Related
I want to compute the gradient (direction and magnitude) of an overdefined plane (> 3 points), such as e.g. four x, y, z coordinates:
[0, 0, 1], [1, 0, 0], [0, 1, 1], [1, 1, 0]
My code for computing the gradient looks like this (using singular value decomposition from this post, modified):
import numpy as np
def regr(X):
y = np.average(X, axis=0)
Xm = X - y
cov = (1./X.shape[0])*np.matmul(Xm.T,Xm) # Covariance
u, s, v = np.linalg.svd(cov) # Singular Value Decomposition
return u[:,1]
However as a result I get:
[0, 1, 0]
which does not represent the gradient nor the normal vector. The result should look like this:
[sqrt(2)/2, 0, -sqrt(2)/2]
Any ideas why I am getting a wrong vector?
You can use the function numpy.gradient for that. Here is some math background how it works.
In short:
The directional derivative will be the dot product of the gradient with the (unit normalized) vector of the direction.
An easier solution is to use numdifftools, especially the convenience function directionaldiff.
An example can be found here. And if you look at the source code you can see the relation to the math mentioned above.
I am trying to get an average value for parameters to then plot with a given function. I think I have to somehow fill a 3-column array and then take the average of values of that array. I want to create 1000 values for popt[0] , popt[1] , and popt[2] and then take the average of all those values and then plot them.
for n in range(0,1000):
params=np.zeros(3,1000)
y3=y2+np.random.normal(loc=0.0,scale=0.1*y2)
popt,pcov=optimize.curve_fit(fluxmeasureMW,bands,y3)
params.append(popt[0],popt[1],popt[2])
a_avg=st.mean(params[0:])
b_avg=st.mean(params[1:])
e_avg=st.mean(params[2:])
The final goal is to plot:
fluxmeasureMW(bands,a_avg,b_avg,e_avg)
I am just not sure how to iterate the fitting function to then output 1000 values. 1000 is arbitrary, I just want a good sample size. The values for y2 and bands are already defined and can be plotted without issue, as well as the function fluxmeasureMW.
Say your function is like this
def fluxmeasureMW(x,f,g,h):
return result_of_calc
Just run the fit in a loop; accumulate the popts in a list then take the mean
from scipy import optimize
import numpy as np
n = 1000
t = []
for i in range(n):
y3 = y2 + np.random.normal(loc=0.0,scale=.1*y2)
popt,pcov = optimize.curve_fit(fluxmeasureMW,bands,y3)
t.append(popt)
f,g,h = np.mean(t,0)
t will be a list of lists...
[[f,g,h],
[f,g,h],
...]
np.mean(t,0) will average the values over the columns.
You could also use
import statistics
a = [[0, 1, 2],
[1, 2, 3],
[2, 3, 4],
[3, 4, 5]]
for column in zip(*a):
#print(column)
print(statistics.mean(column))
I'm working on an optimization problem, but to avoid getting into the details, I'm going to provide a simple example of a bug that's been giving me headaches for a few days.
Say I have a 2D numpy array with observed x-y coordinates:
from scipy.optimize import distance
x = np.array([1,2], [2,3], [4,5], [5,6])
I also have a list of x-y coordinates to compare to these points (y):
y = np.array([11,13], [12, 14])
I have a function that takes the sum of manhattan differences between a value of x and all of the values in y:
def find_sum(ref_row, comp_rows):
modeled_counts = []
y = ref_row * len(comp_rows)
res = list(map(distance.cityblock, ref_row, comp_rows))
modeled_counts.append(sum(res))
return sum(modeled_counts)
Essentially, what I would like to do is find the sum of manhattan distances for every item in y with each item in x (so basically for each item in x, find the sum of the Manhattan distances between that (x,y) pair and every (x,y) pair in y).
I've tried this out with the following line of code:
z = list(map(find_sum, x, y))
However, z is of length 2 (like y), and not 4 like x. Is there a way to ensure that z is the result of consecutive one-to-all calculations? That is, I'd like to calculate the sum of all of the manhattan differences between x[0] and every set in y, and so on and so forth, so the length of z should be equal to the length of x.
Is there a simple way to do this without a for loop? My data is rather large (~ 4 million rows), so I'd really appreciate fast solutions. I'm fairly new to Python programming, so any explanations about why the solution works and is fast would be appreciated as well, but definitely isn't required!
Thanks!
This solution implements the distance in numpy, as I think it is a good example of broadcasting, which is a very useful thing to know if you need to use arrays and matrices.
By definition of Manhattan distance, you need to evaluate the sum of the absolute value of difference between each column. However, the first column of x, x[:, 0], has shape (4,) and the first column of y, y[:, 0], has shape (2,), so they are not compatible in the sense of applying subtraction: the broadcasting property says that each shape is compared starting with the trailing dimensions and two dimensions are compatible when they are equal or one of them is 1. Sadly, none of them are true for your columns.
However, you can add a new dimension of value 1 using np.newaxis, so
x[:, 0]
is array([1, 2, 4, 5]), but
x[:, 0, np.newaxis]
is
array([[1],
[2],
[4],
[5]])
and its shape is (4 ,1). Now, a matrix of shape (4, 1) subtracted by an array of shape 2 results in a matrix of shape (4, 2), by numpy's broadcasting treatment:
4 x 1
2
= 4 x 2
You can obtain the differences for each column:
first_column_difference = x[:, 0, np.newaxis] - y[:, 0]
second_column_difference = x[:, 1, np.newaxis] - y[:, 1]
and evaluate the sum of their absolute values:
np.abs(first_column_difference) + np.abs(second_column_difference)
which results in a (4, 2) matrix. Now, you want to sum the values for each row, so that you have 4 values:
np.sum(np.abs(first_column_difference) + np.abs(second_column_difference), axis=1)
which results in array([73, 69, 61, 57]). The rule is simple: the parameter axis will eliminate that dimension from the result, therefore using axis=1 for a (4, 2) matrix generates 4 values -- if you use axis=0, it will generate 2 values.
So, this will solve your problem:
x = np.array([[1, 2], [2, 3], [4, 5], [5, 6]])
y = np.array([[11, 13], [12, 43]])
first_column_difference = x[:, 0, np.newaxis] - y[:, 0]
second_column_difference = x[:, 1, np.newaxis] - y[:, 1]
z = np.abs(first_column_difference) + np.abs(second_column_difference)
print(np.sum(z, axis=1))
You can also skip the intermediate steps for each column and evaluate everything at once (it is a little bit harder to understand, so I prefer the method described above to explain what is happening):
print(np.abs(x[:, np.newaxis] - y).sum(axis=(1, 2)))
It is a general case for an n-dimensional Manhattan distance: if x is (u, n) and y is (v, n), it generates u rows by broadcasting (u, 1, n) by (v, n) = (u, v, n), then applying sum to eliminate the second and third axis.
Here is how you can do it using numpy broadcast with simplified explanation
Adjust Shape For Broadcasting
import numpy as np
start_points = np.array([[1,2], [2,3], [4,5], [5,6]])
dest_points = np.array([[11,13], [12, 14]])
## using np.newaxis as index add a new dimension at that position
## : give all the elements on that dimension
start_points = start_points[np.newaxis, :, :]
dest_points = dest_points[:, np.newaxis, :]
## Now lets check he shape of the point arrays
print('start_points.shape: ', start_points.shape) # (1, 4, 2)
print('dest_points.shape', dest_points.shape) # (2, 1, 2)
Lets try to understand
last element of shape represent x and y of a point, size 2
we can think of start_points as having 1 row and 4 columns of points
we can think of dest_points as having 2 rows and 1 columns of points
We can think start_points and dest_points as matrix or a table of points of size (1X4) and (2X1)
We clearly see that size are not compatible. What will happen if we perform arithmatic
operation between them? Here is where a smart part of numpy comes, called broadcast.
It will repeat rows of start_points to match that of dest_point making matrix of (2X4)
It will repeat columns of dest_point to match that of start_points making matrix of (2X4)
Result is arithmetic operation between every pair of elements on start_points and dest_points
Calculate the distance
diff_x_y = start_points - dest_points
print(diff_x_y.shape) # (2, 4, 2)
abs_diff_x_y = np.abs(start_points - dest_points)
man_distance = np.sum(abs_diff_x_y, axis=2)
print('man_distance:\n', man_distance)
sum_distance = np.sum(man_distance, axis=0)
print('sum_distance:\n', sum_distance)
Oneliner
start_points = np.array([[1,2], [2,3], [4,5], [5,6]])
dest_points = np.array([[11,13], [12, 14]])
np.sum(np.abs(start_points[np.newaxis, :, :] - dest_points[:, np.newaxis, :]), axis=(0,2))
Here is more detail explanation of broadcasting if you want to understand it more
With so many rows you can make substantial savings by using a smart algorithm. Let us for simplicity assume there is just one dimension; once we have established the algorithm, getting back to the general case is a simple matter of summing over coordinates.
The naive algorithm is O(mn) where m,n are the sizes of sets X,Y. Our algorithm is O((m+n)log(m+n)) so it scales much better.
We first have to sort the union of X and Y by coordinate and then form the cumsum over Y. Next, we find for each x in X the number YbefX of y in Y to its left and use it to look up the corresponding cumsum item YbefXval. The summed distances to all y to the left of x are YbefX times coordinate of x minus YbefXval, the distances to all y to the right are sum of all y coordinates minus YbefXval minus n - YbefX times coordinate of x.
Where does the saving come from? Sorting coordinates enables us to recycle the summations we have done before, instead of starting each time from scratch. This uses the fact that up to a sign we always sum the same y coordinates and going from left to right the signs flip one by one.
Code:
import numpy as np
from scipy.spatial.distance import cdist
from timeit import timeit
def pp(X,Y):
(m,k),(n,k) = X.shape,Y.shape
XY = np.concatenate([X.T,Y.T],1)
idx = XY.argsort(1)
Xmsk = idx<m
Ymsk = ~Xmsk
Xidx = np.arange(k)[:,None],idx[Xmsk].reshape(k,m)
Yidx = np.arange(k)[:,None],idx[Ymsk].reshape(k,n)
YbefX = Ymsk.cumsum(1)[Xmsk].reshape(k,m)
YbefXval = XY[Yidx].cumsum(1)[np.arange(k)[:,None],YbefX-1]
YbefXval[YbefX==0] = 0
XY[Xidx] = ((2*YbefX-n)*XY[Xidx]) - 2*YbefXval + Y.sum(0)[:,None]
return XY[:,:m].sum(0)
def summed_cdist(X,Y):
return cdist(X,Y,"minkowski",p=1).sum(1)
# demo
m,n,k = 1000,500,10
X,Y = np.random.randn(m,k),np.random.randn(n,k)
print("same result:",np.allclose(pp(X,Y),summed_cdist(X,Y)))
print("sort :",timeit(lambda:pp(X,Y),number=1000),"ms")
print("scipy cdist:",timeit(lambda:summed_cdist(X,Y),number=100)*10,"ms")
Sample run, comparing smart algo "sort" to naive algo implemented using cdist library function:
same result: True
sort : 1.4447695480193943 ms
scipy cdist: 36.41934019047767 ms
Here is the code.
from sklearn.neighbors import NearestNeighbors
import numpy as np
X = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])
nbrs = NearestNeighbors(n_neighbors=2, algorithm='ball_tree').fit(X)
distances, indices = nbrs.kneighbors(X)
>indices
>array([[0, 1],[1, 0],[2, 1],[3, 4],[4, 3],[5, 4]])
>distances
>array([[0. , 1. ],[0. , 1. ],[0. , 1.41421356], [0. , 1. ],[0. , 1. ],[0. , 1.41421356]])
I don't really understand the shape of 'indices' and 'distances'. How do I understand what these numbers mean?
Its pretty straightforward actually. For each data sample in the input to kneighbors() (X here), it will show 2 neighbors. (Because you have specified n_neighbors=2. The indices will give you the index of training data (again X here) and distances will give you the distance for the corresponding data point in training data (to which the indices are referring).
Take an example of single data point. Assuming X[0] as the first query point, the answer will be indices[0] and distances[0]
So for X[0],
the index of first nearest neighbor in training data is indices[0, 0] = 0 and distance is distances[0, 0] = 0. You can use this index value to get the actual data sample from the training data.
