Map reduce for python higher-order function usage

map()function

Receive two parameters, one is a function, and the other is Iterableto mapapply the passed function to each element of the sequence in turn, and return the result as a new one Iterator.

>>> 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]

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

The map() function can simplify programming

reduce() function

>>> 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
...
>>> 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

reduce() function

reduce (func, seq[, init()]) The
reduce() function is a repeated iteration function. Its execution process is: for each iteration, the previous iteration result and the next value in the sequence are used as the input of the function

init is optional. If specified, it is used as the first iteration. If not specified, the first element in seq is taken.

Use mapand reducewrite a str2floatfunction to '123.456'convert a string to a floating point number 123.456:

from functools import reduce

def str2float(s):
    loc = s.find('.')
    s1 = s.replace('.','')
    num = reduce(fn, map(char2num, s1))
    for n in range(len(s)-loc-1):
        num = num/10.0
    return num
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]
def fn(x, y):
    return x * 10 + y

Reference source: https://www.liaoxuefeng.com/wiki/1016959663602400/1017329367486080