Question

如果我想迭代存储在元组中的字典值。

我需要返回持有＆＃34; CI＆＃34;的对象。值，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。

如果有更快的方式这样做，那将非常有帮助

Answer 1

迭代字典

的方法

首先，有几种方法可以循环遍历字典。

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。

循环使用字典的值：

>>> 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]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

List comprehensions很棒。对于仅搜索CI＆＃39;

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目的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']

如果后两者看起来有点过于复杂，那么你可以将最后一位重写为一个单独的函数。

>>> 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 for 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

试试这个：

>>> z = {'x':(123,'SE',2,1),'z':(124,'CI',1,1)}
>>> list(filter(lambda x:'CI' in z.get(x),z))
['z']

Answer 4

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

Answer 5

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

迭代dict值

5 个答案:

迭代字典

列表理解