原创转载请注明出处:http://agilestyle.iteye.com/blog/2330300
map()
map()函数接收两个参数,一个是函数,一个是Iterable,map将传入的函数依次作用到序列的每个元素,并把结果作为新的Iterator返回。
使用map实现一个f(x) = x * x的功能
def f(x): return x * x m = map(f, list(range(1, 10))) # [1, 4, 9, 16, 25, 36, 49, 64, 81] print(list(m))
另外可以使用lambda函数简化
# [1, 4, 9, 16, 25, 36, 49, 64, 81] print(list(map(lambda x: x * x, list(range(1, 10)))))
Console Output
Note:
可以看出
lambda x: x * x
实际上就是:
def f(x): return x * x
把list中的所有数字转为字符串
# ['1', '2', '3', '4', '5', '6', '7', '8', '9'] print(list(map(str, list(range(1, 10)))))
Console Output
reduce()
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,其效果就是:
reduce(f, [x1, x2, x3, x4]) = f(f(f(x1, x2), x3), x4)
使用reduce对一个序列求和
from functools import reduce def add(x, y): return x + y # 55 print(reduce(add, list(range(1, 11)))) # 55 print(reduce(lambda x, y: x + y, list(range(1, 11))))
Console Output
把序列[1, 2, 3, 4, 5, 6, 7, 8, 9]变换成整数123456789
from functools import reduce def fn(x, y): return x * 10 + y # 123456789 print(reduce(fn, list(range(1, 10)))) # 123456789 print(reduce(lambda x, y: x * 10 + y, list(range(1, 10))))
Console Output
map()和reduce()整合,把str转换位int
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] # '13579' => 13579 print(reduce(fn, map(char2num, '13579'))) # '13579' => 13579 print(reduce(lambda x, y: x * 10 + y, map(char2num, '13579'))) 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)) print(str2int('13579'))
Console Output
参考资料: