RGB图像中最主要的颜色 - OpenCV / NumPy / Python

时间:2018-06-17 19:11:06

标签: python python-3.x numpy opencv

我有一个python图像处理功能,它使用try来获得图像的主色。我使用了https://github.com/tarikd/python-kmeans-dominant-colors/blob/master/utils.py

中找到的函数

它有效,但不幸的是我不太明白它的作用,我了解到np.histogram相当慢,我应该使用cv2.calcHist,因为它的速度要快40倍这个:https://docs.opencv.org/trunk/d1/db7/tutorial_py_histogram_begins.html

我想了解如何更新代码以使用cv2.calcHist,或者更好,我必须输入哪些值。

我的功能

def centroid_histogram(clt):
    # grab the number of different clusters and create a histogram
    # based on the number of pixels assigned to each cluster
    num_labels = np.arange(0, len(np.unique(clt.labels_)) + 1)
    (hist, _) = np.histogram(clt.labels_, bins=num_labels)

    # normalize the histogram, such that it sums to one
    hist = hist.astype("float")
    hist /= hist.sum()

    # return the histogram
    return hist

pprint clt就是这个,不确定这是否有帮助

KMeans(algorithm='auto', copy_x=True, init='k-means++', max_iter=300,
    n_clusters=1, n_init=10, n_jobs=1, precompute_distances='auto',
    random_state=None, tol=0.0001, verbose=0)

我的代码可以在这里找到:https://github.com/primus852/python-movie-barcode

我是一个非常初学者,所以任何帮助都非常感谢。

根据要求:

示例图像

Sample

最主要的颜色:

rgb(22,28,37)

直方图的计算时间:

0.021515369415283203s

3 个答案:

答案 0 :(得分:5)

可以建议使用np.uniquenp.bincount两种方法来获得最主要的颜色。此外,在链接页面中,它将bincount作为更快的替代方案进行讨论,因此这可能是最佳选择。

方法#1

def unique_count_app(a):
    colors, count = np.unique(a.reshape(-1,a.shape[-1]), axis=0, return_counts=True)
    return colors[count.argmax()]

方法#2

def bincount_app(a):
    a2D = a.reshape(-1,a.shape[-1])
    col_range = (256, 256, 256) # generically : a2D.max(0)+1
    a1D = np.ravel_multi_index(a2D.T, col_range)
    return np.unravel_index(np.bincount(a1D).argmax(), col_range)

1000 x 1000彩色图像在[0,9)密集范围内进行验证和计时,以获得可重复的结果 -

In [28]: np.random.seed(0)
    ...: a = np.random.randint(0,9,(1000,1000,3))
    ...: 
    ...: print unique_count_app(a)
    ...: print bincount_app(a)
[4 7 2]
(4, 7, 2)

In [29]: %timeit unique_count_app(a)
1 loop, best of 3: 820 ms per loop

In [30]: %timeit bincount_app(a)
100 loops, best of 3: 11.7 ms per loop

进一步提升

在大数据利用multi-core with numexpr module时进一步提升 -

import numexpr as ne

def bincount_numexpr_app(a):
    a2D = a.reshape(-1,a.shape[-1])
    col_range = (256, 256, 256) # generically : a2D.max(0)+1
    eval_params = {'a0':a2D[:,0],'a1':a2D[:,1],'a2':a2D[:,2],
                   's0':col_range[0],'s1':col_range[1]}
    a1D = ne.evaluate('a0*s0*s1+a1*s0+a2',eval_params)
    return np.unravel_index(np.bincount(a1D).argmax(), col_range)

计时 -

In [90]: np.random.seed(0)
    ...: a = np.random.randint(0,9,(1000,1000,3))

In [91]: %timeit unique_count_app(a)
    ...: %timeit bincount_app(a)
    ...: %timeit bincount_numexpr_app(a)
1 loop, best of 3: 843 ms per loop
100 loops, best of 3: 12 ms per loop
100 loops, best of 3: 8.94 ms per loop

答案 1 :(得分:2)

cv2.calcHist()的等效代码是替换:

(hist, _) = np.histogram(clt.labels_, bins=num_labels)  

dmin, dmax, _, _ = cv2.minMaxLoc(clt.labels_)

if np.issubdtype(data.dtype, 'float'): dmax += np.finfo(data.dtype).eps
else: dmax += 1

hist = cv2.calcHist([clt.labels_], [0], None, [num_labels], [dmin, dmax]).flatten()

请注意,cv2.calcHist仅接受uint8float32作为元素类型。

更新

似乎opencv和numpy的binning彼此不同,因为如果bin的数量没有映射值范围,直方图会有所不同:

import numpy as np
from matplotlib import pyplot as plt
import cv2

#data = np.random.normal(128, 1, (100, 100)).astype('float32')
data = np.random.randint(0, 256, (100, 100), 'uint8')
BINS = 20

np_hist, _ = np.histogram(data, bins=BINS)

dmin, dmax, _, _ = cv2.minMaxLoc(data)
if np.issubdtype(data.dtype, 'float'): dmax += np.finfo(data.dtype).eps
else: dmax += 1

cv_hist = cv2.calcHist([data], [0], None, [BINS], [dmin, dmax]).flatten()

plt.plot(np_hist, '-', label='numpy')
plt.plot(cv_hist, '-', label='opencv')
plt.gcf().set_size_inches(15, 7)
plt.legend()
plt.show()

答案 2 :(得分:1)

@Divakar给出了一个很好的答案。但是如果你想将自己的代码移植到OpenCV,那么:

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         <Button Name="RemoveButton" Content="Remove" Click="Rem_Notification" Style="{StaticResource ButtonStyle}" Background="{StaticResource RemBrushStyle}"/>
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您的图片的结果:

  

显性颜色是:bgr([41 31 23])

时间:0.10798478126525879秒