返回原來的價值，有蟒蛇

Question

我這個選擇的代碼：

def obsfucate_letters(letters): 

    letters[0] = obsfucate_function() 
    letters[1] = obsfucate_function() 
    letters[2] = obsfucate_function() 
    letters[3] = obsfucate_function() 

    return letters 

def obsfucate_function(): 

    import random as r 
    randomnumber = r.random() * 100 
    if randomnumber <= 16: 
     letters1 = "!" 
     return letters1 
    elif randomnumber > 16 and randomnumber <= 30: 
     letters1 = "%" 
     return letters1 
    elif randomnumber > 30 and randomnumber <= 45: 
     letters1 = "&" 
     return letters1 
    elif randomnumber > 45 and randomnumber <= 60: 
     letters1 = "*" 
     return letters1 
    elif randomnumber > 60 and randomnumber <= 75: 
     letters1 = "#" 
     return letters1 
    elif randomnumber > 75 and randomnumber <= 90: 
     letters1 = "@" 
     return letters1 

現在，此刻如果值爲隨機數超過90它只會自動給出值None與None取代例如letters[0] 。

但我想要它做的是，如果數量在90和100之間，我希望它顯示原始值。所以最初存儲在letters[0]。

真的不知道該怎麼做。希望我已經清楚解釋清楚了。

Answer 1

你可以拋出一個異常，而不是，然後通過捕獲該異常防止letters[0]進行更換：

def obsfucate_letters(letters): 
    try:  
     letters[0] = obsfucate_function() 
    except ValueError: 
     # ignore the exception, letters[0] will be unchanged 
     pass 

    try: 
     letters[1] = obsfucate_function() 
    except ValueError: 
     pass 

    try: 
     letters[2] = obsfucate_function() 
    except ValueError: 
     pass 

    try: 
     letters[3] = obsfucate_function() 
    except ValueError: 
     pass 

    return letters 

def obsfucate_function(): 
    import random as r 
    randomnumber = r.random() * 100 
    if randomnumber <= 16: 
     letters1 = "!" 
     return letters1 
    elif randomnumber > 16 and randomnumber <= 30: 
     letters1 = "%" 
     return letters1 
    elif randomnumber > 30 and randomnumber <= 45: 
     letters1 = "&" 
     return letters1 
    elif randomnumber > 45 and randomnumber <= 60: 
     letters1 = "*" 
     return letters1 
    elif randomnumber > 60 and randomnumber <= 75: 
     letters1 = "#" 
     return letters1 
    elif randomnumber > 75 and randomnumber <= 90: 
     letters1 = "@" 
     return letters1 
    else: 
     raise ValueError('random number out of range') 

你也電流值傳遞給obfuscate_function()功能，如果隨機值返回大於90：

def obsfucate_letters(letters): 
    letters[0] = obsfucate_function(letters[0]) 
    letters[1] = obsfucate_function(letters[1]) 
    letters[2] = obsfucate_function(letters[2]) 
    letters[3] = obsfucate_function(letters[3]) 

    return letters 

def obsfucate_function(default): 
    import random as r 
    randomnumber = r.random() * 100 
    if randomnumber <= 16: 
     letters1 = "!" 
     return letters1 
    elif randomnumber > 16 and randomnumber <= 30: 
     letters1 = "%" 
     return letters1 
    elif randomnumber > 30 and randomnumber <= 45: 
     letters1 = "&" 
     return letters1 
    elif randomnumber > 45 and randomnumber <= 60: 
     letters1 = "*" 
     return letters1 
    elif randomnumber > 60 and randomnumber <= 75: 
     letters1 = "#" 
     return letters1 
    elif randomnumber > 75 and randomnumber <= 90: 
     letters1 = "@" 
     return letters1 
    else: 
     return default 

你可以儘量避免重複自己;使用循環在obfuscate_letters例如：

def obsfucate_letters(letters): 
    for i in range(len(letters)): 
     try: 
      letters[i] = obfuscate_function() 
     except ValueError: 
      # don't replace a letter when the random value was out of range 
      pass 

    return letters 

或傳遞一個的當前值：

def obsfucate_letters(letters): 
    for i in range(len(letters)): 
     letters[i] = obfuscate_function(letters[i]) 

    return letters 

obsfucate_function()你可以太簡化;或者通過使用random.choice()來進行角色選擇，並且單獨的10％機率不作任何改變，或者通過使用bisect來選擇角色或基於累積權重的默認值。

使用random.choice()：

import random 

def obsfucate_function(default): 
    if random.random() < 0.1: 
     # 10% chance the default is picked 
     return default 
    return random.choice('!%&*#@') 

使用二分法：

import random 
import bisect 

def obsfucate_function(default): 
    cumulative_weights = [16, 30, 45, 60, 75, 90, 100] 
    values = ['!', '%', '&', '*', '#', '@', default] 
    return values[bisect.bisect(cumulative_weights, random.randrange(100))] 

的random.choice()選項並不完全複製你的原始功能，選擇一個!是16％的機率，而其餘角色有15％的機會被挑選。平分法將選擇具有完全相同概率的字符。

Answer 2

它沒有返回值，因爲你的值沒有返回值。換句話說，你沒有定義90到100之間範圍的最終elif條件。

Answer 3

最簡單的方法是使用letters[i] = obsfucate_function() or letters[i]。當obsfucate_function返回無（randomnumber> 90）時，它會自動使用字母[i]的值。

順便說一句，你可能想使用混淆而不是晦澀難懂。