在字典中设置默认值(!!)通常是os。环境字典:

# Set the venv dir if it isn't already overridden:
os.environ.setdefault('VENV_DIR', '/my/default/path')

简单点说，是这样的:

# Set the venv dir if it isn't already overridden:
if 'VENV_DIR' not in os.environ:
    os.environ['VENV_DIR'] = '/my/default/path')

值得注意的是，你也可以使用结果变量:

venv_dir = os.environ.setdefault('VENV_DIR', '/my/default/path')

但与违约之前相比，这已经不那么必要了。

除了上述建议之外，如果您不想修改已经设置的值，setdefault可能会很有用。例如，当你有重复的数字，你想把它们当作一组。在这种情况下，如果遇到已设置的重复键，则不会更新该键的值。您将保留第一次遇到的值。就好像你只迭代/更新重复的键一次。

下面是一个记录排序列表中键/元素索引的代码示例:

nums = [2,2,2,2,2]
d = {}
for idx, num in enumerate(sorted(nums)):
    # This will be updated with the value/index of the of the last repeated key
    # d[num] = idx # Result (sorted_indices): [4, 4, 4, 4, 4]
    # In the case of setdefault, all encountered repeated keys won't update the key.
    # However, only the first encountered key's index will be set 
    d.setdefault(num,idx) # Result (sorted_indices): [0, 0, 0, 0, 0]

sorted_indices = [d[i] for i in nums]

在CPython中setdefault的另一个用例是，它在所有情况下都是原子的，而defaultdict将不是原子的，如果你使用从lambda创建的默认值。

cache = {}

def get_user_roles(user_id):
    if user_id in cache:
        return cache[user_id]['roles']

    cache.setdefault(user_id, {'lock': threading.Lock()})

    with cache[user_id]['lock']:
        roles = query_roles_from_database(user_id)
        cache[user_id]['roles'] = roles

如果两个线程执行缓存。同时设置default，它们中只有一个能够创建默认值。

如果你使用defaultdict:

cache = defaultdict(lambda: {'lock': threading.Lock()}

这将导致竞态条件。在我上面的例子中，第一个线程可以创建一个默认锁，第二个线程可以创建另一个默认锁，然后每个线程可以锁定自己的默认锁，而不是每个线程试图锁定单个锁的预期结果。

从概念上讲，setdefault的基本行为是这样的(如果你使用空列表、空dict、int或其他不是用户python代码(如lambda)的默认值，defaultdict也会这样表现):

gil = threading.Lock()

def setdefault(dict, key, value_func):
    with gil:
        if key not in dict:
            return
       
        value = value_func()

        dict[key] = value

从概念上讲，defaultdict的基本行为是这样的(只有在使用lambda这样的python代码时-如果使用空列表则不是这样):

gil = threading.Lock()

def __setitem__(dict, key, value_func):
    with gil:
        if key not in dict:
            return

    value = value_func()

    with gil:
        dict[key] = value