2018-08-11

reduce用法：

reduce把一个函数作用在一个序列[x1, x2, x3, ...]上，这个函数必须接收两个参数，reduce把结果继续和序列的下一个元素做累积计算

效果是：reduce(f, [x1, x2, x3, x4]) = f(f(f(x1, x2), x3), x4)

---

求和序列采用reduce：

from functools import reduce

>>> def add(x,y):

...    return x + y

...

>>> reduce(add,[1,3,5,7,9])

25

---

如果要把序列[1, 3, 5, 7, 9]变换成整数13579

from functools import reduce

>>> def fn(x,y):

...    return x * 10 + y

...

>>> reduce(fn,[1,3,5,7,9])

13579

---

考虑到字符串str也是一个序列，配合map()，我们就可以写出把str转换为int的函数：

from functools import reduce

>>> def fn(x,y):

...    return x * 10 + y

...

>>> def char2num(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(char2num,'13579'))

13579

---

整理成一个str2int的函数就是：

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 char2num(s):

...        return DIGITS[s]

...    return reduce(fn,map(char2num,s))

---

lambda函数进行简化：

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 char2num(s):

    return DIGITS[s]

def str2int(s):

    return reduce(lambda x, y: x * 10 + y, map(char2num, s))

---

lambda函数的定义：

b = lambda a:a+1

>>> print(b(3))

4

>>>

>>>

>>> def fun(a):

...    return a+1

...

>>> print(fun(3))

4