可以尝试的小事:

将两个图像重新采样为小的缩略图(例如64 x 64)，并将缩略图与某个阈值逐像素进行比较。如果原始图像几乎相同，重新采样的缩略图将非常相似，甚至完全相同。这种方法可以处理特别是在低光场景中可能出现的噪音。如果你调成灰度，效果可能会更好。

给出的大多数答案都不涉及照明水平。

在进行比较之前，我首先将图像归一化到标准的光照水平。

import os
from PIL import Image
from PIL import ImageFile
import imagehash
  
#just use to the size diferent picture
def compare_image(img_file1, img_file2):
    if img_file1 == img_file2:
        return True
    fp1 = open(img_file1, 'rb')
    fp2 = open(img_file2, 'rb')

    img1 = Image.open(fp1)
    img2 = Image.open(fp2)

    ImageFile.LOAD_TRUNCATED_IMAGES = True
    b = img1 == img2

    fp1.close()
    fp2.close()

    return b





#through picturu hash to compare
def get_hash_dict(dir):
    hash_dict = {}
    image_quantity = 0
    for _, _, files in os.walk(dir):
        for i, fileName in enumerate(files):
            with open(dir + fileName, 'rb') as fp:
                hash_dict[dir + fileName] = imagehash.average_hash(Image.open(fp))
                image_quantity += 1

    return hash_dict, image_quantity

def compare_image_with_hash(image_file_name_1, image_file_name_2, max_dif=0):
    """
    max_dif: The maximum hash difference is allowed, the smaller and more accurate, the minimum is 0.
    recommend to use
    """
    ImageFile.LOAD_TRUNCATED_IMAGES = True
    hash_1 = None
    hash_2 = None
    with open(image_file_name_1, 'rb') as fp:
        hash_1 = imagehash.average_hash(Image.open(fp))
    with open(image_file_name_2, 'rb') as fp:
        hash_2 = imagehash.average_hash(Image.open(fp))
    dif = hash_1 - hash_2
    if dif < 0:
        dif = -dif
    if dif <= max_dif:
        return True
    else:
        return False


def compare_image_dir_with_hash(dir_1, dir_2, max_dif=0):
    """
    max_dif: The maximum hash difference is allowed, the smaller and more accurate, the minimum is 0.

    """
    ImageFile.LOAD_TRUNCATED_IMAGES = True
    hash_dict_1, image_quantity_1 = get_hash_dict(dir_1)
    hash_dict_2, image_quantity_2 = get_hash_dict(dir_2)

    if image_quantity_1 > image_quantity_2:
        tmp = image_quantity_1
        image_quantity_1 = image_quantity_2
        image_quantity_2 = tmp

        tmp = hash_dict_1
        hash_dict_1 = hash_dict_2
        hash_dict_2 = tmp

    result_dict = {}

    for k in hash_dict_1.keys():
        result_dict[k] = None

    for dif_i in range(0, max_dif + 1):
        have_none = False

        for k_1 in result_dict.keys():
            if result_dict.get(k_1) is None:
                have_none = True

        if not have_none:
            return result_dict

        for k_1, v_1 in hash_dict_1.items():
            for k_2, v_2 in hash_dict_2.items():
                sub = (v_1 - v_2)
                if sub < 0:
                    sub = -sub
                if sub == dif_i and result_dict.get(k_1) is None:
                    result_dict[k_1] = k_2
                    break
    return result_dict


def main():
    print(compare_image('image1\\815.jpg', 'image2\\5.jpg'))
    print(compare_image_with_hash('image1\\815.jpg', 'image2\\5.jpg', 7))
    r = compare_image_dir_with_hash('image1\\', 'image2\\', 10)
    for k in r.keys():
        print(k, r.get(k))


if __name__ == '__main__':
    main()

输出: 假 真正的 image2 jpg image1 5. \ \ 815. jpg image2 jpg image1 6. \ \ 819. jpg image2 jpg image1 7. \ \ 900. jpg image2 jpg image1 8. \ \ 998. jpg image2 jpg image1 9. \ \ 1012. jpg 示例图片: 815. jpg 5. jpg

可以尝试的小事:

将两个图像重新采样为小的缩略图(例如64 x 64)，并将缩略图与某个阈值逐像素进行比较。如果原始图像几乎相同，重新采样的缩略图将非常相似，甚至完全相同。这种方法可以处理特别是在低光场景中可能出现的噪音。如果你调成灰度，效果可能会更好。

I have been having a lot of luck with jpg images taken with the same camera on a tripod by (1) simplifying greatly (like going from 3000 pixels wide to 100 pixels wide or even fewer) (2) flattening each jpg array into a single vector (3) pairwise correlating sequential images with a simple correlate algorithm to get correlation coefficient (4) squaring correlation coefficient to get r-square (i.e fraction of variability in one image explained by variation in the next) (5) generally in my application if r-square < 0.9, I say the two images are different and something happened in between.

这是强大的和快速的在我的实现(Mathematica 7)

这是值得玩转的部分，你感兴趣的图像，并通过裁剪所有的图像到那个小区域，否则一个远离相机但重要的变化将被错过。

我不知道如何使用Python，但我确信它也有相关性，不是吗?

你可以计算这两个图像的直方图，然后计算Bhattacharyya系数，这是一个非常快速的算法，我已经用它来检测板球视频中的镜头变化(在C中使用openCV)