What does IFFT return in Python? - python

I need the inverse Fourier transform of a complex array. ifft should return a real array, but it returns another complex array.
In MATLAB,
a=ifft(fft(a)), but in Python it does not work like that.
a = np.arange(6)
m = ifft(fft(a))
m # Google says m should = a, but m is complex
Output :
array([0.+0.00000000e+00j, 1.+3.70074342e-16j, 2.+0.00000000e+00j,
3.-5.68396583e-17j, 4.+0.00000000e+00j, 5.-3.13234683e-16j])

The imaginary part is result floating precision number calculation error. If it is very small, it rather can be dropped.
Numpy has built-in function real_if_close, to do so:
>>> np.real_if_close(np.fft.ifft(np.fft.fft(a)))
array([0., 1., 2., 3., 4., 5.])
You can read about floating system limitations here:
https://docs.python.org/3.8/tutorial/floatingpoint.html

if the imaginary part is close to zero you could discard it:
import numpy as np
arr = np.array(
[
0.0 + 0.00000000e00j,
1.0 + 3.70074342e-16j,
2.0 + 0.00000000e00j,
3.0 - 5.68396583e-17j,
4.0 + 0.00000000e00j,
5.0 - 3.13234683e-16j,
]
)
if all(np.isclose(arr.imag, 0)):
arr = arr.real
# [ 0. 1. 2. 3. 4. 5.]
(that's what real_if_close does in one line as in R2RT's answer).

You can test like this:
import numpy as np
from numpy import fft
a = np.arange(6)
print(a)
f = np.fft.fft(a)
print(f)
m = np.fft.ifft(f)
print(m)
[0 1 2 3 4 5]
[15.+0.j -3.+5.19615242j -3.+1.73205081j -3.+0.j
-3.-1.73205081j -3.-5.19615242j]
[0.+0.j 1.+0.j 2.+0.j 3.+0.j 4.+0.j 5.+0.j]
To get the real part only you can use:
print(m.real) # [0. 1. 2. 3. 4. 5.]

You are mistaken in "Ifft should return a real array". If you want a real valued output (i.e. you have the fft of real data and now want to perform the ifft) you should use irfft.
See this example from the docs:
>>> np.fft.ifft([1, -1j, -1, 1j])
array([ 0.+0.j, 1.+0.j, 0.+0.j, 0.+0.j]) #Output is complex which is correct
>>> np.fft.irfft([1, -1j, -1])
array([ 0., 1., 0., 0.]) #Output is real valued

Related

Python Optimization: Using vector technique to find power of each matrix in an numpy array

3D numpy array A contains a series (in this example, I am choosing 3) of 2D numpy array D of shape 2 x 2. The D matrix is as follows:
D = np.array([[1,2],[3,4]])
A is initialized and assigned as below:
idx = np.arange(3)
A = np.zeros((3,2,2))
A[idx,:,:] = D # This gives A = [[[1,2],[3,4]],[[1,2],[3,4]],\
# [[1,2],[3,4]]]
# In mathematical notation: A = {D, D, D}
Now, essentially what I require after the execution of the codes is:
Mathematically, A = {D^0, D^1, D^2} = {D0, D1, D2}
where D0 = [[1,0],[0,1]], D1 = [[1,2],[3,4]], D2=[[7,10],[15,22]]
Is it possible to apply power to each matrix element in A without using a for-loop? I would be doing larger matrices with more in the series.
I had defined, n = np.array([0,1,2]) # corresponding to powers 0, 1 and 2 and tried
Result = np.power(A,n) but I do not get the desired output.
Is there are an efficient way to do it?
Full code:
D = np.array([[1,2],[3,4]])
idx = np.arange(3)
A = np.zeros((3,2,2))
A[idx,:,:] = D # This gives A = [[[1,2],[3,4]],[[1,2],[3,4]],\
# [[1,2],[3,4]]]
# In mathematical notation: A = {D, D, D}
n = np.array([0,1,2])
Result = np.power(A,n) # ------> Not the desired output.
A cumulative product exists in numpy, but not for matrices. Therefore, you need to make your own 'matcumprod' function. You can use np.dot for this, but np.matmul (or #) is specialized for matrix multiplication.
Since you state your powers always go from 0 to some_power, I suggest the following function:
def matcumprod(D, upto):
Res = np.empty((upto, *D.shape), dtype=A.dtype)
Res[0, :, :] = np.eye(D.shape[0])
Res[1, :, :] = D.copy()
for i in range(1,upto):
Res[i, :, :] = Res[i-1,:,:] # D
return Res
By the way, a loop often times outperforms a built-in numpy function if the latter uses a lot of memory, so don't fret over it if your powers stay within bounds...
Alright, i spent a lot of time on this problem but could not seem to find a vectorized solution in the way you'd like. So i would like to instead first propose a basic solution, and then perhaps an optimization if you require finding continuous powers.
The function you're looking for is called numpy.linalg.matrix_power
import numpy as np
D = np.matrix([[1,2],[3,4]])
idx = np.arange(3)
A = np.zeros((3,2,2))
A[idx,:,:] = D # This gives A = [[[1,2],[3,4]],[[1,2],[3,4]],\
# [[1,2],[3,4]]]
# In mathematical notation: A = {D, D, D}
np.zeros(A.shape)
n = np.array([0,1,2])
result = [np.linalg.matrix_power(D, i) for i in n]
np.array(result)
#Output:
array([[[ 1, 0],
[ 0, 1]],
[[ 1, 2],
[ 3, 4]],
[[ 7, 10],
[15, 22]]])
However, if you notice, you end up calculating multiple powers for the same base matrix. We could instead utilize the intermediate results and go from there, using numpy.linalg.multi_dot
def all_powers_arr_of_matrix(A):
result = np.zeros(A.shape)
result[0] = np.linalg.matrix_power(A[0], 0)
for i in range(1, A.shape[0]):
result[i] = np.linalg.multi_dot([result[i - 1], A[i]])
return result
result = all_powers_arr_of_matrix(A)
#Output:
array([[[ 1., 0.],
[ 0., 1.]],
[[ 1., 2.],
[ 3., 4.]],
[[ 7., 10.],
[15., 22.]]])
Also, we can avoid creating the matrix A entirely, saving some time.
def all_powers_matrix(D, *rangeargs): #end exclusive
''' Expects 2D matrix.
Use as all_powers_matrix(D, end) or
all_powers_matrix(D, start, end)
'''
if len(rangeargs) == 1:
start = 0
end = rangeargs[0]
elif len(rangeargs) == 2:
start = rangeargs[0]
end = rangeargs[1]
else:
print("incorrect args")
return None
result = np.zeros((end - start, *D.shape))
result[0] = np.linalg.matrix_power(A[0], start)
for i in range(start + 1, end):
result[i] = np.linalg.multi_dot([result[i - 1], D])
return result
return result
result = all_powers_matrix(D, 3)
#Output:
array([[[ 1., 0.],
[ 0., 1.]],
[[ 1., 2.],
[ 3., 4.]],
[[ 7., 10.],
[15., 22.]]])
Note that you'd need to add error handling if you decide to use these functions as-is.
To calculate power of matrix D, one way could be to find the eigenvalues and right eigenvectors of it with np.linalg.eig and then raise the power of the diagonal matrix as it is easier, then after some manipulation, you can use two np.einsum to calculate A
#get eigvalues and eigvectors
eigval, eigvect = np.linalg.eig(D)
# to check how it works, you can do:
print (np.dot(eigvect*eigval,np.linalg.inv(eigvect)))
#[[1. 2.]
# [3. 4.]]
# so you get back on D
#use power as ufunc of outer with n on the eigenvalues to get all the one you want
arrp = np.power.outer( eigval, n).T
#apply_along_axis to create the diagonal matrix along the last axis
diagp = np.apply_along_axis( np.diag, axis=-1, arr=arrp)
#finally use two np.einsum to calculate with the subscript to get what you want
A = np.einsum('lij,jk -> lik',
np.einsum('ij,kjl -> kil',eigvect,diagp), np.linalg.inv(eigvect)).round()
print (A)
print (A.shape)
#[[[ 1. 0.]
# [-0. 1.]]
#
# [[ 1. 2.]
# [ 3. 4.]]
#
# [[ 7. 10.]
# [15. 22.]]]
#
#(3, 2, 2)
I don't have a full solution, but there are some things I wanted to mention which are a bit too long for the comments.
You might first look into addition chain exponentiation if you are computing big powers of big matrices. This is basically asking how many matrix multiplications are required to compute A^k for a given k. For instance A^5 = A(A^2)^2 so you need to only three matrix multiplies: A^2 and (A^2)^2 and A(A^2)^2. This might be the simplest way to gain some efficiency, but you will probably still have to use explicit loops.
Your question is also related to the problem of computing Ax, A^2x, ... , A^kx for a given A and x. This is an active area of research right now (search "matrix powers kernel"), since computing such a sequence efficiently is useful for parallel/communication avoiding Krylov subspace methods. If you're looking for a very efficient solution to your problem it might be worth looking into some of the results about this.

How using "dot" (or "matmul") function for iterative multiplication in Python

I need obtain a "W" matrix of multiples matrix multiplications (all multiplications result in column vectors).
from numpy import matrix
from numpy import transpose
from numpy import matmul
from numpy import dot
# Iterative matrix multiplication
def iterativeMultiplication(X, Y):
W = [] # Matrix of matricial products
X = matrix(X) # same number of rows
Y = matrix(Y) # same number of rows
h = 0
while (h < X.shape[1]):
W.append([])
W[h] = dot(transpose(X), Y) # using "dot" function
h += 1
return W
But, unexpectedly, I obtain a list of objects with their respective data types.
X = [[0., 0., 1.], [1.,0.,0.], [2.,2.,2.], [2.,5.,4.]]
Y = [[-0.2], [1.1], [5.9], [12.3]] # Edit Y column
iterativeMultiplication( X, Y )
Results in:
[array([[37.5],[73.3],[60.8]]),
array([[37.5],[73.3],[60.8]]),
array([[37.5],[73.3],[60.8]])]
I need any method for obtain only the numerical values for the matrix conversion.
W = matrix(W) # Results in error
It is the same using "matmul" function. Thx for your time.
If you want to stack multiple matrices, you can use numpy.vstack:
W = numpy.vstack(W)
Edit: There seems to be a discrepancy between your function, X and Y versus the "result" list in your question. But based on your comments below, what you're actually looking for is numpy.hstack (horizontal stack) which will give you the desired 3x3 matrix based on your "result" list.
W = numpy.hstack(W)
Of course you are going to get a list. You initial W as a list, and append the same calculation to it 3 times.
But your 3 element arrays don't make sense with this data, array([[ 3.36877336],[ 3.97112615],[ 3.8092797 ]]).
If I make Xm=np.matrix(X), etc:
In [162]: Xm
Out[162]:
matrix([[ 0., 0., 1.],
[ 1., 0., 0.],
[ 2., 2., 2.],
[ 2., 5., 4.]])
In [163]: Ym
Out[163]:
matrix([[ 0.1, -0.2],
[ 0.9, 1.1],
[ 6.2, 5.9],
[ 11.9, 12.3]])
In [164]: Xm.T.dot(Ym)
Out[164]:
matrix([[ 37.1, 37.5],
[ 71.9, 73.3],
[ 60.1, 60.8]])
In [165]: Xm.T*Ym # matrix interprets * as .dot
Out[165]:
matrix([[ 37.1, 37.5],
[ 71.9, 73.3],
[ 60.1, 60.8]])
You need to edit the question, to have both valid Python code (missing def and :), and results that match the inputs.
===============
In [173]: Y = [[-0.2], [1.1], [5.9], [12.3]]
In [174]: Ym=np.matrix(Y)
Out[176]:
matrix([[ 37.5],
[ 73.3],
[ 60.8]])
=====================
This iteration is clumsy:
h = 0
while (h < X.shape[1]):
W.append([])
W[h] = dot(transpose(X), Y) # using "dot" function
h += 1
A more Pythonic approach
for h in range(X.shape[1]):
W.append(np.dot(...))
Or even
W = [np.dot(....) for h in range(X.shape[1])]

Implement this situation in numpy

Given these two arrays:
E = [[16.461, 17.015, 14.676],
[15.775, 18.188, 14.459],
[14.489, 18.449, 14.756],
[14.171, 19.699, 14.406],
[14.933, 20.644, 13.839],
[16.233, 20.352, 13.555],
[16.984, 21.297, 12.994],
[16.683, 19.056, 13.875],
[17.918, 18.439, 13.718],
[17.734, 17.239, 14.207]]
S = [[0.213, 0.660, 1.287],
[0.250, 2.016, 1.509],
[0.016, 2.995, 0.619],
[0.142, 4.189, 1.194],
[0.451, 4.493, 2.459],
[0.681, 3.485, 3.329],
[0.990, 3.787, 4.592],
[0.579, 2.170, 2.844],
[0.747, 0.934, 3.454],
[0.520, 0.074, 2.491]]
The problem states that I should get the 3x3 covariance matrix (C) between S and E using the following formula:
C = (1/(n-1))[S'E - (1/10)S'i i'E]
Here n is 10, and i is an n x 1 column vector consisting of only ones. S' and i' are the transpose of matrix S and column vector i, respectively.
So far, I can't get C because I don't understand the meaning of i (and i') and its implementation in the formula. Using numpy, so far I do:
import numpy as np
tS = numpy.array(S).T
C = (1.0/9.0)*(np.dot(tS, E)-((1.0/10.0)*np.dot(tS, E))) #Here is where I lack the i and i' implementation.
I will really appreciate your help to understand and implement i and i' in the formula. The output should be:
C= [[0.2782, 0.2139, -0.1601],
[-1.4028, 1.9619, -0.2744],
[1.0443, 0.9712, -0.6610]]
It looks like the only part you're missing is making i:
>>> i = np.ones((N, 1))
>>> i
array([[ 1.],
[ 1.],
[ 1.],
[ 1.],
[ 1.],
[ 1.],
[ 1.],
[ 1.],
[ 1.],
[ 1.]])
After that, we get
>>> C = (1.0/(N-1)) * (S.T.dot(E) - (1.0/N) * S.T.dot(i) * i.T.dot(E))
>>> C
array([[ 0.27842301, 0.21388842, -0.16011839],
[-1.4017267 , 1.96193373, -0.27441417],
[ 1.04532836, 0.97120807, -0.66095656]])
Note that this doesn't quite produce the array you expected, which is more obvious if you round it, but maybe there are some minor typos in your data?
>>> C.round(4)
array([[ 0.2784, 0.2139, -0.1601],
[-1.4017, 1.9619, -0.2744],
[ 1.0453, 0.9712, -0.661 ]])
This is what you want I guess:
S = numpy.array(S)
E = numpy.array(E)
ones = np.ones((10,1))
C = (1.0/9)*(np.dot(S.T, E)-((1.0/10)* (np.dot(np.dot(np.dot(S.T,ones),ones.T),E))))
My output is :
array([[ 0.27842301, 0.21388842, -0.16011839],
[-1.4017267 , 1.96193373, -0.27441417],
[ 1.04532836, 0.97120807, -0.66095656]])

reshape list of numpy arrays and then reshape back

I have a list which consists of several numpy arrays with different shapes.
I want to reshape this list of arrays into a numpy vector and then change each element in the vector and then reshape it back to the original list of arrays.
For example:
input
[numpy.zeros((2,2)), numpy.ones((3,3))]
First
To vector
[0,0,0,0,1,1,1,1,1,1,1,1,1]
Second
every time change only one element. for example change the 1st element 0 to 2
[0,2,0,0,1,1,1,1,1,1,1,1,1]
Last
convert it back to
[array([[0,2],[0,0]]),array([[1,1,1],[1,1,1],[1,1,1]])]
Is there any fast implementation? Thanks very much.
It seems like converting to a list and back will be inefficient. Instead, why not figure out which array to index (and where) and then just update that index? e.g.
def change_element(arr1, arr2, ix, value):
which = ix >= arr1.size
arr = [arr1, arr2][which]
ix = ix - arr1.size if which else ix
arr.ravel()[ix] = value
And here's some example usage:
>>> arr1 = np.zeros((2, 2))
>>> arr2 = np.ones((3, 3))
>>> change_element(arr1, arr2, 1, 2)
>>> change_element(arr1, arr2, 6, 3.14)
>>> arr1
array([[ 0., 2.],
[ 0., 0.]])
>>> arr2
array([[ 1. , 1. , 3.14],
[ 1. , 1. , 1. ],
[ 1. , 1. , 1. ]])
>>> change_element(arr1, arr2, 7, 3.14)
>>> arr1
array([[ 0., 2.],
[ 0., 0.]])
>>> arr2
array([[ 1. , 1. , 3.14],
[ 3.14, 1. , 1. ],
[ 1. , 1. , 1. ]])
A few notes -- This updates the arrays in place. It doesn't create new arrays. If you really need to create new arrays, I suppose you could np.copy them and return. Also, this relies on the arrays sharing memory before and after the ravel. I don't remember the exact circumstances where ravel would return a new array rather than a view into the original array . . .
Generalizing to more arrays is actually quite easy. We just need to walk down the list of arrays and see if ix is less than the array size. If it is, we've found our array. If it isn't, we need to subtract the array's size from ix to represent the number of elements we've traversed thus far:
def change_element(arrays, ix, value):
for arr in arrays:
if ix < arr.size:
arr.ravel()[ix] = value
return
ix -= arr.size
And you can call this similar to before:
change_element([arr1, arr2], 6, 3.14159)
#mgilson probably has the best answer for you, but if you absolutely have to convert to a flat list first and then go back again (perhaps because you need to do something else with the flat list as well), then you can do this with list comprehensions:
lst = [numpy.zeros((2,4)), numpy.ones((3,3))]
tlist = [e for a in lst for e in a.ravel()]
tlist[1] = 2
i = 0
lst2 = []
dims = [a.shape for a in lst]
for n, m in dims:
lst2.append(np.array(tlist[i:i+n*m]).reshape(n,m))
i += n*m
lst2
[array([[ 0., 2.],
[ 0., 0.]]), array([[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.]])]
Of course, you lose the information about your array sizes when you flatten, so you need to store them somewhere (here, in dims).

Scipy filter with multi-dimensional (or non-scalar) output

Is there a filter similar to ndimage's generic_filter that supports vector output? I did not manage to make scipy.ndimage.filters.generic_filter return more than a scalar. Uncomment the line in the code below to get the error: TypeError: only length-1 arrays can be converted to Python scalars.
I'm looking for a generic filter that process 2D or 3D arrays and returns a vector at each point. Thus the output would have one added dimension. For the example below I'd expect something like this:
m.shape # (10,10)
res.shape # (10,10,2)
Example Code
import numpy as np
from scipy import ndimage
a = np.ones((10, 10)) * np.arange(10)
footprint = np.array([[1,1,1],
[1,0,1],
[1,1,1]])
def myfunc(x):
r = sum(x)
#r = np.array([1,1]) # uncomment this
return r
res = ndimage.generic_filter(a, myfunc, footprint=footprint)
The generic_filter expects myfunc to return a scalar, never a vector.
However, there is nothing that precludes myfunc from also adding information
to, say, a list which is passed to myfunc as an extra argument.
Instead of using the array returned by generic_filter, we can generate our vector-valued array by reshaping this list.
For example,
import numpy as np
from scipy import ndimage
a = np.ones((10, 10)) * np.arange(10)
footprint = np.array([[1,1,1],
[1,0,1],
[1,1,1]])
ndim = 2
def myfunc(x, out):
r = np.arange(ndim, dtype='float64')
out.extend(r)
return 0
result = []
ndimage.generic_filter(
a, myfunc, footprint=footprint, extra_arguments=(result,))
result = np.array(result).reshape(a.shape+(ndim,))
I think I get what you're asking, but I'm not completely sure how does the ndimage.generic_filter work (how abstruse is the source!).
Here's just a simple wrapper function. This function will take in an array, all the parameters ndimage.generic_filter needs. Function returns an array where each element of the former array is now represented by an array with shape (2,), result of the function is stored as the second element of that array.
def generic_expand_filter(inarr, func, **kwargs):
shape = inarr.shape
res = np.empty(( shape+(2,) ))
temp = ndimage.generic_filter(inarr, func, **kwargs)
for row in range(shape[0]):
for val in range(shape[1]):
res[row][val][0] = inarr[row][val]
res[row][val][1] = temp[row][val]
return res
Output, where res denotes just the generic_filter and res2 denotes generic_expand_filter, of this function is:
>>> a.shape #same as res.shape
(10, 10)
>>> res2.shape
(10, 10, 2)
>>> a[0]
array([ 0., 1., 2., 3., 4., 5., 6., 7., 8., 9.])
>>> res[0]
array([ 3., 8., 16., 24., 32., 40., 48., 56., 64., 69.])
>>> print(*res2[0], sep=", ") #this is just to avoid the vertical default output
[ 0. 3.], [ 1. 8.], [ 2. 16.], [ 3. 24.], [ 4. 32.], [ 5. 40.], [ 6. 48.], [ 7. 56.], [ 8. 64.], [ 9. 69.]
>>> a[0][0]
0.0
>>> res[0][0]
3.0
>>> res2[0][0]
array([ 0., 3.])
Of course you probably don't want to save the old array, but instead have both fields as new results. Except I don't know what exactly you had in mind, if the two values you want stored are unrelated, just add a temp2 and func2 and call another generic_filter with the same **kwargs and store that as the first value.
However if you want an actual vector quantity that is calculated using multiple inarr elements, meaning that the two new created fields aren't independent, you are just going to have to write that kind of a function, one that takes in an array, idx, idy indices and returns a tuple\list\array value which you can then unpack and assign to the result.

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