函数式编程

Posted 2020-11-27 waynew

高阶函数

传入函数

既然变量可以指向函数，函数的参数能接收变量，那么一个函数就可以接收另一个函数作为参数，这种函数称为高阶函数

def add(x, y, f):
    return f(x) + f(y)
print(add(-3,5,abs))

map/reduce

map(func, *iterables) --> map object

Make an iterator that computes the function using arguments from each of the iterables. Stops when the shortest iterable is exhausted.
map()接收两个参数，一个是函数，一个是Iterbale，map将传入的函数依次作用于序列的每个元素，并把结果作为新的Iterator返回

def f(x):
    return x * x
r = map(f, [1,2,3,4,5])
list(r)

把list所有数字转为字符串

list(map(str, [1,2,3,4,5]))

reduce

reduce(function, sequence[, initial]) -> value
Apply a function of two arguments cumulatively to the items of a sequence,from left to right, so as to reduce the sequence to a single value.For example, reduce(lambda x, y: x+y, [1, 2, 3, 4, 5]) calculates ((((1+2)+3)+4)+5). If initial is present, it is placed before the items of the sequence in the calculation, and serves as a default when the sequence is empty.

reduce()接收两个参数，函数和序列，其中函数也要有两个参数，reduce把结果和序列剩下的部分依次进行计算，效果就是

reduce(f,[x1,x2,x3,x4,x5])=f(f(f(f(x1,x2),x3),x4),x5)

如求和

from functools import reduce
def add(x,y):
    return x+y
reduce(add,[1,2,3,4,5])

把序列变换成整数

from functools import reduce
def fn(x,y):
    return x * 10 + y
reduce(fn,[1,2,3,4,5])

把str转换成int

from functools import reduce
def fn(x,y):
    return x * 10 + y
def chr2num(s):
    digits = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}
    return digits[s]
reduce(fn, map(chr2num,'12345'))

整理成str2int的函数

from functools import reduce
DIGITS = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}
def str2int(s):
    def fn(x, y):
        return x * 10 + y
    def chr2num(s):
        return DIGITS[s]
    return reduce(fn, map(chr2num, s))

用lambda函数精简

from functools import reduce
DIGITS = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}
def chr2num(s):
    return DIGITS[s]
def str2int(s):
    return reduce(lambda x, y: x * 10 + y, map(chr2num, s))