我正在尝试使用python中的主成分分析(PCA)进行手势识别(类似于人脸识别)。我有一个测试图像,我想从一组训练图像中获得最接近的匹配。
这是我的代码:
import os, sys
import numpy as np
import PIL.Image as Image
def read_images(path, sz=None):
c = 0
X,y = [], []
for dirname, dirnames, filenames in os.walk(path):
for subdirname in dirnames:
subject_path = os.path.join(dirname, subdirname)
for filename in os.listdir(subject_path):
try:
im = Image.open(os.path.join(subject_path, filename))
im = im.convert("L")
# resize to given size (if given)
if (sz is not None):
im = im.resize(sz, Image.ANTIALIAS)
X.append(np.asarray(im, dtype=np.uint8))
y.append(c)
except IOError:
print "I/O error({0}): {1}".format(errno, strerror)
except:
print "Unexpected error:", sys.exc_info()[0]
raise
c = c+1
return [X,y]
def asRowMatrix(X):
if len(X) == 0:
return np.array([])
mat = np.empty((0, X[0].size), dtype=X[0].dtype)
for row in X:
mat = np.vstack((mat, np.asarray(row).reshape(1,-1)))
return mat
def asColumnMatrix(X):
if len(X) == 0:
return np.array([])
mat = np.empty((X[0].size, 0), dtype=X[0].dtype)
for col in X:
mat = np.hstack((mat, np.asarray(col).reshape(-1,1)))
return mat
def pca(X, y, num_components=0):
[n,d] = X.shape
if (num_components <= 0) or (num_components>n):
num_components = n
mu = X.mean(axis=0)
X = X - mu
if n>d:
C = np.dot(X.T,X)
[eigenvalues,eigenvectors] = np.linalg.eigh(C)
else:
C = np.dot(X,X.T)
[eigenvalues,eigenvectors] = np.linalg.eigh(C)
eigenvectors = np.dot(X.T,eigenvectors)
for i in xrange(n):
eigenvectors[:,i] = eigenvectors[:,i]/np.linalg.norm(eigenvectors[:,i])
# or simply perform an economy size decomposition
# eigenvectors, eigenvalues, variance = np.linalg.svd(X.T, full_matrices=False)
# sort eigenvectors descending by their eigenvalue
idx = np.argsort(-eigenvalues)
eigenvalues = eigenvalues[idx]
eigenvectors = eigenvectors[:,idx]
# select only num_components
eigenvalues = eigenvalues[0:num_components].copy()
eigenvectors = eigenvectors[:,0:num_components].copy()
return [eigenvalues, eigenvectors, mu, X]
#Get eigenvalues, eigenvectors, mean and shifted images (Training)
[a, b] = read_images('C:\\Users\\Karim\\Desktop\\Training & Test images\\AT&T\\att_faces', (90,90))
[evalues, evectors, mean_image, shifted_images] = pca(asRowMatrix(a), b)
#Input(Test) image
input_image = Image.open('C:\\Users\\Karim\\Desktop\\Training & Test images\\AT&T\\Test\\4.pgm').convert('L').resize((90, 90))
input_image = np.asarray(input_image).flatten()
#Normalizing input image
shifted_in = input_image - mean_image
#Finding weights
w = evectors.T * shifted_images
w = np.asarray(w)
w_in = evectors.T * shifted_in
w_in = np.asarray(w_in)
#Euclidean distance
df = np.asarray(w - w_in) # the difference between the images
dst = np.sqrt(np.sum(df**2, axis=1)) # their euclidean distances
现在我有一个距离dst
数组,其中包含测试图像与训练图像集中每个图像之间的欧氏距离。
如何获取最近(最小)距离及其路径(或子目录名称)的图像?不是数组dst
答案 0 :(得分:3)
dst.argmin()
会告诉您dst
中最小的元素索引。
所以最近的图像是
idx = dst.argmin()
closest = a[idx]
因为a
是表示训练面的数组列表。
要显示最近的图像,您可以使用:
img = Image.fromarray(closest, 'L')
img.show()
要查找最近图像的文件路径,我会改变read_images
以返回所有文件路径的列表,因此可以将其编入索引,就像图像列表一样。
def read_images(path, sz=None):
X, y = [], []
for dirname, dirnames, filenames in os.walk(path):
for filename in filenames:
subject_path = os.path.join(dirname, filename)
try:
im = Image.open(subject_path)
except IOError as err:
print "I/O error: {e}: {f}".format(e=err, f=subject_path)
except:
print "Unexpected error:", sys.exc_info()[0]
raise
else:
im = im.convert("L")
# resize to given size (if given)
if (sz is not None):
im = im.resize(sz, Image.ANTIALIAS)
X.append(np.asarray(im, dtype=np.uint8))
y.append(subject_path)
return [X, y]
下面,请这样称呼:
images, paths = read_images(TRAINING_DIR, (90, 90))
然后,您可以使用
获取最近图像的完整路径idx = dst.argmin()
filename = paths[idx]
如果您只想要子目录的路径,请使用
os.path.dirname(filename)
对于子目录的名称,请使用
os.path.basename(os.path.dirname(filename))