This makes sense, because you used the same data for training and testing, so the first nearest neighbor for each point is itself and the distance is 0.
the index of second nearest neigbor is indices[0, 1] = 1 and distance is distances[0, 1] = 1
Similarly for all other points. The first dimension in indices and distances correspond to the query points and second dimension to the number of neighbors asked.
Maybe a little sketch will help
As an example, the closest point to the training sample with index 0 is 1, and since you are using n_neighbors = 2 (two neighbors) you would expect to see this pair in the results. And indeed you see that the pair [0, 1] appears in the output.
I will comment to the aforementioned, how you can get the "n_neighbors=2" neighbors using the indices array, in a pandas dataframe. So,
import pandas as pd
df = pd.DataFrame([X.iloc[indices[row,col]] for row in range(indices.shape[0]) for col in range(indices.shape[1])])
How to identify the linearly independent rows from a matrix? For instance,
The 4th rows is independent.
First, your 3rd row is linearly dependent with 1t and 2nd row. However, your 1st and 4th column are linearly dependent.
Two methods you could use:
Eigenvalue
If one eigenvalue of the matrix is zero, its corresponding eigenvector is linearly dependent. The documentation eig states the returned eigenvalues are repeated according to their multiplicity and not necessarily ordered. However, assuming the eigenvalues correspond to your row vectors, one method would be:
import numpy as np
matrix = np.array(
[
[0, 1 ,0 ,0],
[0, 0, 1, 0],
[0, 1, 1, 0],
[1, 0, 0, 1]
])
lambdas, V = np.linalg.eig(matrix.T)
# The linearly dependent row vectors
print matrix[lambdas == 0,:]
Cauchy-Schwarz inequality
To test linear dependence of vectors and figure out which ones, you could use the Cauchy-Schwarz inequality. Basically, if the inner product of the vectors is equal to the product of the norm of the vectors, the vectors are linearly dependent. Here is an example for the columns:
import numpy as np
matrix = np.array(
[
[0, 1 ,0 ,0],
[0, 0, 1, 0],
[0, 1, 1, 0],
[1, 0, 0, 1]
])
print np.linalg.det(matrix)
for i in range(matrix.shape[0]):
for j in range(matrix.shape[0]):
if i != j:
inner_product = np.inner(
matrix[:,i],
matrix[:,j]
)
norm_i = np.linalg.norm(matrix[:,i])
norm_j = np.linalg.norm(matrix[:,j])
print 'I: ', matrix[:,i]
print 'J: ', matrix[:,j]
print 'Prod: ', inner_product
print 'Norm i: ', norm_i
print 'Norm j: ', norm_j
if np.abs(inner_product - norm_j * norm_i) < 1E-5:
print 'Dependent'
else:
print 'Independent'
To test the rows is a similar approach.
Then you could extend this to test all combinations of vectors, but I imagine this solution scale badly with size.
With sympy you can find the linear independant rows using: sympy.Matrix.rref:
>>> import sympy
>>> import numpy as np
>>> mat = np.array([[0,1,0,0],[0,0,1,0],[0,1,1,0],[1,0,0,1]]) # your matrix
>>> _, inds = sympy.Matrix(mat).T.rref() # to check the rows you need to transpose!
>>> inds
[0, 1, 3]
Which basically tells you the rows 0, 1 and 3 are linear independant while row 2 isn't (it's a linear combination of row 0 and 1).
Then you could remove these rows with slicing:
>>> mat[inds]
array([[0, 1, 0, 0],
[0, 0, 1, 0],
[1, 0, 0, 1]])
This also works well for rectangular (not only for quadratic) matrices.
