函数式编程

2017-11-06  本文已影响0人  抽象语法树

高阶函数

变量可以指向函数

>>> abs(-1)
1
>>> f = abs
>>> f(-1)
1
>>> 

函数名也是变量

>>> abs = 1
>>> abs (-1)
Traceback (most recent call last):
  File "<pyshell#4>", line 1, in <module>
    abs (-1)
TypeError: 'int' object is not callable
>>> abs = f
>>> abs(-1)
1
>>> 

传入函数

def add(a, b, abs):
    return a + abs(b)

print(add(2, -3, abs))
==============================
5

map/reduce

map函数

def f(a):
    return a*a

r=[1, 2, 3, 4, 5, 6, 7, 8, 9]
r = map(f, r)
print(list(r))
============================
[1, 4, 9, 16, 25, 36, 49, 64, 81]

reduce函数

reduce(f, [x1, x2, x3, x4]) = f(f(f(x1, x2), x3), x4)
import functools
def f(a, b):
    return a*10 + b

print(functools.reduce(f,[1,2,3,4]))
=============================
1234

filter

def is_odd(x):
    return x % 2 == 1


L1 = range(0, 5)
L2 = filter(is_odd, L1)
print(list(L2))
=========================
[1, 3]

sorted

L1 = [1, -2, 9, -3, 4]
L1 = sorted(L1, key=abs)
print(L1)
============================
[1, -2, -3, 4, 9]
L2 = ["Jack", "Mike", "jason", "luke"]
L2 = sorted(L2, key=str.lower, reverse=True)
print(L2)
==============================
['Mike', 'luke', 'jason', 'Jack']

返回函数

def lazy_sum(*args):
    def cal_sum():
        ax = 0
        for n in args:
            ax = n + ax
        return ax
    return cal_sum


f1 = lazy_sum(1, 2, 3, 4, 5, 6, 7, 8)
f2 = lazy_sum(1, 2, 3, 4, 5, 6, 7, 8)
print(f1())
print(f1 == f2)
print(f1() == f2())

函数lazy_sum返回的是一个内部定义的cal_sum函数,所以只有调用f()才会显示结果。
当我们调用lazy_sum()时,每次调用都会返回一个新的函数,即使是传入相同的参数
即f1()和f2()的调用结果互不影响。

闭包

def count():
    fs = []
    for i in range(1, 4):
        def f():
            return i*i
        fs.append(f)
    return fs


f1, f2, f3 = count()
print(f1())
print(f2())
print(f3())
=================================
9
9
9

可以看到上面的输出全为9,而不是1,4,9,是因为在调用f1(),f2(),f3()时,i已经变成了3

def count():
    def f(j):
        def g():
            return j*j
        return g
    fs = []
    for i in range(1,4):
        fs.append(f(i))
    return fs
=========================
1
4
9

匿名函数

L = list(map(lambda x: x*x, [1, 2, 3, 4, 5]))
print(L)
===================================
[1, 4, 9, 16, 25]
f = lambda x: x*x
print(f(5))
======================
25
def get_sum(x, y):
    return lambda: x+y

f = get_sum(1,2)
print(f())

装饰器

def log(func):
    def wrapper(*args, **kw):
        print('call %s()' % func.__name__)
        return func(*args, **kw)
    return wrapper
@log()
def now():
    print(time.strftime("%y %m %d", time.localtime()))
call now()
17 11 07
now = log(now)
>>> now.__name__
'wrapper'
def log(*text):
    def decorator(func):
        @functools.wraps(func)
        def wrapper(*args, **kw):
            print("begin call")
            if text:
                print('%s %s' % (text[0], func.__name__))
            else:
                print('call %s()' % func.__name__)
            func(*args, **kw)
            print("end call")
            return
        return wrapper
    return decorator

该函数经过了一些修改

偏函数

import functools
int2 = functools.partial(int, base=2)
print(int2('100000'))
print(int2('100000', base = 8))
int2 = functools.partial(int, base=2)

相当于:

kw = { 'base': 2 }
int('10010', **kw)
``
当传入:

max2 = functools.partial(max, 10)

实际上会把10作为*args的一部分自动加到左边
相当于:

args = (10, 5, 6, 7)
max(*args)






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