for循環的控制流程

Question

我正在寫一個僞代碼，我希望重複下面兩個「for」循環，直到data_changes和modem_changes中的所有鍵都相同，即不應該存在任何存在data_changes的鍵，但是不在modem_changes中，反之亦然。我應該可以在這之後編寫Python實現;任何人可以提供投入？ 編輯：

我更感興趣的

1. 如何調用該存在於一個，而不是其他

2.repeating for循環，直到鍵鍵FUNC1和FUNC2在data_changes相同，modem_changes

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

請嘗試以下方法不斷重複的for循環：

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

既然你沒有解釋究竟你的意思是「密鑰是相同的」我會認爲func1和func2是你已經寫和使用的功能他們在我的答案中是這樣的。我還假設這些函數採用他們正在處理的字典/鍵。如果不是這種情況，您可以隨時在您的代碼中進行更改。

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. 

上面的方法將在下面的表格參數，以你的例子說明：

method(data_changes, func1, modem_changes, func2) 

---

如果你的意思是不在兩個字典的所有鍵都應該被刪除，然後一個下面的代碼片段應該爲你做：

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) 

只需選擇一個最適合你的你的情況，如果你有幾個類似的鍵，那麼使用第一個，如果你有幾個不相似的鍵，那麼使用第二個。