什么是优雅的pythonic方式来实施cumsum?
或者 - 如果there'a已经内置的方式做到这一点,那将是更好的,当然...优雅的pythonic cumsum
回答
它的问世在Numpy:
>>> import numpy as np
>>> np.cumsum([1,2,3,4,5])
array([ 1, 3, 6, 10, 15])
或者使用itertools.accumulate
因为Python 3.2 :
>>> from itertools import accumulate
>>> list(accumulate([1,2,3,4,5]))
[ 1, 3, 6, 10, 15]
如果numpy的是不是一种选择,一台发电机循环将是最完美的解决方案,我能想到的:
def cumsum(it):
total = 0
for x in it:
total += x
yield total
Ex。
>>> list(cumsum([1,2,3,4,5]))
>>> [1, 3, 6, 10, 15]
不错,简单,收益率总计:) – 2012-02-13 10:16:40
for循环Python的
def cumsum(vec):
r = [vec[0]]
for val in vec[1:]:
r.append(r[-1] + val)
return r
@larsmans:在这种情况下,会有'r [-1]'是什么? – WolframH 2013-02-25 00:05:06
a=[1,2,3,4,5]
def cumsum(a):
a=iter(a)
cc=[next(a)]
for i in a:
cc.append(cc[-1]+i)
return cc
print cumsum(a)
"[1, 3, 6, 10, 15]"
到位:
a=[1,2,3,4,5]
def cumsum(a):
for i in range(1,len(a)):
a[i]+=a[i-1]
cumsum(a)
print a
"[1, 3, 6, 10, 15]"
a=[1,2,3,4,5]
def cumsum(a):
b=a[:]
for i in range(1,len(a)):
b[i]+=b[i-1]
return b
print cumsum(a)
"[1, 3, 6, 10, 15]"
a = [1, 2, 3 ,4, 5]
# Using list comprehention
cumsum = [sum(a[:i+1]) for i in range(len(a))] # [1, 3, 6, 10, 15]
# Using map()
cumsum = map(lambda i: sum(a[:i+1]), range(len(a))) # [1, 3, 6, 10, 15]
虽然这很好,很容易,但要警惕的是,在一个大的列表中,这将非常缓慢! :) – 2012-02-14 07:32:14
我的想法是用减少功能方式:
from operator import iadd
reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), [1, 2, 3, 4, 5], [0])[1:]
>>> [1, 3, 6, 10, 15]
来自操作员模块的iadd具有进行原地添加并返回目的地的独特属性。
如果[1:]副本让你畏缩,你可以这样做:
from operator import iadd
reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm),
[1, 2, 3, 4, 5], ([], 0))[0]
>>> [1, 3, 6, 10, 15]
但我发现,在当地的第一个例子是更快,IMO发电机比像函数式编程更Python化“减少“:
reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm), values_ten, ([], 0))[0]
Average: 6.4593828736e-06
reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm), values_mil, ([], 0))[0]
Average: 0.727404361961
reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), values_ten, [0])[1:]
Average: 5.16271911336e-06
reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), values_mil, [0])[1:]
Average: 0.524223491301
cumsum_rking(values_ten)
Average: 1.9828751369e-06
cumsum_rking(values_mil)
Average: 0.234241141632
list(cumsum_larsmans(values_ten))
Average: 2.02786211569e-06
list(cumsum_larsmans(values_mil))
Average: 0.201473119335
这里的基准测试脚本,情况因人而异:
from timeit import timeit
def bmark(prog, setup, number):
duration = timeit(prog, setup=setup, number=number)
print prog
print 'Average:', duration/number
values_ten = list(xrange(10))
values_mil = list(xrange(1000000))
from operator import iadd
bmark('reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm), \
values_ten, ([], 0))[0]',
setup='from __main__ import iadd, values_ten', number=1000000)
bmark('reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm), \
values_mil, ([], 0))[0]',
setup='from __main__ import iadd, values_mil', number=10)
bmark('reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), \
values_ten, [0])[1:]',
setup='from __main__ import iadd, values_ten', number=1000000)
bmark('reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), \
values_mil, [0])[1:]',
setup='from __main__ import iadd, values_mil', number=10)
def cumsum_rking(iterable):
values = list(iterable)
for pos in xrange(1, len(values)):
values[pos] += values[pos - 1]
return values
bmark('cumsum_rking(values_ten)',
setup='from __main__ import cumsum_rking, values_ten', number=1000000)
bmark('cumsum_rking(values_mil)',
setup='from __main__ import cumsum_rking, values_mil', number=10)
def cumsum_larsmans(iterable):
total = 0
for value in iterable:
total += value
yield total
bmark('list(cumsum_larsmans(values_ten))',
setup='from __main__ import cumsum_larsmans, values_ten', number=1000000)
bmark('list(cumsum_larsmans(values_mil))',
setup='from __main__ import cumsum_larsmans, values_mil', number=10)
这是我的Python版本字符串:
Python 2.7 (r27:82525, Jul 4 2010, 09:01:59) [MSC v.1500 32 bit (Intel)] on win32
+1进行彻底的基准测试 – Jonathan 2013-02-25 14:16:27
def cumsum(a):
return map(lambda x: sum(a[0:x+1]), range(0, len(a)))
cumsum([1,2,3])
> [1, 3, 6]
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cumsum在http://docs.scipy.org/doc/numpy/reference/generated/numpy.cumsum.html? – 2012-02-13 10:06:37
http://docs.python.org/dev/library/itertools.html#itertools.accumulate – ChessMaster 2012-02-13 10:20:24