迭代dict值

Question

如果我想迭代存储在元组中的字典值.

我需要返回保存"CI"值的对象,我假设我需要某种for循环:

z = {'x':(123,SE,2,1),'z':(124,CI,1,1)}
for i, k in db.z:
    for k in db.z[i]:
        if k == 'CI':
            return db.z[k]

我可能在这里遗漏了一些东西,一个参考点会很好.

如果有更快的方式,那么这一切都会有很大的帮助

Answer 1

迭代字典的方法

首先,有几种方法可以循环字典.

直接在字典上循环:

>>> z = {'x':(123,'SE',2,1),'z':(124,'CI',1,1)}
>>> for key in z:
...     print key,
...
'x' 'z'

请注意,循环遍历字典时返回的循环变量是键,而不是与这些键关联的值.

循环遍历字典的值:

>>> z = {'x':(123,'SE',2,1),'z':(124,'CI',1,1)}
>>> for value in z.values(): # Alternatively itervalues() for memory-efficiency (but ugly)
...     print value,
...
(123,'SE',2,1) (124,'CI',1,1)

循环键和值:

>>> z = {'x':(123,'SE',2,1),'z':(124,'CI',1,1)}
>>> for key, value in z.items(): # Again, iteritems() for memory-efficiency
...     print key, value,
...
'x' (123,'SE',2,1) 'z' (124,'CI',1,1)

后两者比循环键和运行z [key]以获得值更有效.它也可以说更具可读性.

建立在这些...

列表理解

列表理解很棒.对于仅搜索"CI"的简单情况:

>>> z = {'x':(123,'SE',2,1),'z':(124,'CI',1,1)}
>>> [key for key, value in z.items() if 'CI' in value]
['z']

用于查找包含多个搜索项的dict键:

>>> z = {'x':(123,'SE',2,1),'z':(124,'CI',1,1)}
>>> search_items = ('CI', 1) # Only keys that hold both CI and 1 will match
>>> [key for key, value in z.items() if all(item in value for item in search_items)]
['z']

用于查找包含多个搜索项中的任何一个的dict键:

>>> z = {'x':(123,'SE',2,1),'z':(124,'CI',1,1)}
>>> search_items = ('CI', 'SE', 'JP') # Keys that hold any of the three items will match
>>> [key for key, value in z.items() if any(item in value for item in search_items)]
['x', 'z']

如果后两个看起来有点过于复杂,那么您可以将最后一位重写为单独的函数.

>>> z = {'x':(123,'SE',2,1),'z':(124,'CI',1,1)}
>>> search_items = ('CI', 'SE', 'JP') # Keys that hold any of the three items will match
>>> def match_any(dict_value, search_items):
...     return any(item in dict_value for item in search_items)
...
>>> [key for key, value in z.items() if match_any(value, search_items)]
['x', 'z']

一旦习惯了[x for x in iterable if condition(x)]语法,该格式应该非常容易阅读和遵循.

Answer 2

z = {'x':(123,"SE",2,1),'q':(124,"CI",1,1)}
for i in z.keys(): #reaching the keys of dict
    for x in z[i]: #reaching every element in tuples
        if x=="CI": #if match found..
            print ("{} holding {}.".format(i,x)) #printing it..

这可能会解决您的问题.

输出:

>>> 
q holding CI.
>>> 

编辑您的评论:

def func(*args):
    mylist=[]
    z = {'x':(123,"SE",2,1),'q':(124,"CI",1,1)}
    for x,y in z.items():
        for t in args:
            if t in y:
                mylist.append(x)
    return mylist
print (func(1,"CI"))

输出:

>>> 
['q', 'q', 'x']
>>>  

希望这是你想要的,否则第一种方法已经打印了所有键,例如输出:

if x==1 or x=="CI":

>>> 
x holding 1.
q holding CI.
q holding 1.
q holding 1.
>>> 

Answer 3

如果您只对值感兴趣，则无需检索密钥：

在 Python 2.x 中：

z = {'x':(123,"SE",2,1),'q':(124,"CI",1,1)}
for value in z.itervalues():
    if 'CI' in value:
        return value

在 Python 3.x 中：

z = {'x':(123,"SE",2,1),'q':(124,"CI",1,1)}
for value in z.values():
    if 'CI' in value:
        return value