Python 什么';Welford的公式是';批量更新的方差/Std的s算法?
我想扩展Welford的在线算法,使其能够以多个数字(批量)更新,而不是一次只更新一个数字: 我尝试从wiki页面更新算法,如下所示:Python 什么';Welford的公式是';批量更新的方差/Std的s算法?,python,algorithm,statistics,variance,batch-updates,Python,Algorithm,Statistics,Variance,Batch Updates,我想扩展Welford的在线算法,使其能够以多个数字(批量)更新,而不是一次只更新一个数字: 我尝试从wiki页面更新算法,如下所示: # my attempt. def update1(existingAggregate, newValues): (count, mean, M2) = existingAggregate count += len(newValues) delta = np.sum(np.subtract(newValues, [mean] * len
# my attempt.
def update1(existingAggregate, newValues):
(count, mean, M2) = existingAggregate
count += len(newValues)
delta = np.sum(np.subtract(newValues, [mean] * len(newValues)))
mean += delta / count
delta2 = np.sum(np.subtract(newValues, [mean] * len(newValues)))
M2 += delta * delta2
return (count, mean, M2)
# The original two functions from wikipedia.
def update(existingAggregate, newValue):
(count, mean, M2) = existingAggregate
count += 1
delta = newValue - mean
mean += delta / count
delta2 = newValue - mean
M2 += delta * delta2
def finalize(existingAggregate):
(count, mean, M2) = existingAggregate
(mean, variance, sampleVariance) = (mean, M2/count, M2/(count - 1))
if count < 2:
return float('nan')
else:
return (mean, variance, sampleVariance)
注意a=(2,2.0,2.0)意味着我们有2个观察值,它们的平均值为2.0
# update one at a time.
temp = update(a, newValues[0])
result_single = update(temp, newValues[1])
print(finalize(result_single))
# update with my faulty batch function.
result_batch = update1(a, newValues)
print(finalize(result_batch))
正确的输出应该是两次应用单个数字更新的输出:
(3.0, 2.0, 2.6666666666666665)
(3.0, 2.5, 3.3333333333333335)
关于正确的差异更新,我遗漏了什么?我是否也需要以某种方式更新finalize函数
我之所以需要这样做,是因为我处理的是非常大的月度文件(具有不同数量的观察值),我需要了解年度均值和方差。我对Python不太熟悉,所以我宁愿坚持使用数学符号 要更新平均值,您必须执行以下操作:
s = sum of new values
c = number of new values
newMean = oldMean + sum_i (newValue[i] - oldMean) / newCount
对于M2
,您必须添加另一个总和:
newM2 = oldM2 + sum_i ((newValue[i] - newMean) * (newValue[i] - oldMean))
我不确定你是否真的在批量更新中保存了任何东西,因为你的内部仍然有一个循环。多亏了Nico的澄清,我找到了它! 问题是我对delta求和,然后相乘得到M2,但必须对delta的乘积求和。 以下是能够接受单个编号和批次的正确批处理函数:
# https://en.wikipedia.org/wiki/Algorithms_for_calculating_variance
def update(existingAggregate, newValues):
if isinstance(newValues, (int, float, complex)):
# Handle single digits.
newValues = [newValues]
(count, mean, M2) = existingAggregate
count += len(newValues)
# newvalues - oldMean
delta = np.subtract(newValues, [mean] * len(newValues))
mean += np.sum(delta / count)
# newvalues - newMeant
delta2 = np.subtract(newValues, [mean] * len(newValues))
M2 += np.sum(delta * delta2)
return (count, mean, M2)
def finalize(existingAggregate):
(count, mean, M2) = existingAggregate
(mean, variance, sampleVariance) = (mean, M2/count, M2/(count - 1))
if count < 2:
return float('nan')
else:
return (mean, variance, sampleVariance)
而且它确实更快:
import timeit
x = random.sample(range(1, 10000), 1000)
# ...
b = random.sample(range(1, 10000), 1000)
start_time = timeit.default_timer()
result_batch = update(a, b)
print(f'{timeit.default_timer() - start_time:.4f}')
print(*(f'{x:.2f}' for x in finalize(result_batch)))
start_time = timeit.default_timer()
for i in b:
a = update1(a, i)
print(f'{timeit.default_timer() - start_time:.4f}')
print(*(f'{x:.2f}' for x in finalize(result_batch)))
结果:
0.0010
5008.36 8423224.68 8427438.40
0.0031
5008.36 8423224.68 8427438.40
谢谢,这非常有帮助!一个数字改进(以避免灾难性的取消)是在减法后求和以得到新均值,即``cOldMean=c*oldMean newMean=oldMean+sum_i(新值[i]-cOldMean)/newCount`````(特别是,我们不使用
s
),它确实更快:``导入timeit x=random.sample(范围(1,10000),1000)`。。。b=random.sample(范围(1,10000),1000)start_time=timeit.default_timer()result_batch=update(a,b)print(f'{timeit.default_timer()-start_time:.4f}')print(*(f'{x:.2f})finalize(result_batch))中x的开始时间=timeit{x:.2f}'用于最终确定(结果批次))``结果:``0.0010 5008.36 8423224.68 8427438.40 0.0031 5008.36 8423224.68 8427438.40```
import timeit
x = random.sample(range(1, 10000), 1000)
# ...
b = random.sample(range(1, 10000), 1000)
start_time = timeit.default_timer()
result_batch = update(a, b)
print(f'{timeit.default_timer() - start_time:.4f}')
print(*(f'{x:.2f}' for x in finalize(result_batch)))
start_time = timeit.default_timer()
for i in b:
a = update1(a, i)
print(f'{timeit.default_timer() - start_time:.4f}')
print(*(f'{x:.2f}' for x in finalize(result_batch)))
0.0010
5008.36 8423224.68 8427438.40
0.0031
5008.36 8423224.68 8427438.40