一种方法是创建一个简单的信噪比估计，使用像素强度和地面实况图像作为掩码。这是我在 Python 中的意思的一个例子：

import numpy as np

# Sample data 
# Indices [0, 1, 2, 6] are background, [3, 4, 5] are object
original  = np.array([1, 2, 2, 5, 6, 8, 2])
truth     = np.array([0, 0, 0, 1, 1, 1, 0])
back_mask = 1 - truth
result    = np.array([0, 1, 0, 4, 5, 7, 1]) # Result of background subtraction

def non_masked_mean(input_array, mask):
    """mean of non-masked elements in the array"""
    return np.ma.masked_array(input_array, mask).mean()

def snr(input_array, back_mask):
    """ratio of non-masked to masked mean intensities"""
    with np.errstate(all='ignore'):
        snr = non_masked_mean(input_array, back_mask) / non_masked_mean(input_array, (1-back_mask))
    return 0 if np.isnan(snr) else snr

snr_before = snr(original, back_mask)  # (mean of objects / mean of background), before
snr_after  = snr(result, back_mask)    # (mean of objects / mean of background), after
snr_ratio  = snr_after / snr_before

if snr_after == np.inf:
    print "Perfect SNR in result! Either great or suspect..."
elif snr_after == 0:
    print "Image was flattened :("
else:
    print "SNR changed by a multiple of %.2f" % (snr_ratio)

这样做的问题是，您通过偏爱严厉的背景减法获得了完美的分数。即使对象只有 1 个像素幸存，此 SNR 也将是无限的。另一种方法是分别考虑背景和信号，并对两者的结果进行加权。然后，您可以决定是否更重要的是展平背景，还是保留原件。

def non_masked_mean_ratio(input_array, result_array, mask):
    """ratio of two array means, with the same mask"""
    return non_masked_mean(result_array, mask) / non_masked_mean(input_array, mask) 

signal_ratio     = non_masked_mean_ratio(original, result, back_mask)     # (mean of objects after) / (mean of objects before)
background_ratio = non_masked_mean_ratio(original, result, truth)         # (mean of background after) / (mean of background before)
weight           = 0.5   # Higher weight favours preservation of objects

score = (weight*signal_ratio) + ((1-weight) * (1-background_ratio))

这为分数提供了大约 0-1 的良好预期范围，其中 1 是理想的结果。但它并不完美，因为您可以通过增加对象的强度来增强它。

如果您期望二进制结果，您可以使用类似的策略，除了使用非零像素数而不是平均值：

def nonzero_count_ratio(result_array, ideal_array):
    """ratio of actual number of non-zero pixels vs ideal non-zero pixels"""
    return float(np.count_nonzero(result_array*ideal_array)) / np.count_nonzero(ideal_array)

true_obj_ratio   = nonzero_count_ratio(result, truth)     # Number of correct object pixels vs number of potential correct
false_back_ratio = nonzero_count_ratio(result, back_mask) # Number of false background pixels vs number of potential false
weight           = 0.5

score = (weight*true_obj_ratio) + ((1-weight) * (1-false_back_ratio))