3.6.4.4阶段四：高阶函数.md

# 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
代码首先是给人看的
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 ***"
```