Variance/Std 批量更新的 Welford 算法的公式是什么？

Question

我想扩展 Welford 的在线算法，以便能够更新多个数字（批量），而不是一次只更新一个：https://en.wikipedia.org/wiki/Algorithms_for_calculating_variance

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(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)

不过，我一定没有理解正确，因为结果是错误的：

# example x that might have led to an a = (2, 2.0, 2.0).
x = [1.0, 3.0]
mean = np.mean(x)
count = len(x)
m2 = np.sum(np.subtract(x, [mean] * count)**2)

a = (count, mean, m2)
print(a)
# new batch of values.
b = [5, 3]

注意 a = (2, 2.0, 2.0) 表示我们有 2 个观测值，它们的平均值是 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结果：

(3.0, 2.0, 2.6666666666666665)
(3.0, 2.5, 3.3333333333333335)

关于正确的方差更新，我遗漏了什么？我是否也需要以某种方式更新 finalize 函数？

我需要这样做的原因是因为我正在处理非常大的月度文件（具有不同数量的观察值）并且我需要获得年度均值和方差。

Answer 1

我不太熟悉Python，所以我宁愿坚持数学符号。

要更新均值，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，您必须添加另一个求和：

newM2 = oldM2 + sum_i ((newValue[i] - newMean) * (newValue[i] - oldMean))

我不确定你是否真的用这个批量更新保存了任何东西，因为你里面还有一个循环。

Answer 2

感谢 Nico 的澄清，我明白了！问题是我对增量求和然后乘以得到 M2，但必须对增量的乘积求和。这是能够接受单个数字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)

示例用法：

x = [1.0, 3.0]
mean = np.mean(x)
count = len(x)
m2 = np.sum(np.subtract(x, [mean] * count)**2)

a = (count, mean, m2)
print(a)
# new batch of values.
b = [5, 3]

result_batch = update(a, b)
result_batch1 = update(a, b[0])

print(finalize(result_batch))
print(finalize(result_batch1))

而且确实更快：

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

Answer 3

加上前面的答案 derivation for the generalized case (calculate s_{n+k} given s_n) follows similarly to the proof for the k=1 version，得到：

$eq1$

其中 u_{n+k} 也是 calculated batch-wise 通过：

$eq2$

您还可以考虑一种基于计算 mean/variance of c = A union B from the mean/variance of A and B, that is easily parallelizable (Chan's algorithm) 的方法。

Variance/Std 批量更新的 Welford 算法的公式是什么？

What's the formula for Welford's Algorithm for Variance/Std with Batch Updates?

python

algorithm

statistics

variance

batch-updates