如何在Python中对列表元素进行数学运算？

Question

假设我有一个数字列表 [3, 51, 34]。我想要把每个元素都加上它之前所有元素的和，然后返回一个新的列表，里面是这些新的值。比如说，结果应该是 [3, 54, 88]。那么，怎么才能对任意大小的输入列表做到这一点呢？代码的最后一行应该能处理已知大小的列表。

indices1 = range(len(list1))

indices1.sort(key=lambda x: list2[x])
list1 = map(lambda i: list1[i], indices1)
labelled = zip(list1, ascii_uppercase)
sorted_data = sorted(labelled, key=itemgetter(0)) 

labels = [pair[1] for pair in sorted_data]
newlist, = [ 0, list1[0], list1[1] + list1[2], list1[0] + list[1] + list[2]]

Answer 1

基于生成器的解决方案

In [25]: ll = [3,51,34]

In [26]: def acc(myiter):                                                                                                                                   
   ....:     it = iter(myiter)
   ....:     total = it.next()
   ....:     yield total
   ....:     for element in it:
   ....:         total = total + element                                                                                                                    
   ....:         yield total
   ....: 

In [27]: acc(ll)
Out[27]: <generator object acc at 0x1ec9e10>

In [28]: [x for x in acc(ll)]                                                                                                                               
Out[28]: [3, 54, 88]

因为这是最快的方案：

In [29]: %timeit [x for x in acc(ll)]
1000000 loops, best of 3: 1.26 µs per loop

In [30]: import numpy as np

In [31]: np.cumsum(ll)                                                                                                                                      
Out[31]: array([ 3, 54, 88])

In [32]: %timeit np.cumsum(ll)
100000 loops, best of 3: 15.8 µs per loop

In [33]: %timeit reduce(lambda x, y: [y] if not x else x + [y + x[-1]], ll, None)                                                                           
1000000 loops, best of 3: 1.87 µs per loop                                                                                                                  

In [34]: %timeit  [n + sum(ll[:i]) for i, n in enumerate(ll)]                                                                                               
100000 loops, best of 3: 1.99 µs per loop 

Answer 2

这里有一个简单的、按步骤的解决方案，供大家参考：

a = [3,51,34]

def cumsum(numbers):
  accum = 0
  result = []
  for n in numbers:
    accum += n
    result.append(accum)
  return result

print cumsum(a)

它会输出 [3, 54, 88]

Answer 3

这里提到了一种叫做“列表推导”的方法，因为这种方法总是很有趣：

>>> data = [3, 51, 34]
>>> result = [n + sum(data[:i]) for i, n in enumerate(data)]
>>> result
[3, 54, 88]

Answer 4

numpy.cumsum 可能是处理这种情况的一个不错选择。

In [1]: import numpy as np

In [2]: a = [3,51,34]

In [3]: np.cumsum(a)
Out[3]: array([ 3, 54, 88])

Answer 5

一个简单的归约操作：

nums = [3,51,34]
reduce(lambda x, y: [y] if not x else x + [y + x[-1]], nums, None)
# [3, 54, 88]