Is it possible to generate random numbers that are almost equally spaced which shouldnot be exactly same as numpy.linspace output
I look into the numpy.random.uniform function but it doesnot give the required results.
Moreover the the summation of the values generated by the function should be same as the summation of the values generated by numpy.linspace function.
code
import random
import numpy as np
random.seed(42)
data=np.random.uniform(2,4,10)
print(data)
You might consider drawing random samples around the output of numpy.linspace. Setting these numbers as the mean of the normal distribution and setting the variance not too high would generate numbers close to the output of numpy.linspace. For example,
>>> import numpy as np
>>> exact_numbers = np.linspace(2.0, 10.0, num=5)
>>> exact_numbers
array([ 2., 4., 6., 8., 10.])
>>> approximate_numbers = np.random.normal(exact_numbers, np.ones(5) * 0.1)
>>> approximate_numbers
array([2.12950013, 3.9804745 , 5.80670316, 8.07868932, 9.85288221])
Maybe this trick by combining numpy.linspace and numpy.random.uniform and random choice two indexes and increase one of them and decrease other help you:
(You can change size=10, threshold=0.1 for how random numbers are bigger or smaller)
import numpy as np
size = 10
theroshold = 0.1
r = np.linspace(2,4,size) # r.sum()=30
# array([2. , 2.22222222, 2.44444444, 2.66666667, 2.88888889,
# 3.11111111, 3.33333333, 3.55555556, 3.77777778, 4. ])
c = np.random.uniform(0,theroshold,size)
# array([0.02246768, 0.08661081, 0.0932445 , 0.00360563, 0.06539992,
# 0.0107167 , 0.06490493, 0.0558159 , 0.00268924, 0.00070247])
s = np.random.choice(range(size), size+1)
# array([5, 5, 8, 3, 6, 4, 1, 8, 7, 1, 7])
for idx, (i,j) in enumerate(zip(s, s[1:])):
r[i] += c[idx]
r[j] -= c[idx]
print(r)
print(r.sum())
Output:
[2. 2.27442369 2.44444444 2.5770278 2.83420567 3.19772192
3.39512762 3.50172642 3.77532244 4. ]
30
I am trying to implement FT of cross correlation to see if the convolution theorem holds true.
I am stuck, because right now my convolution and my fourier transform multiplication are not equal, but they contain the same elements. Just in a different order.
Can somebody help me out? I tried flipping the filter, I tried conjugating it and I tried transposing it.
My inspiration is from https://dsp.stackexchange.com/questions/29761/convolution-theorem-for-cross-correlation
Code
conv: tensor([[[[ 0.6908, 1.2291, -0.7020, 1.0186, -1.2585],
[ 0.3773, 1.8541, 0.8134, 0.2672, -0.5912],
[ 0.1427, 0.6185, 1.0000, -0.8145, -0.4732],
[ 1.9047, -0.0119, 0.8849, -1.2321, -0.5614],
[ 0.0910, 1.0923, -0.6255, -1.9881, 1.9961]]]])
FT result: [[-1.2321 -0.5614 1.9047 -0.0119 0.8849]
[-1.9881 1.9961 0.091 1.0923 -0.6255]
[ 1.0186 -1.2585 0.6908 1.2291 -0.702 ]
[ 0.2672 -0.5912 0.3773 1.8541 0.8134]
[-0.8145 -0.4732 0.1427 0.6185 1. ]]
added code:
import torch
import torch.nn.functional as F
from scipy.fft import fft2, fftshift, ifft2
import numpy as np
nb_channels = 1
h, w = 5, 5
x = [[[[ 0.6908, 1.2291, -0.7020, 1.0186, -1.2585],
[ 0.3773, 1.8541, 0.8134, 0.2672, -0.5912],
[ 0.1427, 0.6185, 1.000, -0.8145, -0.4732],
[ 1.9047, -0.0119, 0.8849, -1.2321, -0.5614],
[ 0.0910, 1.0923, -0.6255, -1.9881, 1.9961]]]]
x = torch.tensor(x)
x = x.view(1, nb_channels, h, w)
weights1 = torch.tensor([[0., 0., 0.],
[0., 1., 0.],
[0., 0., 0.]])
weights = weights1.view(1, 1, 3, 3).repeat(1, nb_channels, 1, 1)
output = F.conv2d(x, weights, padding='same')
print("conv:", output)
################### FT numpy
a = x.numpy()
mac = a.copy()
img = np.squeeze(mac)
fft_in1 = fft2(img)
w_numpy = weights1.numpy()
weights_numpy = w_numpy.copy()
pad_weights = np.pad(weights_numpy, ((1, 1), (1, 1)), mode='constant', constant_values=(0, 0))
fft_weights = fft2(pad_weights)
pad_weights = fft_weights
out1 = fft_in1 * pad_weights
output_n = out1
output_numpy = ifft2(output_n)
output_numpy = output_numpy.real
output_numpy = np.around(output_numpy, decimals = 4)
print("after IFFT:", output_numpy)
I have two numpy arrays of different shapes, but with the same length (leading dimension). I want to shuffle each of them, such that corresponding elements continue to correspond -- i.e. shuffle them in unison with respect to their leading indices.
This code works, and illustrates my goals:
def shuffle_in_unison(a, b):
assert len(a) == len(b)
shuffled_a = numpy.empty(a.shape, dtype=a.dtype)
shuffled_b = numpy.empty(b.shape, dtype=b.dtype)
permutation = numpy.random.permutation(len(a))
for old_index, new_index in enumerate(permutation):
shuffled_a[new_index] = a[old_index]
shuffled_b[new_index] = b[old_index]
return shuffled_a, shuffled_b
For example:
>>> a = numpy.asarray([[1, 1], [2, 2], [3, 3]])
>>> b = numpy.asarray([1, 2, 3])
>>> shuffle_in_unison(a, b)
(array([[2, 2],
[1, 1],
[3, 3]]), array([2, 1, 3]))
However, this feels clunky, inefficient, and slow, and it requires making a copy of the arrays -- I'd rather shuffle them in-place, since they'll be quite large.
Is there a better way to go about this? Faster execution and lower memory usage are my primary goals, but elegant code would be nice, too.
One other thought I had was this:
def shuffle_in_unison_scary(a, b):
rng_state = numpy.random.get_state()
numpy.random.shuffle(a)
numpy.random.set_state(rng_state)
numpy.random.shuffle(b)
This works...but it's a little scary, as I see little guarantee it'll continue to work -- it doesn't look like the sort of thing that's guaranteed to survive across numpy version, for example.
Your can use NumPy's array indexing:
def unison_shuffled_copies(a, b):
assert len(a) == len(b)
p = numpy.random.permutation(len(a))
return a[p], b[p]
This will result in creation of separate unison-shuffled arrays.
X = np.array([[1., 0.], [2., 1.], [0., 0.]])
y = np.array([0, 1, 2])
from sklearn.utils import shuffle
X, y = shuffle(X, y, random_state=0)
To learn more, see http://scikit-learn.org/stable/modules/generated/sklearn.utils.shuffle.html
Your "scary" solution does not appear scary to me. Calling shuffle() for two sequences of the same length results in the same number of calls to the random number generator, and these are the only "random" elements in the shuffle algorithm. By resetting the state, you ensure that the calls to the random number generator will give the same results in the second call to shuffle(), so the whole algorithm will generate the same permutation.
If you don't like this, a different solution would be to store your data in one array instead of two right from the beginning, and create two views into this single array simulating the two arrays you have now. You can use the single array for shuffling and the views for all other purposes.
Example: Let's assume the arrays a and b look like this:
a = numpy.array([[[ 0., 1., 2.],
[ 3., 4., 5.]],
[[ 6., 7., 8.],
[ 9., 10., 11.]],
[[ 12., 13., 14.],
[ 15., 16., 17.]]])
b = numpy.array([[ 0., 1.],
[ 2., 3.],
[ 4., 5.]])
We can now construct a single array containing all the data:
c = numpy.c_[a.reshape(len(a), -1), b.reshape(len(b), -1)]
# array([[ 0., 1., 2., 3., 4., 5., 0., 1.],
# [ 6., 7., 8., 9., 10., 11., 2., 3.],
# [ 12., 13., 14., 15., 16., 17., 4., 5.]])
Now we create views simulating the original a and b:
a2 = c[:, :a.size//len(a)].reshape(a.shape)
b2 = c[:, a.size//len(a):].reshape(b.shape)
The data of a2 and b2 is shared with c. To shuffle both arrays simultaneously, use numpy.random.shuffle(c).
In production code, you would of course try to avoid creating the original a and b at all and right away create c, a2 and b2.
