我个人更喜欢这样的简单生成器：
def gen(lst):
cumulative = 0
for item in lst:
if item:
cumulative += item
else:
cumulative = 0
yield cumulative
没什么神奇的（当您知道yield 的工作原理时），易于阅读并且应该相当快。
如果您需要更高的性能，您甚至可以将其包装为 Cython 扩展类型（我在这里使用 IPython）。因此你失去了“易于理解”的部分，它需要“高度依赖”：
%load_ext cython
%%cython
cdef class Cumulative(object):
cdef object it
cdef object cumulative
def __init__(self, it):
self.it = iter(it)
self.cumulative = 0
def __iter__(self):
return self
def __next__(self):
cdef object nxt = next(self.it)
if nxt:
self.cumulative += nxt
else:
self.cumulative = 0
return self.cumulative
两者都需要消耗，例如使用list 来提供所需的输出：
>>> list_a = [1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1]
>>> list(gen(list_a))
[1, 2, 3, 0, 1, 2, 0, 1, 0, 1, 2, 3]
>>> list(Cumulative(list_a))
[1, 2, 3, 0, 1, 2, 0, 1, 0, 1, 2, 3]
不过，既然你问到速度，我想分享一下我的计时结果：
import pandas as pd
import numpy as np
import random
import pandas as pd
from itertools import takewhile
from itertools import groupby, accumulate, chain
def MSeifert(lst):
return list(MSeifert_inner(lst))
def MSeifert_inner(lst):
cumulative = 0
for item in lst:
if item:
cumulative += item
else:
cumulative = 0
yield cumulative
def MSeifert2(lst):
return list(Cumulative(lst))
def original1(list_a):
list_b = []
for i, x in enumerate(list_a):
if x == 0:
list_b.append(x)
else:
sum_value = 0
for j in list_a[i::-1]:
if j != 0:
sum_value += j
else:
break
list_b.append(sum_value)
def original2(list_a):
return [sum(takewhile(lambda x: x != 0, list_a[i::-1])) for i, d in enumerate(list_a)]
def Coldspeed1(data):
data = data.copy()
for i in range(1, len(data)):
if data[i]:
data[i] += data[i - 1]
return data
def Coldspeed2(data):
s = pd.Series(data)
return s.groupby(s.eq(0).cumsum()).cumsum().tolist()
def Chris_Rands(list_a):
return list(chain.from_iterable(accumulate(g) for _, g in groupby(list_a, bool)))
def EvKounis(list_a):
cum_sum = 0
list_b = []
for item in list_a:
if not item: # if our item is 0
cum_sum = 0 # the cumulative sum is reset (set back to 0)
else:
cum_sum += item # otherwise it sums further
list_b.append(cum_sum) # and no matter what it gets appended to the result
def schumich(list_a):
list_b = []
s = 0
for a in list_a:
s = a+s if a !=0 else 0
list_b.append(s)
return list_b
def jbch(seq):
return list(jbch_inner(seq))
def jbch_inner(seq):
s = 0
for n in seq:
s = 0 if n == 0 else s + n
yield s
# Timing setup
timings = {MSeifert: [],
MSeifert2: [],
original1: [],
original2: [],
Coldspeed1: [],
Coldspeed2: [],
Chris_Rands: [],
EvKounis: [],
schumich: [],
jbch: []}
sizes = [2**i for i in range(1, 20, 2)]
# Timing
for size in sizes:
print(size)
func_input = [int(random.random() < 0.75) for _ in range(size)]
for func in timings:
if size > 10000 and (func is original1 or func is original2):
continue
res = %timeit -o func(func_input) # if you use IPython, otherwise use the "timeit" module
timings[func].append(res)
%matplotlib notebook
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure(1)
ax = plt.subplot(111)
baseline = MSeifert2 # choose one function as baseline
for func in timings:
ax.plot(sizes[:len(timings[func])],
[time.best / ref.best for time, ref in zip(timings[func], timings[baseline])],
label=func.__name__) # you could also use "func.__name__" here instead
ax.set_ylim(0.8, 1e4)
ax.set_yscale('log')
ax.set_xscale('log')
ax.set_xlabel('size')
ax.set_ylabel('time relative to {}'.format(baseline)) # you could also use "func.__name__" here instead
ax.grid(which='both')
ax.legend()
plt.tight_layout()
如果您对确切的结果感兴趣，我将它们放入 this gist。
这是一个对数图，相对于 Cython 的答案。简而言之：越低越快，两个主要刻度之间的范围代表一个数量级。
因此，除了您拥有的解决方案之外，所有解决方案都倾向于在一个数量级内（至少在列表很大时）。奇怪的是，与纯 Python 方法相比，pandas 解决方案相当慢。然而，Cython 解决方案比所有其他方法高出 2 倍。