ceshi

技术资源汇总(杭电支持版)/3.编程思维体系构建/3.6.4.4阶段四：高阶函数.md

@@ -0,0 +1,152 @@
+# 阶段四：高阶函数
+
+::: warning 🐱 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
+
+<font size=5><strong>代码首先是给人看的</strong></font>
+
+机器看的永远只是你的机器码。
+
+可参考教程 [Lambda](https://zhuanlan.zhihu.com/p/80960485)
+:::
+
+## 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
+
+```python
+>>> 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)
+______
+```
+
+```python
+>>> 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？
+
+```python
+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)。
+
+```python
+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 ***"
+```