Develop Reference

Shape of face encodings differ

I'm trying to make a face recognition program but the problem is the face encoding shape of some encodings are bigger than the others and thus im getting the error
ValueError: setting an array element with a sequence.
Here's my code to generate the encodings
class FaceEncoder():
def __init__(self, files, singleton = False, model_path='./models/lbpcascade_animeface.xml', scale_factor=1.1, min_neighbours=1):
self.singleton = singleton
self.files = files
self.model = model_path
self.scale_factor = scale_factor
self.min_neighbours = min_neighbours
def encode(self, singleton=False):
if self.singleton == False:
encodings = []
labels = []
for file in self.files:
cascade = cv2.CascadeClassifier(self.model)
image = cv2.imread(file)
rgb = cv2.cvtColor(image, cv2.COLOR_BGR2RGB)
faces = cascade.detectMultiScale(rgb, self.scale_factor, self.min_neighbours)
if len(faces) > 0:
print('Found face in '+file)
encodings.append(faces.flatten())
labels.append(file.split('/')[2])
else:
print('Couldnt find face in '+file)
return encodings, labels
Here are some of the encodings
[204 96 211 211]
[525 168 680 680]
[205 11 269 269]
[ 165 31 316 316 1098 181 179 179]
[ 113 422 1371 1371]
[ 71 86 183 183]
[209 19 33 33 88 27 60 60 133 80 65 65 68 117 52 52]
[117 77 149 149]
[ 63 77 284 284]
[370 222 490 490]
[433 112 114 114 183 98 358 358]
[ 44 35 48 48 192 34 48 48]
[210 82 229 229]
[429 90 153 153]
[318 50 174 174 118 142 120 120]

you should not put several found rects into the same list entry.
if there are many faces found, put each on its own row, and add a label per face found (not per image)
then, what you have now, are NOT "encodings", just mere boxes / rectangles.
read up on how to get real encodings (facenet, spherenet ?), then you need to:
crop the face region fom the image
resize it to the nn input size (e.g. 96x96)
run it through the nn to receive the encoding
save that along with a label to a db/list

Find Minimum Of Function That Takes Matrix As Input

Given a numpy array of shape (64,64) (=an image) and an arbitrary function that takes that image as an input, I want to find the image that minimizes the function. Let's say the function computes the contrast.
Example:
import numpy as np
def contrast(X):
vmin, vmax = int(np.min(X)), int(np.max(X))
num = vmax - vmin
denom = vmax + vmin
if denom == 0:
return 0
else:
return num / denom
img = np.random.randint(256, size=(64,64), dtype=np.uint8)
res = contrast(img)
Scipy offers fmin(), but that function would not work with such a large input. Any ideas how to find
the image that minimizes the function?

Run the code in google colab.
It is by no means perfect, but you can at least get close to a local minimum¹ with a simple gradient descent optimization and automatic differentiation in e.g. autograd. In order for the automatic gradient to work, you most likely will have to convert the image data to floats, do the optimization, and subsequently convert and cast back to ints. This might in principle cause you to miss minima or find wrong ones or get stuck in local ones.
1: Note that this in no way guarantees that you find a global minimum in any case where such a thing exists, this will find a minimum.
import autograd.numpy as np
from autograd import elementwise_grad
def michelson_contrast(image):
vmin, vmax = np.min(image), np.max(image)
if (vmax + vmin) > 1e-15:
return (vmax - vmin) / (vmax + vmin)
return 0
For the specific function you specified, the Michelson contrast, the optimization converges extremely slowly,
f = michelson_contrast
df = elementwise_grad(f)
img = np.random.randint(256, size=(100, 100)).astype(np.float64)
# Simple gradient descent.
for i in range(1, (max_iterations := 100000) + 1):
img -= 10**3 * df(img)
# Round and cast the image back to integer values.
img = np.round(img).astype(int)
but a 100 x 100 random test converges on my laptop in about a minute.
iter. function
--------------------------------------
0 1.0000000000
10000 0.6198908490
20000 0.4906918649
30000 0.3968742592
40000 0.3204002330
50000 0.2539835041
60000 0.1942016682
70000 0.1386916909
80000 0.0863448569
90000 0.0361678029
100000 0.0003124169
Rounded back to integers, the answer is an exact minimum with f = 0, but there of course exists many (256 of them to be exact):
[[146 146 146 ... 146 146 146]
[146 146 146 ... 146 146 146]
[146 146 146 ... 146 146 146]
...
[146 146 146 ... 146 146 146]
[146 146 146 ... 146 146 146]
[146 146 146 ... 146 146 146]]
A different example, the RMS contrast, converges much faster (less than a second)
def rms_contrast(image):
N = image.size
image_mean = np.mean(image)
return np.sum((image - image_mean)**2) / N
f = rms_contrast
df = elementwise_grad(f)
img = np.random.randint(256, size=(100, 100)).astype(np.float64)
for i in range(1, (max_iterations := 100) + 1):
img -= 10**3 * df(img)
img = np.round(img).astype(int)
with
iter. function
--------------------------------------
0 5486.3646543900
10 63.2534779216
20 0.7292629494
30 0.0084078294
40 0.0000969357
50 0.0000011176
60 0.0000000129
70 0.0000000001
80 0.0000000000
90 0.0000000000
100 0.0000000000
and resulting image (again a perfect minimum after casting back to integers).
[[126 126 126 ... 126 126 126]
[126 126 126 ... 126 126 126]
[126 126 126 ... 126 126 126]
...
[126 126 126 ... 126 126 126]
[126 126 126 ... 126 126 126]
[126 126 126 ... 126 126 126]]
Unless the function is very complicated or computationally expensive, or the input image is enormous, this approach should at least get you somewhat closer to your answer.

Why is Python automatically removing negative values from image array?

I am trying to apply the below convolve method below on the cameraman image. The kernel applied to the image is a 3x3 filter populated with -1/9. I print the values of the cameraman image before applying the convolve method, and all I get are positive values. Next, when I apply the 3x3 negative kernel on the Image, I still get positive values when I print the values of the cameraman image after convolution.
The convolving function:
def convolve2d(image, kernel):
# This function which takes an image and a kernel
# and returns the convolution of them
# Args:
# image: a numpy array of size [image_height, image_width].
# kernel: a numpy array of size [kernel_height, kernel_width].
# Returns:
# a numpy array of size [image_height, image_width] (convolution output).
output = np.zeros_like(image) # convolution output
# Add zero padding to the input image
padding = int(len(kernel)/2)
image_padded=np.pad(image,((padding,padding),(padding,padding)),'constant')
for x in range(image.shape[1]): # Loop over every pixel of the image
for y in range(image.shape[0]):
# element-wise multiplication of the kernel and the image
output[y,x]=(kernel*image_padded[y:y+3,x:x+3]).sum()
return output
And here is the filter I am applying to the image:
filter2= [[-1/9,-1/9,-1/9],[-1/9,-1/9,-1/9],[-1/9,-1/9,-1/9]]
Finally, these are the intial values of the images, and the values after convolution respectively:
[[156 159 158 ... 151 152 152]
[160 154 157 ... 154 155 153]
[156 159 158 ... 151 152 152]
...
[114 132 123 ... 135 137 114]
[121 126 130 ... 133 130 113]
[121 126 130 ... 133 130 113]]
After convolution:
[[187 152 152 ... 154 155 188]
[152 99 99 ... 104 104 155]
[152 99 100 ... 103 103 154]
...
[175 133 131 ... 127 130 174]
[174 132 124 ... 125 130 175]
[202 173 164 ... 172 173 202]]
This is how I call the convolve2d method:
convolved_camManImage= convolve2d(camManImage,filter2)

This might be caused by how numpy dtypes work. As numpy.zeros_like's help says:
Return an array of zeros with the same shape and type as a given
array.
Thus your output might be dtype uint8, which use modulo arithmetics. To check if this is case add print(output.dtype) immediately after output = np.zeros_like(image) line

How to filter clusters produced by DBSCAN based on size?

I have applied DBSCAN to perform clustering on a dataset consisting of X, Y and Z coordinates of each point in a point cloud. I want to plot only the clusters which have less than 100 points. This is what I have so far:
clustering = DBSCAN(eps=0.1, min_samples=20, metric='euclidean').fit(only_xy)
plt.scatter(only_xy[:, 0], only_xy[:, 1],
c=clustering.labels_, cmap='rainbow')
clusters = clustering.components_
#Store the labels
labels = clustering.labels_
#Then get the frequency count of the non-negative labels
counts = np.bincount(labels[labels>=0])
print(counts)
Output:
[1278 564 208 47 36 30 191 54 24 18 40 915 26 20
24 527 56 677 63 57 61 1544 512 21 45 187 39 132
48 55 160 46 28 18 55 48 35 92 29 88 53 55
24 52 114 49 34 34 38 52 38 53 69]
So I have found the number of points in each cluster, but I'm not sure how to select only the clusters which have less than 100 points.

You may find indexes of the labels where you have counts less than 100:
ls, cs = np.unique(labels,return_counts=True)
dic = dict(zip(ls,cs))
idx = [i for i,label in enumerate(labels) if dic[label] <100 and label >= 0]
Then you may apply resulting index to your DBSCAN results and labels like (more or less):
plt.scatter(only_xy[idx, 0], only_xy[idx, 1],
c=clustering.labels_[idx], cmap='rainbow')

I think if you run this code, you can get the labels, and cluster components of the cluster with size more than 100:
from collections import Counter
labels_with_morethan100=[label for (label,count) in Counter(clustering.labels_).items() if count>100]
clusters_biggerthan100= clustering.components_[np.isin(clustering.labels_[clustering.labels_>=0], labels_with_morethan100)]

Speeding up array iteration time in python

Currently I am iterating over one array and for each value in this array I am looking for the closest value at the corresponding point in another array that is within a region surrounding the corresponding point.
In summary: For any point in an array, how far away from a corresponding point in another array do you need to go to get the same value.
The code seems to work well for small arrays, however I am working now with 1024x768 arrays, leading me to wait a long time for each run....
Any help or advice would be greatly appreciated as I have been on this for a while!!
Example matrix in format Im using: np.array[[1,2],[3,4]]
#Distance to agreement
#Used later to define a region of pixels around a corresponding point
#to iterate over:
DTA = 26
#To account for noise in pixels - doesnt have to find the exact value,
#just one within +/-130 of it.
limit = 130
#Containers for all pixel value matches and also the smallest distance
#to pixel match
Dist = []
Dist_min = []
#Continer matrix for gamma pass/fail values
Dist_to_agree = np.zeros((i_size,j_size))
#i,j indexes the reference matrix (x), ii,jj indexes the measured
#matrix(y). Finds a match within the limits,
#appends the distance to the match into Dist.
#Then find the minimum distance to a match for that pixel and append it
#to dist_min
for i, k in enumerate(x):
for j, l in enumerate(k):
#added 10 packing to y matrix, so need to shift it by 10 in i&j
for ii in range((i+10)-DTA,(i+10)+DTA):
for jj in range((j+10)-DTA,(j+10)+DTA):
#If the pixel value is within a range to account for noise,
#let it be "found"
if (y[ii,jj]-limit) <= x[i,j] <= (y[ii,jj]+limit):
#Calculating distance
dist_eu = sqrt(((i)-(ii))**2 + ((j) - (jj))**2)
Dist.append(dist_eu)
#If a value cannot be found within the noise range,
#append 10 = instant fail.
else:
Dist.append(10)
try:
Dist_min.append(min(Dist))
Dist_to_agree[i,j] = min(Dist)
except ValueError:
pass
#Need to reset container or previous values will also be
#accounted for when finding minimum
Dist = []
print Dist_to_agree

First, you are getting the elements of x in k and l, but then throwing that away and indexing x again. So in place of x[i,j], you could just use l, which would be much faster (although l isn't a very meaningful name, something like xi and xij might be better).
Second, you are recomputing y[ii,jj]-limit and y[ii,jj]+limitevery time. If you have enough memory, you can-precomputer these:ym = y-limitandyp = y+limit`.
Third, appending to a list is slower than creating an array and setting the values for long lists vs. long arrays. You can also skip the entire else clause by pre-setting the default value.
Fourth, you are computing min(dist) twice, and further may be using the python version rather than the numpy version, the latter being faster for arrays (which is another reason to make dist and array).
However, the biggest speedup would be to vectorize the inner two loops. Here is my tests, with x=np.random.random((10,10)) and y=np.random.random((100,100)):
Your version takes 623 ms.
Here is my version, which takes 7.6 ms:
dta = 26
limit = 130
dist_to_agree = np.zeros_like(x)
dist_min = []
ym = y-limit
yp = y+limit
for i, xi in enumerate(x):
irange = (i-np.arange(i+10-dta, i+10+dta))**2
if not irange.size:
continue
ymi = ym[i+10-dta:i+10+dta, :]
ypi = yp[i+10-dta:i+10+dta, :]
for j, xij in enumerate(xi):
jrange = (j-np.arange(j+10-dta, j+10+dta))**2
if not jrange.size:
continue
ymij = ymi[:, j+10-dta:j+10+dta]
ypij = ypi[:, j+10-dta:j+10+dta]
imesh, jmesh = np.meshgrid(irange, jrange, indexing='ij')
dist = np.sqrt(imesh+jmesh)
dist[ymij > xij or xij < ypij] = 10
mindist = dist.min()
dist_min.append(mindist)
dist_to_agree[i,j] = mindist
print(dist_to_agree)

#Ciaran
Meshgrid is kinda a vectorized equivalent of two nested loops. Below are two equivalent ways of calculating the dist. One with loops and one with meshgrid+numpy vector operations. The second one is six times faster.
DTA=5
i=100
j=200
def func1():
dist1=np.zeros((DTA*2,DTA*2))
for ii in range((i+10)-DTA,(i+10)+DTA):
for jj in range((j+10)-DTA,(j+10)+DTA):
dist1[ii-((i+10)-DTA),jj-((j+10)-DTA)] =sqrt(((i)-(ii))**2 + ((j) - (jj))**2)
return dist1
def func2():
dist2=np.zeros((DTA*2,DTA*2))
ii, jj = meshgrid(np.arange((i+10)-DTA,(i+10)+DTA),
np.arange((j+10)-DTA,(j+10)+DTA))
dist2=np.sqrt((i-ii)**2+(j-jj)**2)
return dist2
This is how ii and jj matrices look after meshgrid operation
ii=
[[105 106 107 108 109 110 111 112 113 114]
[105 106 107 108 109 110 111 112 113 114]
[105 106 107 108 109 110 111 112 113 114]
[105 106 107 108 109 110 111 112 113 114]
[105 106 107 108 109 110 111 112 113 114]
[105 106 107 108 109 110 111 112 113 114]
[105 106 107 108 109 110 111 112 113 114]
[105 106 107 108 109 110 111 112 113 114]
[105 106 107 108 109 110 111 112 113 114]
[105 106 107 108 109 110 111 112 113 114]]
jj=
[[205 205 205 205 205 205 205 205 205 205]
[206 206 206 206 206 206 206 206 206 206]
[207 207 207 207 207 207 207 207 207 207]
[208 208 208 208 208 208 208 208 208 208]
[209 209 209 209 209 209 209 209 209 209]
[210 210 210 210 210 210 210 210 210 210]
[211 211 211 211 211 211 211 211 211 211]
[212 212 212 212 212 212 212 212 212 212]
[213 213 213 213 213 213 213 213 213 213]
[214 214 214 214 214 214 214 214 214 214]]

for loops are very slow in pure python and you have four nested loops which will be very slow. Cython does wonders to the for loop speed. You can also try vectorization. While I'm not sure I fully understand what you are trying to do, you may try to vectorize at last some of the operations. Especially the last two loops.
So instead of two ii,jj cycles over
y[ii,jj]-limit) <= x[i,j] <= (y[ii,jj]+limit)
you can do something like
ii, jj = meshgrid(np.arange((i+10)-DTA,(i+10)+DTA), np.arange((j+10)-DTA,(j+10)+DTA))
t=(y[(i+10)-DTA,(i+10)+DTA]-limit>=x[i,j]) & (y[(i+10)-DTA,(i+10)+DTA]+limit<=x[i,j])
Dist=np.sqrt((i-ii)**2)+(j-jj)**2))
np.min(Dist[t]) will have your minimum distance for element i,j

The numbapro compiler offers gpu Acceleration. Unfortunately it isn't free.
http://docs.continuum.io/numbapro/