Question

我有这样的代码选择：

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

现在，如果值随机数超过90，它将自动赋予值None并将letters[0]替换为None。

但我想要做的是，如果数字在90到100之间，我希望它显示原始值。所以最初存储在letters[0]中的内容。

真的不知道该怎么做。希望我能够清楚地解释清楚。

Answer 1

您可以改为引发异常，然后通过捕获该异常来阻止letters[0]被替换：

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')

你也可以将当前值传递给obfuscate_function()函数，如果随机值大于90则返回：

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

你可以尝试避免重复自己;例如在obfuscate_letters中使用循环：

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

或传入当前值：

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

    return letters

您的obsfucate_function()也可以简化;通过仅使用random.choice()进行角色选择，单独使用10％的几率进行更改，或使用bisect选择字符或基于累计权重的默认值。

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))]

random.choice()选项并不完全复制原始函数，因为选择!的机会是16％，而其余字符有15％的机会被选中。二分法将选择具有完全相同概率的字符。

Answer 2

它没有返回，因为你的值没有返回值。换句话说，你没有为90到100之间的范围定义最终的elif条件

Answer 3

最简单的方法是使用letters[i] = obsfucate_function() or letters[i]。当obsfucate_function返回None（randomnumber> 90）时，它将自动使用字母[i]的值。

顺便说一句，你可能想要使用混淆而不是黑曜石。

返回原始值，python

3 个答案: