Question

我正在编写一个伪代码，我想要重复下面的两个“for”循环，直到data_changes和modem_changes中的所有键都相同，即不应该有任何键存在data_changes而不是modem_changes和版本反之亦然。我应该能够在此之后编写Python实现;谁能提供投入？编辑：

对

更感兴趣

如何为一个而不是另一个

2.重复for循环，直到data_changes和modem_changes中的键相同

data_changes = {
    '253036': [''], 
    '313115': ['313113']
    }

modem_changes = {'305403': [], 
                 '311957': ['253036', '312591']
                 }

s1 = set(data_changes.keys())
s2 = set(modem_changes.keys())
value1 = s2.difference(s1)
print value1
value2 = s1.difference(s2)
print value2

def func1 (data_changes,key):
    if key == '311957':
        output = ''
    if key == '305403':
        output = ''
        return output

def func2 (modem_changes,key):
    if key == '313115':
        output =''
    if key == '253036':
        output=''
    return output

def method(d1, f1, d2, f2):
    s1 = set(d1.keys())
    s2 = set(d2.keys())
    for k in s2.difference(s1):#set(['311957', '305403'])
        f1(d1, k) # k in d2 not in d1.
    for k in s1.difference(s2):#set(['313115', '253036'])
        f2(d2, k) # k in d1 not in d2.

while(True):
    method(data_changes, func1, modem_changes, func2)
    value = set(data_changes.keys()) - set(modem_changes.keys()) 
    print value
    if value == set([]):
        break;


EXPECTED OUTPUT:-

data_changes = {
    '253036': [''], 
    '313115': ['313113']
    '305403':[''] 
    '311957':['']
    }

modem_changes = {'305403': [], 
                 '311957': ['253036', '312591']
                 '253036':[]
                 '313115':[] 
                 }

Answer 1

In [8]: keys = set(data_changes.keys()) & set(modem_changes.keys())

In [9]: data_changes = {k:data_changes[k] for k in keys}

In [10]: modem_changes = {k:modem_changes[k] for k in keys}

In [11]: data_changes
Out[11]: {'253036': ['']}

In [12]: modem_changes
Out[12]: {'253036': ['311957', '312994', '312591']}

Answer 2

尝试以下操作以继续重复for循环：

while(True):
    for key in data_changes:
        if key not in modem_changes:
            func1()
    for key in modem_changes:
        if key not in data_changes:
            func2()
    if(True): #logic to check wether to run for loops again
        break;

Answer 3

由于您没有通过“密钥相同”来解释您的确切含义，我将假设func1和func2是您已编写的函数，并在我的答案中使用它们。我还假设函数采用他们正在处理的字典/密钥。如果不是这种情况，您可以随时在代码中进行更改。

def method(d1, f1, d2, f2):
    s1 = set(d1.keys())
    s2 = set(d2.keys())
    for k in s2.difference(s1):
        f1(d1, k) # k in d2 not in d1.
    for k in s1.difference(s2):
        f2(d2, k) # k in d1 not in d2.

上述方法采用以下形式的参数，例如：

method(data_changes, func1, modem_changes, func2)

如果您的意思是删除所有不在两个词典中的键，则以下代码段之一应该为您执行：

def keep_similar(d1, d2):
    '''creates new dicts from shared keys'''
    shared_keys = set(d1.keys()).intersection(set(d2.keys()))
    r1 = dict((k, d1[k]) for k in shared_keys)
    r2 = dict((k, d2[k]) for k in shared_keys)
    return (r1, r2)

def eliminate_diffs(d1, d2):
    '''removes non-shared keys from dicts'''
    s1 = set(d1.keys())
    s2 = set(d2.keys())
    for k in s2.difference(s1):
        del d2[k]
    for k in s1.difference(s2):
        del d1[k]
    return (d1, d2)

只需选择最适合您情况的那个，如果您的钥匙很少，那么使用第一个，如果你有几个不相似的钥匙，那么使用第二个。

控制for循环的流程

3 个答案: