Python - 如何检查用户输入是否为复杂类型输入答案

【问题标题】：Python - How to check if user input is a Complex type inputPython - 如何检查用户输入是否为复杂类型输入
【发布时间】：2021-11-24 03:53:28
【问题描述】：

我想根据输入的类型打印一条消息，但是每次我输入一个复数时，例如（5j）它都会被检测为字符串输入。请问我该如何解决？谢谢。

while True:
    a = input("a ? ")
    if (isinstance(a, complex)):
        print("Valid number, please not Complex!")  
    try:
        a = float(a)
    except ValueError:
        print ('please input a int or float')
        if (type(a)==str):
            print("Valid number, please not String!")
        continue
    if 0.5 <= a <= 100:
        break
    elif 0 <= a < 0.5:
        print ('bigger number, please: 0.5-100')
    elif a < 0:
        print ('positive number, please')
    elif a > 100:
        print ('smaller number, please: 0.5-100')

执行示例：

a ? 5j
please input a int or float
Valid number, please not String!

我试过这样做：

while True:
    try:
        a = input("a ? ")
        if ('j' in a):
            print("Valid number, please not Complex!")
        a = float(a)
    except ValueError:
        print ('please input a int or float')
        if (type(a)==str and 'j' not in a):
            print("Valid number, please not String!")
        continue
    if 0.5 <= a <= 100:
        break
    elif 0 <= a < 0.5:
        print ('bigger number, please: 0.5-100')
    elif a < 0:
        print ('positive number, please')
    elif a > 100:
        print ('smaller number, please: 0.5-100')

但这不是“完美”

【问题讨论】：

标签： python string input user-input complex-numbers

【解决方案1】：

你可以将第一个代码块添加到 try 块中

像这样-

while True:
    try:
        a = input("a ? ")
        if (isinstance(a, complex)):
            print("Valid number, please not Complex!")  
        a = float(a)
    except ValueError:
        print ('please input a int or float')
        if (type(a)==str):
            print("Valid number, please not String!")
        continue
    if 0.5 <= a <= 100:
        break
    elif 0 <= a < 0.5:
        print ('bigger number, please: 0.5-100')
    elif a < 0:
        print ('positive number, please')
    elif a > 100:
        print ('smaller number, please: 0.5-100')

这是你的意思吗？

【讨论】：

不，遗憾的是在输入复数时它仍然会打印“字符串错误”

【解决方案2】：

您可以改用嵌套的try-except 和内置函数complex()。
所以，你的代码需要是这样的

while True:
    a = input("a? ")
    try:
        a = float(a)
        if 0.5 <= a <= 100:
            break
        elif 0 <= a < 0.5:
            print ('bigger number, please: 0.5-100')
        elif a < 0:
            print ('positive number, please')
        elif a > 100:
            print ('smaller number, please: 0.5-100')
    except ValueError:
        try:
            a = complex(a)
            print("Valid number, please not Complex!")
        except ValueError:
            print ("Valid number, please not String!")
            continue

【讨论】：