python：字符的熵条件答案

【问题标题】：python: entropy conditions for characterspython：字符的熵条件
【发布时间】：2020-11-06 19:52:08
【问题描述】：

我希望它根据波兰字母表和使用的符号计算熵：

如果只有大写或小写字母 32
如果它们对于 64 来说既小又大
如果有数字+10
如果有特殊字符+33

不幸的是，这些条件的第二个公式对我不起作用。当我输入“Apple”时，它会弹出 NameError: 名称 'entropia2' 未定义

import math

def entropy_poland(n):
    print("Znaki nie powtarzają się, więc liczymy ze wzoru Hartleya: ")
    if count_upper == True and count_lower == False and count_other == False and count_number == False:
        entropy2 = math.log2(32)
    elif count_upper == False and count_lower == True and count_other == False and count_number == False:
        entropy2 = math.log2(32)
    elif count_upper == False and count_lower == False and count_other == False and count_number == False:
        entropy2 = math.log2(64)
    elif count_upper == True and count_lower == False and count_other == False and count_number == True:
        entropy2 = math.log2(42)
    elif count_upper == False and count_lower == True and count_other == False and count_number == True:
        entropy2 = math.log2(42)
    elif count_upper == False and count_lower == False and count_other == False and count_number == True:
        entropy2 = math.log2(74)
    elif count_upper == False and count_lower == True and count_other == True and count_number == False:
        entropy2 = math.log2(65)
    elif count_upper == True and count_lower == False and count_other == True and count_number == False:
        entropy2 = math.log2(65)
    elif count_upper == False and count_lower == False and count_other == True and count_number == False:
        entropy2 = math.log2(97)
    elif count_upper == True and count_lower == False and count_other == True and count_number == True:
        entropy2 = math.log2(75)
    elif count_upper == False and count_lower == True and count_other == True and count_number == True:
        entropy2 = math.log2(75)
    elif count_upper == False and count_lower == False and count_other == True and count_number == True:
        entropy2 = math.log2(107)
    return entropy2


count_number = False
count_upper = False
count_lower = False
count_other = False

odp = "Reks"

for ascii in odp:
    if chr(32) <= ascii <= chr(47) or chr(58) <= ascii <= chr(64) or chr(91) <= ascii <= chr(96) or chr(
            123) <= ascii <= chr(126):
        count_other = True
    if chr(48) <= ascii <= chr(57):
        count_number = True
    if chr(65) <= ascii <= chr(90):
        count_upper = True
    if chr(97) <= ascii <= chr(122):
        count_lower = True


print(entropy_poland(odp))
````

【问题讨论】：

我无法复制错误。请发布minimal reproducible example。由于您发布的代码从未尝试使用变量entropia2，因此它无法抛出该错误。在else 子句中的大多数地方你使用entropy2 但在一个地方你使用entropia2 似乎很奇怪。
我试图最小化代码，但我忘了改变熵2

标签： python entropy

【解决方案1】：

如果插入行

print(count_upper,count_lower,count_other,count_number)

在entropy_poland 定义的开头你会看到它打印出来

True True False False

在抛出错误之前。该布尔组合不是您的子句涵盖的组合之一。有 4 个变量的 16 种布尔组合，但您只为其中的 12 个定义了 entropy2。你需要重新考虑你的逻辑，要么添加一个else 子句，要么如果这个布尔组合实际上是一个错误，则可能引发（然后处理）一个错误。

【讨论】：