This solution could be adapted to the case that a and b have different dtypes.
Very simple solution:
randomize = np.arange(len(x))
np.random.shuffle(randomize)
x = x[randomize]
y = y[randomize]
the two arrays x,y are now both randomly shuffled in the same way
James wrote in 2015 an sklearn solution which is helpful. But he added a random state variable, which is not needed. In the below code, the random state from numpy is automatically assumed.
X = np.array([[1., 0.], [2., 1.], [0., 0.]])
y = np.array([0, 1, 2])
from sklearn.utils import shuffle
X, y = shuffle(X, y)
from np.random import permutation
from sklearn.datasets import load_iris
iris = load_iris()
X = iris.data #numpy array
y = iris.target #numpy array
# Data is currently unshuffled; we should shuffle
# each X[i] with its corresponding y[i]
perm = permutation(len(X))
X = X[perm]
y = y[perm]
Shuffle any number of arrays together, in-place, using only NumPy.
import numpy as np
def shuffle_arrays(arrays, set_seed=-1):
"""Shuffles arrays in-place, in the same order, along axis=0
Parameters:
-----------
arrays : List of NumPy arrays.
set_seed : Seed value if int >= 0, else seed is random.
"""
assert all(len(arr) == len(arrays[0]) for arr in arrays)
seed = np.random.randint(0, 2**(32 - 1) - 1) if set_seed < 0 else set_seed
for arr in arrays:
rstate = np.random.RandomState(seed)
rstate.shuffle(arr)
And can be used like this
a = np.array([1, 2, 3, 4, 5])
b = np.array([10,20,30,40,50])
c = np.array([[1,10,11], [2,20,22], [3,30,33], [4,40,44], [5,50,55]])
shuffle_arrays([a, b, c])
A few things to note:
The assert ensures that all input arrays have the same length along
their first dimension.
Arrays shuffled in-place by their first dimension - nothing returned.
Random seed within positive int32 range.
If a repeatable shuffle is needed, seed value can be set.
After the shuffle, the data can be split using np.split or referenced using slices - depending on the application.
you can make an array like:
s = np.arange(0, len(a), 1)
then shuffle it:
np.random.shuffle(s)
now use this s as argument of your arrays. same shuffled arguments return same shuffled vectors.
x_data = x_data[s]
x_label = x_label[s]
There is a well-known function that can handle this:
from sklearn.model_selection import train_test_split
X, _, Y, _ = train_test_split(X,Y, test_size=0.0)
Just setting test_size to 0 will avoid splitting and give you shuffled data.
Though it is usually used to split train and test data, it does shuffle them too.
From documentation
Split arrays or matrices into random train and test subsets
Quick utility that wraps input validation and
next(ShuffleSplit().split(X, y)) and application to input data into a
single call for splitting (and optionally subsampling) data in a
oneliner.
This seems like a very simple solution:
import numpy as np
def shuffle_in_unison(a,b):
assert len(a)==len(b)
c = np.arange(len(a))
np.random.shuffle(c)
return a[c],b[c]
a = np.asarray([[1, 1], [2, 2], [3, 3]])
b = np.asarray([11, 22, 33])
shuffle_in_unison(a,b)
Out[94]:
(array([[3, 3],
[2, 2],
[1, 1]]),
array([33, 22, 11]))
One way in which in-place shuffling can be done for connected lists is using a seed (it could be random) and using numpy.random.shuffle to do the shuffling.
# Set seed to a random number if you want the shuffling to be non-deterministic.
def shuffle(a, b, seed):
np.random.seed(seed)
np.random.shuffle(a)
np.random.seed(seed)
np.random.shuffle(b)
That's it. This will shuffle both a and b in the exact same way. This is also done in-place which is always a plus.
EDIT, don't use np.random.seed() use np.random.RandomState instead
def shuffle(a, b, seed):
rand_state = np.random.RandomState(seed)
rand_state.shuffle(a)
rand_state.seed(seed)
rand_state.shuffle(b)
When calling it just pass in any seed to feed the random state:
a = [1,2,3,4]
b = [11, 22, 33, 44]
shuffle(a, b, 12345)
Output:
>>> a
[1, 4, 2, 3]
>>> b
[11, 44, 22, 33]
Edit: Fixed code to re-seed the random state
Say we have two arrays: a and b.
a = np.array([[1,2,3],[4,5,6],[7,8,9]])
b = np.array([[9,1,1],[6,6,6],[4,2,0]])
We can first obtain row indices by permutating first dimension
indices = np.random.permutation(a.shape[0])
[1 2 0]
Then use advanced indexing.
