import math # 内置函数 abs(-200) max(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12) # 数据类型转换 int('123') float('123.123') str(1.23) bool(1) bool('') a = abs x = a(123.45) print(x) n1 = 255 n2 = 1000 print(hex(n1)) def my_abs(x): if not isinstance(x, (int, float)): raise TypeError('x must be an integer or float') if x >= 0: return x else: return -x def nop(age): if age >= 18: pass def move(x, y, step, angle=0): nx = x + step * math.cos(angle) ny = y + step * math.sin(angle) return nx, ny r = move(100, 100, 60, math.pi / 6) def quadratic(a, b, c): if not isinstance(a * b * c, (float, int)): raise TypeError("quadratic must be a float or int") q = b ** 2 - 4 * a * c if q < 0: print('no solution') elif q == 0: print('此方程的解是唯一的') x = (-b) / (2 * a) print(x) else: x1 = ((-b + math.sqrt(q)) / (2 * a)) x2 = ((-b - math.sqrt(q)) / (2 * a)) return x1, x2 def enroll(name, gender, age=6, city='Beijing'): pass def add_end(L=None): if L is None: L = [] L.append('END') return L # 可变参数 # 可变参数在函数调用时自动组装为一个tuple def calc(*numbers): sum = 0 for n in numbers: sum += n ** 2 return sum nums = [1, 2, 3] calc(*nums) # 关键字参数 # 允许你传入0个或任意个含参数名的参数，这些关键字参数在函数内部自动组装为一个dict def person(name, age, **kw): print('name:', name, 'age:', age, 'other:', kw) person('Bob', 35, city='Beijing') # name: Bob age: 35 other: {'city': 'Beijing'} extra = {'city': 'Beijing', 'job': 'Engineer'} person('Bob', 35, **extra) # 命名关键字参数 # 函数的调用者可以传入任意不受限制的关键字参数, *后面的参数被视为命名关键字参数 def person_2(name, age, *, city='Beijing', job): print(name, age, city, job) # 参数组合 def f1(a, b, c=0, *args, **kw): print('a =', a, 'b =', b, 'c =', c, 'args =', args, 'kw =', kw) # 使用命名关键字参数时，要特别注意，如果没有可变参数，就必须加一个*作为特殊分隔符 def f2(a, b, c=0, d=1, *arg, e, **kw): print('a =', a, 'b =', b, 'c =', c, 'e =', e, 'kw =', kw, 'arg = ', arg) f2(1, 2, 3, 4, 56, 7, 8, 9, 0, e=4, f=9) # 练习 def mul(*numbers): if not len(numbers) == 0: sum = 1 for n in numbers: if not isinstance(n, (int, float)): continue # 跳出本次循环，执行下次循环 sum *= n return sum else: raise TypeError('numbers must be a list') # 测试 print('mul(5) =', mul(5)) print('mul(5, 6) =', mul(5, 6)) print('mul(5, 6, 7) =', mul(5, 6, 7)) print('mul(5, 6, 7, 9) =', mul(5, 6, 7, 9)) if mul(5) != 5: print('测试失败!') elif mul(5, 6) != 30: print('测试失败!') elif mul(5, 6, 7) != 210: print('测试失败!') elif mul(5, 6, 7, 9, 'test') != 1890: print('测试失败!') else: try: mul() print('测试失败!') except TypeError: print('测试成功!') # 递归 def fact_1(n): if n == 1: return 1 else: return n * fact(n - 1) # 尾递归优化 def fact(n): return fact_iter(n, 1) def fact_iter(num, product): if num == 1: return product return fact_iter(num - 1, num * product) print('fact_iter', fact(10)) # 汉诺塔问题实现 # a存放起始柱，b存放辅助柱、c存放目标柱 def move(n, a, b, c): if n == 1: print(a, '--->', c) else: move(n - 1, a, c, b) # 借助c把第 num 个以外的圆盘从a移动到b move(1, a, b, c) # 把第num个从a移动到c move(n - 1, b, a, c) # 借助a把第 num 个以外的圆盘从b移动到c move(3, 'A', 'B', 'C') # 返回函数 def lazy_sum(*args): def sum(): ax = 0 for n in args: ax = ax + n return ax return sum f = lazy_sum(1, 2, 3, 4, 5) print(f()) # 闭包 def count(): fs = [] def f(j): def g(): return j * j return g for i in range(1, 4): fs.append(f(i)) return fs f1, f2, f3 = count() print(f1()) print(f2()) print(f3()) # 一个函数可以返回一个计算结果，也可以返回一个函数。 # 返回一个函数时，牢记该函数并未执行，返回函数中不要引用任何可能会变化的变量。 def createCounter(): x = 0 def counter(): nonlocal x # 如果对外层变量赋值，由于Python解释器会把x当作函数fn()的局部变量，它会报错： x = x + 1 return x return counter counterA = createCounter() print(counterA(), counterA(), counterA(), counterA(), counterA()) # lambda函名函数 def is_odd(n): return n % 2 == 1 L = list(filter(lambda n: n % 2 == 1, range(1, 20))) print(L)