Question

我有这样的代码（Python 2.7 +）：

target = "some_string_key1_key3_key5_key6"

some_dict = {
    'key1': 'value1',
    'key2': 'value1',
    'key3': 'value2',
    'key4': 'value2',
    'key5': 'value3',
    'kyy6': 'value4',
}

for k, v in some_dict.iteritems():
    target = target.replace(k, v)

如您所见，有不同的键使用相同的值（key1和key2 / key3和key4）。在我的情况下，所有键和值都是字符串。

是否有可能在字典初始化时使用更少的代码并仍然获得

for k, v in some_dict.iteritems()

行为没有改变，只有标准的python 2.7模块？

我尝试了multi_key_dict，但对我来说它不太方便，因为语法略有不同。

我想要实现的就是：

new_cool_dict = {
    ['key1', 'key2', ]: 'value1',
    ['key3', 'key4', ]: 'value2',
    'key5': 'value3',
    'key6': 'value6',
}

for k, v in new_cool_dict.iteritems():
    target = target.replace(k, v)

Answer 1

我发现这是 less 有用且可读，但你可以在一行上使用多个作业：

some_dict = {
    'key5': 'value3',
    'kyy6': 'value4',
}
some_dict['key1'] = some_dict['key2'] = 'value1'
some_dict['key3'] = some_dict['key4'] = 'value2'

另一种方法是构建一个扩展元组键的工厂函数：

def expand_tuple_keys(d):
    result = {}
    for key, value in d.iteritems():
        if not hasattr(key, '__iter__'):
            key = (key,)
        result.update((k, value) for k in key)
    return result

然后使用：

new_cool_dict = expand_tuple_keys({
    ('key1', 'key2'): 'value1',
    ('key3', 'key4'): 'value2',
    'key5': 'value3',
    'key6': 'value6',
})

演示：

>>> def expand_tuple_keys(d):
...     result = {}
...     for key, value in d.iteritems():
...         if not hasattr(key, '__iter__'):
...             key = (key,)
...         result.update((k, value) for k in key)
...     return result
... 
>>> expand_tuple_keys({
...     ('key1', 'key2'): 'value1',
...     ('key3', 'key4'): 'value2',
...     'key5': 'value3',
...     'key6': 'value6',
... })
{'key3': 'value2', 'key2': 'value1', 'key1': 'value1', 'key6': 'value6', 'key5': 'value3', 'key4': 'value2'}

Answer 2

为方便起见，您也可以反过来定义词典：

my_dict = {
    'value1': ['key1', 'key2'],
    'value2': ['key3', 'key4'],
    'value3': ['key5']
}

然后将其转换为更有用的表示形式：

>>> dict((k, v) for (v, l) in my_dict.iteritems() for k in l)
{'key1': 'value1',
 'key2': 'value1',
 'key3': 'value2',
 'key4': 'value2',
 'key5': 'value3'}

Answer 3

我建议将密钥设为元组并稍微更改for循环，如下所示

target = "some_string_key1_key3_key5_key6"

some_dict = {
    ('key1', 'key2'): 'value1',
    ('key3', 'key4'): 'value2',
    ('key5',): 'value3',
    ('key6',): 'value6',
}

for k, v in some_dict.iteritems():
    target = reduce(lambda x, y: x.replace(y, v), k, target)

print target

<强>输出

some_string_value1_value2_value3_value6

Answer 4

反向键和值; - ）

dict = { 
   'value1': ['key1', 'key2'],
   'value2': ['key3', 'key3'],
   'value3': ['key5'],
   'value4': ['key6']

}



for v, ks in dict.iteritems:
   for k in ks:
       target = target.replace.(k, v)

鲍勃是你的叔叔（不过我可能会将k重命名为v，反之亦然

关于多键字典初始化的代码较少

4 个答案: