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python PCA dimensionality reduction

i'm learning using PCA to finish dimensionality reduction (Python3.6) but i've got very similar but different results when using different methods here's my code
from numpy import *
from sklearn.decomposition import PCA
data_set = [[-1., -2.],
[-1., 0.],
[0., 0.],
[2., 1.],
[0., 1.]]
# 1
pca_sk = PCA(n_components=1)
newmat = pca_sk.fit_transform(data_set)
print(newmat)
# 2
meanVals = mean(data_set, axis=0)
meanRemoved = data_set - meanVals
covMat = cov(meanRemoved, rowvar=0)
eigVals, eigVects = linalg.eig(mat(covMat))
eigValInd = argsort(eigVals)
eigValInd = eigValInd[:-(1 + 1):-1]
redEigVects = eigVects[:, eigValInd]
lowDDataMat = meanRemoved * redEigVects
print(lowDDataMat)
the first one output
[[ 2.12132034]
[ 0.70710678]
[-0. ]
[-2.12132034]
[-0.70710678]]
but anothor output
[[-2.12132034]
[-0.70710678]
[ 0. ]
[ 2.12132034]
[ 0.70710678]]
why dose it happen

Output of projectPoints() function

i used projectPoints() function of OpenCV to do projection from world coordinates to pixel coordinates, the output image points are vector point2f, how can I extract the x,y coordinates from imagePoints?
seconde question, some of the imagePoints are negative like below: I just project 2 points and these are the results
[[[-37.95361728 316.5438248 ]]
[[204.89090594 316.5144533 ]]]
if I show these coordinates without the negative sign on the image it is correct
first why i get a negative sign and how can i solve these issues?
i appreciate any help, thanks
this is my code :
import cv2
import numpy as np
objectPoints = np.array([[ -0.8565132125748637 , 0.18200966481269648 , 0.9606457931958912 ],[-0.2565132125748638 , 0.18200966481269648 , 0.9606457931958912]] , np.float)
tvec = np.matrix([[-0.00016514 ],[ 0.00523247 ], [-0.00371881]])
rvec = np.matrix([[0.99987256 , -0.00294761 , -0.01569025],[0.00261951 , 0.99977833 , -0.02089080],[0.01574835 , 0.02084704 , 0.99965864]])
cameraMatrix = np.matrix([[ 381.58892822265625 , 0 , 313.5216979980469 ],[ 0, 381.1356201171875 , 250.1746826171875],[ 0 , 0 , 1 ]])
distCoeffs = np.array([0, 0, 0, 0, 0], np.float)
imagePoints, jacobian = cv2.projectPoints( objectPoints, rvec, tvec, cameraMatrix, distCoeffs )
print(imagePoints)

How to vectorize a 'for' loop which calls a function (that takes a 2-Dimensional array as argument) over a 3-Dimensional numpy array

I have a numpy array containing the XYZ coordinates of the k-neighboors (k=10) points from a point cloud:
k_neighboors
Out[53]:
array([[[ 2.51508147e-01, 5.60274944e-02, 1.98303187e+00],
[ 2.48552352e-01, 5.95569573e-02, 1.98319519e+00],
[ 2.56611764e-01, 5.36767729e-02, 1.98236740e+00],
...,
[ 2.54520357e-01, 6.23480231e-02, 1.98255634e+00],
[ 2.57603496e-01, 5.19787706e-02, 1.98221457e+00],
[ 2.43914440e-01, 5.68424985e-02, 1.98352253e+00]],
[[ 9.72352773e-02, 2.06699912e-02, 1.99344850e+00],
[ 9.91205871e-02, 2.36056261e-02, 1.99329960e+00],
[ 9.59625840e-02, 1.71508361e-02, 1.99356234e+00],
...,
[ 1.03216261e-01, 2.19752081e-02, 1.99304521e+00],
[ 9.65025574e-02, 1.44127617e-02, 1.99355054e+00],
[ 9.59930867e-02, 2.72080526e-02, 1.99344873e+00]],
[[ 1.76408485e-01, 2.81930678e-02, 1.98819435e+00],
[ 1.78670138e-01, 2.81904750e-02, 1.98804617e+00],
[ 1.80372953e-01, 3.05109434e-02, 1.98791444e+00],
...,
[ 1.81960404e-01, 2.47725621e-02, 1.98785996e+00],
[ 1.74499243e-01, 3.50728296e-02, 1.98826015e+00],
[ 1.83470801e-01, 2.70808022e-02, 1.98774099e+00]],
...,
[[ 1.78178743e-01, -4.60980982e-02, -1.98792374e+00],
[ 1.77953839e-01, -4.73701134e-02, -1.98792756e+00],
[ 1.77889392e-01, -4.75468598e-02, -1.98793030e+00],
...,
[ 1.79924294e-01, -5.08776568e-02, -1.98772371e+00],
[ 1.76720902e-01, -5.11409082e-02, -1.98791265e+00],
[ 1.83644593e-01, -4.64747548e-02, -1.98756230e+00]],
[[ 2.00245917e-01, -2.33091787e-03, -1.98685515e+00],
[ 2.02384919e-01, -5.60011715e-04, -1.98673022e+00],
[ 1.97325528e-01, -1.03301927e-03, -1.98705769e+00],
...,
[ 1.95464164e-01, -6.23105839e-03, -1.98713481e+00],
[ 1.98985338e-01, -8.39920342e-03, -1.98688531e+00],
[ 1.95959195e-01, 2.68006674e-03, -1.98713303e+00]],
[[ 1.28851235e-01, -3.24527062e-02, -1.99127460e+00],
[ 1.26415789e-01, -3.27731185e-02, -1.99143147e+00],
[ 1.25985757e-01, -3.24910432e-02, -1.99146211e+00],
...,
[ 1.28296465e-01, -3.92388329e-02, -1.99117136e+00],
[ 1.34895295e-01, -3.64872888e-02, -1.99083793e+00],
[ 1.29047096e-01, -3.97952795e-02, -1.99111152e+00]]])
With this shape:
k_neighboors.shape
Out[54]: (2999986, 10, 3)
And I have this function which applies a Principal Component Analysis to some data provided as 2-Dimensional array:
def PCA(data, correlation=False, sort=True):
""" Applies Principal Component Analysis to the data
Parameters
----------
data: array
The array containing the data. The array must have NxM dimensions, where each
of the N rows represents a different individual record and each of the M columns
represents a different variable recorded for that individual record.
array([
[V11, ... , V1m],
...,
[Vn1, ... , Vnm]])
correlation(Optional) : bool
Set the type of matrix to be computed (see Notes):
If True compute the correlation matrix.
If False(Default) compute the covariance matrix.
sort(Optional) : bool
Set the order that the eigenvalues/vectors will have
If True(Default) they will be sorted (from higher value to less).
If False they won't.
Returns
-------
eigenvalues: (1,M) array
The eigenvalues of the corresponding matrix.
eigenvector: (M,M) array
The eigenvectors of the corresponding matrix.
Notes
-----
The correlation matrix is a better choice when there are different magnitudes
representing the M variables. Use covariance matrix in any other case.
"""
#: get the mean of all variables
mean = np.mean(data, axis=0, dtype=np.float64)
#: adjust the data by substracting the mean to each variable
data_adjust = data - mean
#: compute the covariance/correlation matrix
#: the data is transposed due to np.cov/corrcoef sintaxis
if correlation:
matrix = np.corrcoef(data_adjust.T)
else:
matrix = np.cov(data_adjust.T)
#: get the eigenvalues and eigenvectors
eigenvalues, eigenvectors = np.linalg.eig(matrix)
if sort:
#: sort eigenvalues and eigenvectors
sort = eigenvalues.argsort()[::-1]
eigenvalues = eigenvalues[sort]
eigenvectors = eigenvectors[:,sort]
return eigenvalues, eigenvectors
So the question is: how can I apply the PCA function mentioned above over each of the 2999986 10x3 arrays in a way that doesn't take for ever like this one:
data = np.empty((2999986, 3))
for i in range(len(k_neighboors)):
w, v = PCA(k_neighboors[i])
data[i] = v[:,2]
break #: I break the loop in order to don't have to wait for ever.
data
Out[64]:
array([[ 0.10530792, 0.01028906, 0.99438643],
[ 0. , 0. , 0. ],
[ 0. , 0. , 0. ],
...,
[ 0. , 0. , 0. ],
[ 0. , 0. , 0. ],
[ 0. , 0. , 0. ]])

Thanks to #Divakar and #Eelco comments.
Using the function that Divakar post on this answer
def vectorized_app(data):
diffs = data - data.mean(1,keepdims=True)
return np.einsum('ijk,ijl->ikl',diffs,diffs)/data.shape[1]
And using what Eelco pointed on his comment, I end up with this.
k_neighboors.shape
Out[48]: (2999986, 10, 3)
#: THE (ASSUMED)VECTORIZED ANSWER
data = np.linalg.eig(vectorized_app(k_neighboors))[1][:,:,2]
data
Out[50]:
array([[ 0.10530792, 0.01028906, 0.99438643],
[ 0.06462 , 0.00944352, 0.99786526],
[ 0.0654035 , 0.00860751, 0.99782177],
...,
[-0.0632175 , 0.01613551, 0.99786933],
[-0.06449399, 0.00552943, 0.99790278],
[-0.06081954, 0.01802078, 0.99798609]])
Wich gives the same results as the for loop, without taking forever (althought still takes a while):
data2 = np.empty((2999986, 3))
for i in range(len(k_neighboors)):
if i > 10:
break #: I break the loop in order to don't have to wait for ever.
w, v = PCA(k_neighboors[i])
data2[i] = v[:,2]
data2
Out[52]:
array([[ 0.10530792, 0.01028906, 0.99438643],
[ 0.06462 , 0.00944352, 0.99786526],
[ 0.0654035 , 0.00860751, 0.99782177],
...,
[ 0. , 0. , 0. ],
[ 0. , 0. , 0. ],
[ 0. , 0. , 0. ]])
I don't know if there could be a better way to do this, so I'm going to keep the question open.

Implementing gradient operator in Python

I'm working on a Computer Vision system and this is giving me a serious headache. I'm having trouble re-implementing an old gradient operator more efficiently, I'm working with numpy and openCV2.
This is what I had:
def gradientX(img):
rows, cols = img.shape
out = np.zeros((rows,cols))
for y in range(rows-1):
Mr = img[y]
Or = out[y]
Or[0] = Mr[1] - Mr[0]
for x in xrange(1, cols - 2):
Or[x] = (Mr[x+1] - Mr[x-1])/2.0
Or[cols-1] = Mr[cols-1] - Mr[cols-2]
return out
def gradient(img):
return [gradientX(img), (gradientX(img.T).T)]
I've tried using numpy's gradient operator but the result is not the same
For this input
array([[ 3, 4, 5],
[255, 0, 12],
[ 25, 15, 200]])
Using my gradient returns
[array([[ 1., 0., 1.],
[-255., 0., 12.],
[ 0., 0., 0.]]),
array([[ 252., -4., 0.],
[ 0., 0., 0.],
[-230., 15., 0.]])]
While using numpy's np.gradient returns
[array([[ 252. , -4. , 7. ],
[ 11. , 5.5, 97.5],
[-230. , 15. , 188. ]]),
array([[ 1. , 1. , 1. ],
[-255. , -121.5, 12. ],
[ -10. , 87.5, 185. ]])]
There are cleary some similarities between the results but they're definitely not the same. So I'm missing something here or the two operators aren't mean to produce the same results. In that case, I wanted to know how to re-implement my gradientX function so it doesn't use that awful looking double loop for traversing the 2-d array using mostly numpy's potency.

I've been working a bit more on this just to find that my mistake.
I was skipping last row and last column when iterating. As #wflynny noted, the result was identical except for a row and a column of zeros.
Provided this, the result could not be the same as np.gradient, but with that change, the results are identical, so there's no need to find any other numpy implementation for this.
Answering my own question, a good numpy's implementation for my gradient algorithm would be
import numpy as np
def gradientX(img):
return np.gradient(img)[::-1]
I'm also posting the working code, just because it shows how numpy's gradient operator works
def computeMatXGradient(img):
rows, cols = img.shape
out = np.zeros((rows,cols))
for y in range(rows):
Mr = img[y]
Or = out[y]
Or[0] = float(Mr[1]) - float(Mr[0])
for x in xrange(1, cols - 1):
Or[x] = (float(Mr[x+1]) - float(Mr[x-1]))/2.0
Or[cols-1] = float(Mr[cols-1]) - float(Mr[cols-2])
return out

Numpy: Avoiding nested loops to operate on matrix-valued images

I am a beginner at python and numpy and I need to compute the matrix logarithm for each "pixel" (i.e. x,y position) of a matrix-valued image of dimension NxMx3x3. 3x3 is the dimensions of the matrix at each pixel.
The function I have written so far is the following:
def logm_img(im):
from scipy import linalg
dimx = im.shape[0]
dimy = im.shape[1]
res = zeros_like(im)
for x in range(dimx):
for y in range(dimy):
res[x, y, :, :] = linalg.logm(asmatrix(im[x,y,:,:]))
return res
Is it ok?
Is there a way to avoid the two nested loops ?

Numpy can do that. Just call numpy.log:
>>> import numpy
>>> a = numpy.array(range(100)).reshape(10, 10)
>>> b = numpy.log(a)
__main__:1: RuntimeWarning: divide by zero encountered in log
>>> b
array([[ -inf, 0. , 0.69314718, 1.09861229, 1.38629436,
1.60943791, 1.79175947, 1.94591015, 2.07944154, 2.19722458],
[ 2.30258509, 2.39789527, 2.48490665, 2.56494936, 2.63905733,
2.7080502 , 2.77258872, 2.83321334, 2.89037176, 2.94443898],
[ 2.99573227, 3.04452244, 3.09104245, 3.13549422, 3.17805383,
3.21887582, 3.25809654, 3.29583687, 3.33220451, 3.36729583],
[ 3.40119738, 3.4339872 , 3.4657359 , 3.49650756, 3.52636052,
3.55534806, 3.58351894, 3.61091791, 3.63758616, 3.66356165],
[ 3.68887945, 3.71357207, 3.73766962, 3.76120012, 3.78418963,
3.80666249, 3.8286414 , 3.8501476 , 3.87120101, 3.8918203 ],
[ 3.91202301, 3.93182563, 3.95124372, 3.97029191, 3.98898405,
4.00733319, 4.02535169, 4.04305127, 4.06044301, 4.07753744],
[ 4.09434456, 4.11087386, 4.12713439, 4.14313473, 4.15888308,
4.17438727, 4.18965474, 4.20469262, 4.21950771, 4.2341065 ],
[ 4.24849524, 4.26267988, 4.27666612, 4.29045944, 4.30406509,
4.31748811, 4.33073334, 4.34380542, 4.35670883, 4.36944785],
[ 4.38202663, 4.39444915, 4.40671925, 4.41884061, 4.4308168 ,
4.44265126, 4.4543473 , 4.46590812, 4.47733681, 4.48863637],
[ 4.49980967, 4.51085951, 4.52178858, 4.53259949, 4.54329478,
4.55387689, 4.56434819, 4.57471098, 4.58496748, 4.59511985]])