Here we are using the same indices to shuffle both arrays in unison.
a_shuffled = a[indices[:,np.newaxis], np.arange(a.shape[1])]
b_shuffled = b[indices[:,np.newaxis], np.arange(b.shape[1])]
This is equivalent to
np.take(a, indices, axis=0)
[[4 5 6]
[7 8 9]
[1 2 3]]
np.take(b, indices, axis=0)
[[6 6 6]
[4 2 0]
[9 1 1]]
If you want to avoid copying arrays, then I would suggest that instead of generating a permutation list, you go through every element in the array, and randomly swap it to another position in the array
for old_index in len(a):
new_index = numpy.random.randint(old_index+1)
a[old_index], a[new_index] = a[new_index], a[old_index]
b[old_index], b[new_index] = b[new_index], b[old_index]
This implements the Knuth-Fisher-Yates shuffle algorithm.
Shortest and easiest way in my opinion, use seed:
random.seed(seed)
random.shuffle(x_data)
# reset the same seed to get the identical random sequence and shuffle the y
random.seed(seed)
random.shuffle(y_data)
most solutions above work, however if you have column vectors you have to transpose them first. here is an example
def shuffle(self) -> None:
"""
Shuffles X and Y
"""
x = self.X.T
y = self.Y.T
p = np.random.permutation(len(x))
self.X = x[p].T
self.Y = y[p].T
With an example, this is what I'm doing:
combo = []
for i in range(60000):
combo.append((images[i], labels[i]))
shuffle(combo)
im = []
lab = []
for c in combo:
im.append(c[0])
lab.append(c[1])
images = np.asarray(im)
labels = np.asarray(lab)
I extended python's random.shuffle() to take a second arg:
def shuffle_together(x, y):
assert len(x) == len(y)
for i in reversed(xrange(1, len(x))):
# pick an element in x[:i+1] with which to exchange x[i]
j = int(random.random() * (i+1))
x[i], x[j] = x[j], x[i]
y[i], y[j] = y[j], y[i]
That way I can be sure that the shuffling happens in-place, and the function is not all too long or complicated.
Just use numpy...
First merge the two input arrays 1D array is labels(y) and 2D array is data(x) and shuffle them with NumPy shuffle method. Finally split them and return.
import numpy as np
def shuffle_2d(a, b):
rows= a.shape[0]
if b.shape != (rows,1):
b = b.reshape((rows,1))
S = np.hstack((b,a))
np.random.shuffle(S)
b, a = S[:,0], S[:,1:]
return a,b
features, samples = 2, 5
x, y = np.random.random((samples, features)), np.arange(samples)
x, y = shuffle_2d(train, test)
I would like to improve the speed of my code by computing a function once on a numpy array instead of a for loop is over a function of this python library. If I have a function as following:
import numpy as np
import galsim
from math import *
M200=1e14
conc=6.9
def func(M200, conc):
halo_z=0.2
halo_pos =[1200., 3769.7]
halo_pos = galsim.PositionD(x=halo_pos_arcsec[0],y=halo_pos_arcsec[1])
nfw = galsim.NFWHalo(mass=M200, conc=conc, redshift=halo_z,halo_pos=halo_pos, omega_m = 0.3, omega_lam =0.7)
for i in range(len(shear_z)):
shear_pos=galsim.PositionD(x=pos_arcsec[i,0],y=pos_arcsec[i,1])
model_g1, model_g2 = nfw.getShear(pos=self.shear_pos, z_s=shear_z[i])
l=np.sum(model_g1-model_g2)/sqrt(np.pi)
return l
While pos_arcsec is a two-dimensional array of 24000x2 and shear_z is a 1D array with 24000 elements as well.
The main problem is that I want to calculate this function on a grid where M200=np.arange(13., 16., 0.01) and conc = np.arange(3, 10, 0.01). I don't know how to broadcast this function to be estimated for this two dimensional array over M200 and conc. It takes a lot to run the code. I am looking for the best approaches to speed up these calculations.
