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map/reduce

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Python内建了map()reduce()函数。

>>> def f(x):
...     return x * x
...
>>> r = map(f, [1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> list(r)
[1, 4, 9, 16, 25, 36, 49, 64, 81]

map()传入的第一个参数是f，即函数对象本身。由于结果r是一个IteratorIterator是惰性序列，因此通过list()函数让它把整个序列都计算出来并返回一个list。

L = []
for n in [1, 2, 3, 4, 5, 6, 7, 8, 9]:
L.append(f(n))
print(L)

>>> list(map(str, [1, 2, 3, 4, 5, 6, 7, 8, 9]))
['1', '2', '3', '4', '5', '6', '7', '8', '9']

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

>>> from functools import reduce
>>> def add(x, y):
...     return x + y
...
>>> reduce(add, [1, 3, 5, 7, 9])
25

>>> from functools import reduce
>>> def fn(x, y):
...     return x * 10 + y
...
>>> reduce(fn, [1, 3, 5, 7, 9])
13579

>>> from functools import reduce
>>> def fn(x, y):
...     return x * 10 + y
...
>>> def char2num(s):
...     return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[s]
...
>>> reduce(fn, map(char2num, '13579'))
13579

from functools import reduce

def str2int(s):
def fn(x, y):
return x * 10 + y
def char2num(s):
return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[s]
return reduce(fn, map(char2num, s))

from functools import reduce

def char2num(s):
return {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9}[s]

def str2int(s):
return reduce(lambda x, y: x * 10 + y, map(char2num, s))

lambda函数的用法在后面介绍。

练习

# -*- coding: utf-8 -*-
----
def normalize(name):
pass
----
# 测试:
L1 = ['adam', 'LISA', 'barT']
L2 = list(map(normalize, L1))
print(L2)

Python提供的sum()函数可以接受一个list并求和，请编写一个prod()函数，可以接受一个list并利用reduce()求积：

# -*- coding: utf-8 -*-

from functools import reduce

def prod(L):
----
pass
----
print('3 * 5 * 7 * 9 =', prod([3, 5, 7, 9]))

# -*- coding: utf-8 -*-

from functools import reduce

def str2float(s):
----
pass
----
print('str2float(\'123.456\') =', str2float('123.456'))

do_map.py

do_reduce.py

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