在 python 中编码斐波那契数列。有错误

Question

我正在为 AI 初学者做一个作业 Python。

创建一个 Class 继承自 int 的 NewInt。它应该有一个实例方法 is_fibonacci () 如果数字是斐波那契数则 returns 为真，否则为假。使用 NewInt 生成从 0 到 1000 的列表。然后使用您创建的 class 和实例方法创建一个仅保留斐波那契数字的列表理解。

[0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987] #任务的预期输出

我为任务编写了这段代码

import math

class NewInt(int):
   
    def is_Perfect_Square(x):
        s = int(math.sqrt(x))
        return s*s == x

    def is_fibonacci(n):

        return is_Perfect_Square(5*n*n + 4) or is_Perfect_Square(5*n*n - 4)

    fibonacci_List = [i for i in range(0,1000)if NewInt().is_fibonacci(i)]
    print(fibonacci_List)

它在几个小时前工作，但我收到这样的错误：

TypeError                                 Traceback (most recent call last)
<ipython-input-8-377e0f9b7814> in <module>
      1 import math
      2 
----> 3 class NewInt(int):
      4 
      5     def is_Perfect_Square(x):

<ipython-input-8-377e0f9b7814> in NewInt()
     11         return is_Perfect_Square(5*n*n + 4) or is_Perfect_Square(5*n*n - 4)
     12 
---> 13     fibonacci_List = [i for i in range(0,1000)if NewInt().is_fibonacci(i)]
     14     print(fibonacci_List)

<ipython-input-8-377e0f9b7814> in <listcomp>(.0)
     11         return is_Perfect_Square(5*n*n + 4) or is_Perfect_Square(5*n*n - 4)
     12 
---> 13     fibonacci_List = [i for i in range(0,1000)if NewInt().is_fibonacci(i)]
     14     print(fibonacci_List)

TypeError: is_fibonacci() takes 1 positional argument but 2 were given

有人可以帮我指出我的错误吗？我是 Python.

的新手

Answer 1

import math


class NewInt(int):

    # In my opinion you should make a constructor
    def __init__(self, number):
        # Declare the attributes
        self.number = number

    def is_Perfect_Square(self):
        # Use that attribute which u have initialised already.
        x = self.number
        s = int(math.sqrt(x))
        return s * s == x

    def is_fibonacci(self):
        n = self.number
        # Create an instance of your class each time u wanna access a method
        # e.g. NewInt(1234) creates an instant of the class NewInt. When you do this, u invoke the init method and that number attribute is intialised.
        return NewInt(5*n*n + 4).is_Perfect_Square() or NewInt(5 * n * n - 4).is_Perfect_Square()


fibonacci_List = [i for i in range(0, 1000) if NewInt(i).is_fibonacci()]

print(fibonacci_List)

输出：

[0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987]

如果您总是在函数的形参中取值，那么 class.

就没有意义了

Answer 2

这是python的 OOP 概念

NewInt().is_fibonacci(i)
当上面的方法被调用时python在内部做这个

NewInt().is_fibonacci(self, i)
self指的是本例中调用方法的对象NewInt()
但在 class 定义中，该方法仅针对一个参数定义，因此

TypeError: is_fibonacci() takes 1 positional argument but 2 were given

import math

class NewInt(int):

    def is_Perfect_Square(self, x):
        s = int(math.sqrt(x))
        return s*s == x

    def is_fibonacci(self, n):
        return self.is_Perfect_Square(5*n*n + 4) or self.is_Perfect_Square(5*n*n - 4)

fibonacci_List = [i for i in range(0,1000)if NewInt().is_fibonacci(i)]
print(fibonacci_List)

在 python 中编码斐波那契数列。有错误

Coding Fibonacci in python. Got errors

python

list

fibonacci