阶段四：高阶函数

🐱 阅读以及完成本部分内容可以帮助你有效减少代码冗余。

让你完成更为优雅的代码

各位要记住的是

机器看的永远只是你的机器码。

可参考教程 Lambda

Lambda 介绍

Lambda 表达式是通过指定两件事来评估函数的表达式：参数和返回表达式。

请尝试阅读以下英文表格，对比函数与 lambda 表达式的不同

Lambda 实验

以下代码 python 会显示什么？通过对这些代码的实验，加深你对代码的学习

提示：当你对解释器输入 x 或 x=none 时什么都没有

>>> lambda x: x  # A lambda expression with one parameter x
______

>>> a = lambda x: x  # Assigning the lambda function to the name a
>>> a(5)
______

>>> (lambda: 3)()  # Using a lambda expression as an operator in a call exp.
______

>>> b = lambda x: lambda: x  # Lambdas can return other lambdas!
>>> c = b(88)
>>> c
______

>>> c()
______

>>> d = lambda f: f(4)  # They can have functions as arguments as well.
>>> def square(x):
...     return x * x
>>> d(square)
______

>>> higher_order_lambda = lambda f: lambda x: f(x)
>>> g = lambda x: x * x
>>> higher_order_lambda(2)(g)  # Which argument belongs to which function call?
______

>>> higher_order_lambda(g)(2)
______

>>> call_thrice = lambda f: lambda x: f(f(f(x)))
>>> call_thrice(lambda y: y + 1)(0)
______

>>> print_lambda = lambda z: print(z)  # When is the return expression of a lambda expression executed?
>>> print_lambda
______

>>> one_thousand = print_lambda(1000)
______

>>> one_thousand
______

任务

P9: 我们发现以下两个函数看起来实现的非常相似，是否可以进行改进，将其整合？

def count_factors(n):
    """Return the number of positive factors that n has.
    >>> count_factors(6)
    4   # 1, 2, 3, 6
    >>> count_factors(4)
    3   # 1, 2, 4
    """
    i, count = 1, 0
    while i <= n:
        if n % i == 0:
            count += 1
        i += 1
    return count

def count_primes(n):
    """Return the number of prime numbers up to and including n.
    >>> count_primes(6)
    3   # 2, 3, 5
    >>> count_primes(13)
    6   # 2, 3, 5, 7, 11, 13
    """
    i, count = 1, 0
    while i <= n:
        if is_prime(i):
            count += 1
        i += 1
    return count

def is_prime(n):
    return count_factors(n) == 2 # only factors are 1 and n

需求：

你需要通过自己写一个函数： count_cond ，来接受一个含有两个参数的函数 condition(n, i) (使用 lambda 表达式)，

且 condition 函数应该满足第一个参数为 N，而第二个参数将会在 condition 函数中遍历 1 to N。

count_cond 将返回一个单参数函数 (ps：一个匿名函数)，此单参数函数将会在被调用时返回 1 to N 中所有满足 condition 的数字的个数 (如：1 到 n 中素数的个数)。

def count_cond(condition):
    """Returns a function with one parameter N that counts all the numbers from
    1 to N that satisfy the two-argument predicate function Condition, where
    the first argument for Condition is N and the second argument is the
    number from 1 to N.

    >>> count_factors = count_cond(lambda n, i: n % i == 0)
    >>> count_factors(2)   # 1, 2
    2
    >>> count_factors(4)   # 1, 2, 4
    3
    >>> count_factors(12)  # 1, 2, 3, 4, 6, 12
    6

    >>> is_prime = lambda n, i: count_factors(i) == 2
    >>> count_primes = count_cond(is_prime)
    >>> count_primes(2)    # 2
    1
    >>> count_primes(3)    # 2, 3
    2
    >>> count_primes(4)    # 2, 3
    2
    >>> count_primes(5)    # 2, 3, 5
    3
    >>> count_primes(20)   # 2, 3, 5, 7, 11, 13, 17, 19
    8
    """
    "*** YOUR CODE HERE ***"