如何让我的 Python 代码更紧凑？答案

【问题标题】：How can i make my Python Code more compact?如何让我的 Python 代码更紧凑？
【发布时间】：2020-06-24 20:21:38
【问题描述】：

如何使我的 Python 代码更紧凑？
任何帮助都会非常有用

print("Welcome to my program")
# This program tells if a number form 1 - 10 is even or odd
try:
    del_number = int(input("Input a number 1 - 10: "))
    if del_number == 2 or del_number == 4 or del_number == 6 or del_number == 8 or del_number == 10:
        print("The number you have entered is an even number")
    elif del_number > 10:
        print("Not a valid number")
    else:
        print("The number you have entered is a odd number")
except:
    print("You have entered a letter/letters not a number 1 - 10")

【问题讨论】：

顺便说一句，将您的异常处理限制为except ValueError。您的异常应该是狭窄的，以避免意外抑制意外错误和隐藏错误。

标签： python python-3.x

【解决方案1】：

你可以缩短偶数校验

print("Welcome to my program")
# This program tells if a number form 1 - 10 is even or odd
try:
    del_number = int(input("Input a number 1 - 10: "))
    if del_number in (2 , 4, 6, 8, 10):
        print("The number you have entered is an even number")
    elif del_number > 10:
        print("Not a valid number")
    else:
        print("The number you have entered is a odd number")
except ValueError:
    print("You have entered a letter/letters not a number 1 - 10")

或者使用低位来决定偶数/奇数（也许从列表中选择名称）

print("Welcome to my program")
# This program tells if a number form 1 - 10 is even or odd
try:
    del_number = int(input("Input a number 1 - 10: "))
    if not 1 < del_number <= 10:
        print("Not a valid number")
    else:
        category = ["even", "odd"][del_number & 1]
        print(f"The number you have entered is a {category} number")
except ValueError:
    print("You have entered a letter/letters not a number 1 - 10")

考虑到这一点，您可以将异常处理限制在整数转换本身，以减少您在以后的代码中意外掩盖错误的机会。假设后面的测试中的一个错误也引发了ValueError。直到现场的愤怒报告你才会知道。

print("Welcome to my program")
# This program tells if a number form 1 - 10 is even or odd
try:
    del_number = int(input("Input a number 1 - 10: "))
except ValueError:
    print("You have entered a letter/letters not a number 1 - 10")
else:
    if not 1 < del_number <= 10:
        print("Not a valid number")
    else:
        category = ["even", "odd"][del_number & 1]
        print(f"The number you have entered is a {category} number")

【讨论】：

如果您使用条件表达式而不是可迭代索引，我认为它会是 pythonic："odd" if del_number & 1 else "even"
这是另一种方式，但不是更多pythonic，恕我直言。我的方法适用于超过 1 位的类别，比如包含 8 个条目和 del_numbers & 3 的列表。索引到列表是正常的。我明白你在说什么，有些人更喜欢你的方法。
"说一个有 8 个条目的列表和del_numbers & 3" 哇，这真的让我大吃一惊。对于这种情况来说会很棒。但是，我仍然觉得它令人困惑，因为如果此表达式的计算结果为 True 等，我总是不得不停下来思考选择哪一个。
@Asocia - 是的，不同的观点。我在位域的低级 C 中度过了太多年。我梦想着比特领域（或者也许是地狱般的噩梦）。我只是将其视为类别选择，但您对真/假的看法同样有效。
哈哈哈，我可以想象，因为你也选择了del_numbers & 1 而不是del_numbers % 2 == 1 :)

【解决方案2】：

您可以使用模运算符 % 而不是那么多串联的 or。有关操作员的信息，请阅读here。

这是要替换的代码行：

if del_number == 2 or del_number == 4 or del_number == 6 or del_number == 8 or del_number == 10:

这将是替代品：

elif del_number % 2 == 0:

if 更改为elif 的原因是我们首先要确保号码是<= 10。因此，我们的第一个条件if 将测试数字是否为> 10。如果它通过了该标准，则带有模运算符的 elif 部分。总而言之：

try:
    del_number = int(input("Input a number 1 - 10: "))
    if del_number > 10 or del_number < 0:
        print("Not a valid number")
    elif del_number % 2 == 0:
        print("The number you have entered is an even number")
    else:
        print("The number you have entered is a odd number")
except:
    print("You have entered a letter/letters not a number 1 - 10")

【讨论】：

不完全是。第一个语句只接受 1-10 范围内的偶数，而第二个语句接受 any 偶数。
并且您需要将> 10 比较移到其上方以首先捕获溢出情况。
我可以轻松解决将 elif 放入并将 >10 切换为 if 的第一个条件的问题。感谢您的建议，编辑答案！

【解决方案3】：

你可以用更紧凑的方式来制作它

if del_number in range(2, 11, 2):  # [2, 4, 6, 8, 10]
    print("The number you have entered is an even number")

或

if del_number % 2 == 0:
    print("The number you have entered is an even number")

如果你想检查del_number是偶数还是奇数

【讨论】：

【解决方案4】：

让我们这样试试吧：

print("Welcome to my program")
# This program tells if a number form 1 - 10 is even or odd
try:
    del_number = int(input("Input a number 1 - 10: "))
except:
    print("You have entered a letter/letters not a number 1 - 10")

if del_number % 2 == 0 and del_number <= 10:
    print("The number you have entered is an even number")
elif del_number > 10:
    print("Not a valid number")
else:
    print("The number you have entered is a odd number")

【讨论】：

【解决方案5】：

我会这样做

print("Welcome to my program")
# This program tells if a number form 1 - 10 is even or odd
try:
    del_number = int(input("Input a number 1 - 10: \n"))
    if del_number % 2 == 0 and 1 < del_number < 10 :
        print("The number you have entered is an even number")
    elif del_number > 10 or del_number < 1:
        print("Not a valid number")
    else:
        print("The number you have entered is a odd number")
except:
    print("You have entered a letter/letters not a number 1 - 10")

【讨论】：

【解决方案6】：

print("Welcome to my program")
# This program tells if a number form 1 - 10 is even or odd
del_number = int(input("Input a number 1 - 10: "))
try:
    {
        0: lambda: print("The number you have entered is an even number"),
        1: lambda: print("The number you have entered is an odd number")
    }[del_number % 2 if 0 < del_number <= 10 else None]()
except:
    print("You have entered a letter/letters not a number 1 - 10")

【讨论】：