I edited the code for Cauchy-Schwartz inequality which scales better with dimension: the inputs are the matrix and its dimension, while the output is a new rectangular matrix which contains along its rows the linearly independent columns of the starting matrix. This works in the assumption that the first column in never null, but can be readily generalized in order to implement this case too. Another thing that I observed is that 1e-5 seems to be a "sloppy" threshold, since some particular pathologic vectors were found to be linearly dependent in that case: 1e-4 doesn't give me the same problems. I hope this could be of some help: it was pretty difficult for me to find a really working routine to extract li vectors, and so I'm willing to share mine. If you find some bug, please report them!!
from numpy import dot, zeros
from numpy.linalg import matrix_rank, norm
def find_li_vectors(dim, R):
r = matrix_rank(R)
index = zeros( r ) #this will save the positions of the li columns in the matrix
counter = 0
index[0] = 0 #without loss of generality we pick the first column as linearly independent
j = 0 #therefore the second index is simply 0
for i in range(R.shape[0]): #loop over the columns
if i != j: #if the two columns are not the same
inner_product = dot( R[:,i], R[:,j] ) #compute the scalar product
norm_i = norm(R[:,i]) #compute norms
norm_j = norm(R[:,j])
#inner product and the product of the norms are equal only if the two vectors are parallel
#therefore we are looking for the ones which exhibit a difference which is bigger than a threshold
if absolute(inner_product - norm_j * norm_i) > 1e-4:
counter += 1 #counter is incremented
index[counter] = i #index is saved
j = i #j is refreshed
#do not forget to refresh j: otherwise you would compute only the vectors li with the first column!!
R_independent = zeros((r, dim))
i = 0
#now save everything in a new matrix
while( i < r ):
R_independent[i,:] = R[index[i],:]
i += 1
return R_independent
I know this was asked a while ago, but here is a very simple (although probably inefficient) solution. Given an array, the following finds a set of linearly independent vectors by progressively adding a vector and testing if the rank has increased:
from numpy.linalg import matrix_rank
def LI_vecs(dim,M):
LI=[M[0]]
for i in range(dim):
tmp=[]
for r in LI:
tmp.append(r)
tmp.append(M[i]) #set tmp=LI+[M[i]]
if matrix_rank(tmp)>len(LI): #test if M[i] is linearly independent from all (row) vectors in LI
LI.append(M[i]) #note that matrix_rank does not need to take in a square matrix
return LI #return set of linearly independent (row) vectors
#Example
mat=[[1,2,3,4],[4,5,6,7],[5,7,9,11],[2,4,6,8]]
LI_vecs(4,mat)
I interpret the problem as finding rows that are linearly independent from other rows.
That is equivalent to finding rows that are linearly dependent on other rows.
Gaussian elimination and treat numbers smaller than a threshold as zeros can do that. It is faster than finding eigenvalues of a matrix, testing all combinations of rows with Cauchy-Schwarz inequality, or singular value decomposition.
See:
https://math.stackexchange.com/questions/1297437/using-gauss-elimination-to-check-for-linear-dependence
Problem with floating point numbers:
http://numpy-discussion.10968.n7.nabble.com/Reduced-row-echelon-form-td16486.html
With regards to the following discussion:
Find dependent rows/columns of a matrix using Matlab?
from sympy import *
A = Matrix([[1,1,1],[2,2,2],[1,7,5]])
print(A.nullspace())
It is obvious that the first and second row are multiplication of each other.
If we execute the above code we get [-1/3, -2/3, 1]. The indices of the zero elements in the null space show independence. But why is the third element here not zero? If we multiply the A matrix with the null space, we get a zero column vector. So what's wrong?
The answer which we are looking for is the null space of the transpose of A.
B = A.T
print(B.nullspace())
Now we get the [-2, 1, 0], which shows that the third row is independent.
Two important notes here:
Consider whether we want to check the row dependencies or the column
dependencies.
Notice that the null space of a matrix is not equal to the null
space of the transpose of that matrix unless it is symmetric.
You can basically find the vectors spanning the columnspace of the matrix by using SymPy library's columnspace() method of Matrix object. Automatically, they are the linearly independent columns of the matrix.
import sympy as sp
import numpy as np
M = sp.Matrix([[0, 1, 0, 0],
[0, 0, 1, 0],
[1, 0, 0, 1]])
for i in M.columnspace():
print(np.array(i))
print()
# The output is following.
# [[0]
# [0]
# [1]]
# [[1]
# [0]
# [0]]
# [[0]
# [1]
# [0]]