map/reduce

reduce

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.

将传入的字符编程规范的首字母大写,其他字母小写

>>> L1 = ['adam', 'LISA', 'barT']

>>> list(map(lambda s:s[0].upper()+s[1:].lower() ,L1))
['Adam', 'Lisa', 'Bart']

请编写一个prod()函数，可以接受一个list并利用reduce()求积

>>> from functools import reduce
>>> reduce(lambda x,y : x * y,[3,5,7,9])
945

利用map和reduce编写一个str2float函数

把字符串’123.456’转换成浮点数123.456：

>>> 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]
...

>>> s = "1234.56700"
reduce(lambda x,y:x*10+y,map(char2num,s.split(".")[0]))+reduce(lambda x,y:x/10 + y,map(char2num,s.split(".")[1][::-1]))/10

def str2float(s):
    digit = {'1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8, '9': 9, '0': 0, '.': '.'}
    l = list(map(lambda c: digit[c], s))
    n = l.index('.')
    l.remove('.')
    return reduce(lambda x, y: 10*x+y, l)/(10**(len(l)-n))