5

我正在尝试编写一个函数，该函数映射列表的元素，以使用 python 以函数样式获取元素和列表中先前元素的总和，例如： 

func([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) = [0, 1, 3, 6, 10, 15, 21, 28, 36, 45]

我尝试过使用递归，但得到RuntimeError: maximum recursion depth exceeded了一个很长的列表。：

def recursion_way(inlist, accu, summ):
    if len(inlist) == 0:
         return accu
    else:
        return recursion_way(inlist[1:], accu + [summ + inlist[0]], summ + inlist[0])
4

6 回答 6

7

理解算不算？

>>> [sum(l[:i]) for i, _ in enumerate(l)]
[0, 0, 1, 3, 6, 10, 15, 21, 28, 36]

或者也许使用reduce：

reduce(
    lambda (sums, last), x: (sums+[x+last], x+last),
    l, ([], 0)
)[0]

或者另一种方式：

reduce(lambda sums,x: sums+[x+sums[-1]], l[1:], l[:1])
于 2012-12-05T17:04:41.903 回答
1

这是以函数式编程风格完成的累积和：

def func(l):
   if len(l) < 2: return l
   sub = func(l[:-1])
   return sub + [sub[-1] + l[-1]]

print func([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
于 2012-12-05T17:05:56.153 回答
1

怎么减？慢，但有趣，imo。

def func(p):
    for i in xrange(1, len(p)+1):
        yield reduce(lambda x, y: x + y, p[0:i])

>>> list(func(p))
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45]
于 2012-12-05T17:08:29.280 回答
1

这是我得到的，使用递归：

def f(L, n=0):
    # If the list is empty, that means we went through all the elements there
    if len(L)>0:
        # n is the last element in the sum list. Add to it the first remaining element
        n = n+L[0]
        # Return a list containing the newest item and those of the remaining elements
        return [n] + f(L[1:], n)
    else:
        # It it is empty, there are no more sums to calculate
        return []
于 2012-12-05T17:16:01.000 回答
0

你可以使用numpy吗？

import numpy
numpy.cumsum([0,1,2,3,4,5,6,7,8,9])
array([ 0,  1,  3,  6, 10, 15, 21, 28, 36, 45])
于 2012-12-05T17:05:41.240 回答
-2 
l = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
reduce(lambda x, y: x+y, l)
于 2012-12-05T17:06:50.333 回答