This here should work when pos is an array of shape (n,2)
import numpy as np
def f(pos, z):
r=np.sqrt(pos[...,0]**2+pos[...,1]**2)
return np.log(r)*(z+1)
Example:
z = np.arange(10)
pos = np.arange(20).reshape(10,2)
f(pos,z)
# array([ 0. , 2.56494936, 5.5703581 , 8.88530251,
# 12.44183436, 16.1944881 , 20.11171117, 24.17053133,
# 28.35353608, 32.64709419])
Use numpy.linalg.norm
If you have an array:
import numpy as np
import numpy.linalg as la
a = np.array([[3, 4], [5, 12], [7, 24]])
then you can determine the magnitude of the resulting vector (sqrt(a^2 + b^2)) by
b = np.sqrt(la.norm(a, axis=1)
>>> print b
array([ 5., 15. 25.])
Given a 3 times 3 numpy array
a = numpy.arange(0,27,3).reshape(3,3)
# array([[ 0, 3, 6],
# [ 9, 12, 15],
# [18, 21, 24]])
To normalize the rows of the 2-dimensional array I thought of
row_sums = a.sum(axis=1) # array([ 9, 36, 63])
new_matrix = numpy.zeros((3,3))
for i, (row, row_sum) in enumerate(zip(a, row_sums)):
new_matrix[i,:] = row / row_sum
There must be a better way, isn't there?
Perhaps to clearify: By normalizing I mean, the sum of the entrys per row must be one. But I think that will be clear to most people.
Broadcasting is really good for this:
row_sums = a.sum(axis=1)
new_matrix = a / row_sums[:, numpy.newaxis]
row_sums[:, numpy.newaxis] reshapes row_sums from being (3,) to being (3, 1). When you do a / b, a and b are broadcast against each other.
You can learn more about broadcasting here or even better here.
Scikit-learn offers a function normalize() that lets you apply various normalizations. The "make it sum to 1" is called L1-norm. Therefore:
from sklearn.preprocessing import normalize
matrix = numpy.arange(0,27,3).reshape(3,3).astype(numpy.float64)
# array([[ 0., 3., 6.],
# [ 9., 12., 15.],
# [ 18., 21., 24.]])
normed_matrix = normalize(matrix, axis=1, norm='l1')
# [[ 0. 0.33333333 0.66666667]
# [ 0.25 0.33333333 0.41666667]
# [ 0.28571429 0.33333333 0.38095238]]
Now your rows will sum to 1.
I think this should work,
a = numpy.arange(0,27.,3).reshape(3,3)
a /= a.sum(axis=1)[:,numpy.newaxis]
In case you are trying to normalize each row such that its magnitude is one (i.e. a row's unit length is one or the sum of the square of each element in a row is one):
import numpy as np
a = np.arange(0,27,3).reshape(3,3)
result = a / np.linalg.norm(a, axis=-1)[:, np.newaxis]
# array([[ 0. , 0.4472136 , 0.89442719],
# [ 0.42426407, 0.56568542, 0.70710678],
# [ 0.49153915, 0.57346234, 0.65538554]])
Verifying:
np.sum( result**2, axis=-1 )
# array([ 1., 1., 1.])
I think you can normalize the row elements sum to 1 by this:
new_matrix = a / a.sum(axis=1, keepdims=1).
And the column normalization can be done with new_matrix = a / a.sum(axis=0, keepdims=1). Hope this can hep.
You could use built-in numpy function:
np.linalg.norm(a, axis = 1, keepdims = True)
it appears that this also works
def normalizeRows(M):
row_sums = M.sum(axis=1)
return M / row_sums
You could also use matrix transposition:
(a.T / row_sums).T
Here is one more possible way using reshape:
a_norm = (a/a.sum(axis=1).reshape(-1,1)).round(3)
print(a_norm)
Or using None works too:
a_norm = (a/a.sum(axis=1)[:,None]).round(3)
print(a_norm)
Output:
array([[0. , 0.333, 0.667],
[0.25 , 0.333, 0.417],
[0.286, 0.333, 0.381]])
Use
a = a / np.linalg.norm(a, ord = 2, axis = 0, keepdims = True)
Due to the broadcasting, it will work as intended.
Or using lambda function, like
>>> vec = np.arange(0,27,3).reshape(3,3)
>>> import numpy as np
>>> norm_vec = map(lambda row: row/np.linalg.norm(row), vec)
each vector of vec will have a unit norm.
We can achieve the same effect by premultiplying with the diagonal matrix whose main diagonal is the reciprocal of the row sums.
A = np.diag(A.sum(1)**-1) # A