I'm trying to do some matrix calculations in python and came across a problem when I tried to speed up my code using stacked arrays instead of simple for loops. I need to create a 2D-array with values (given as 1D-array) on the diagonal, but could't figure out a smart way to do it with stacked arrays.
In the old (loop) version, I used the np.diag() method, which returns exactly what I need (a 2D-array in that case) if I give the values as 1D-array as input. However, when I switched to stacked arrays my input is not a 1D-array anymore, so that the np.diag() method returns a copy of the diagonal of my 2D-input instead.
Old version with 1D input:
import numpy as np
vals = np.array([1,2,3])
mat = np.diag(vals)
print(mat.shape)
Out: (3, 3)
New version with 2D input:
vals_stack = np.repeat(np.expand_dims(vals, axis=0), 5, axis=0)
# btw: is there a better way to repeat/stack my array?
mat_stack = np.diag(vals_stack)
print(mat_stack.shape)
Out: (3,)
So you can see that np.diag() returns a 1D-array (as expected from the documentation), but I actually need a stack of 2D-arrays. So the shape of the mat_stack must be (7,3,3) and not (3,). Is there any function for that in numpy? Or do I have to loop over that additional dimension like this:
def mydiag(stack):
diag = np.zeros([stack.shape[0], stack.shape[1], stack.shape[1]])
for i in np.arange(stack.shape[0]):
diag[i,:,:] = np.diag([stack[i,:].ravel()])
return diag
In numpy you should use apply_along_axis. There is even an example at the end of the doc for your specific case (here). So the answer is :
np.apply_along_axis(np.diag, -1, vals_stack)
A more pythonic way would be something like this:
[np.diag(row) for row in vals_stack]
Is this what you had in mind:
In [499]: x = np.arange(12).reshape(4,3)
In [500]: X = np.zeros((4,3,3),int)
In [501]: X[np.arange(4)[:,None],np.arange(3), np.arange(3)] = x
In [502]: X
Out[502]:
array([[[ 0, 0, 0],
[ 0, 1, 0],
[ 0, 0, 2]],
[[ 3, 0, 0],
[ 0, 4, 0],
[ 0, 0, 5]],
[[ 6, 0, 0],
[ 0, 7, 0],
[ 0, 0, 8]],
[[ 9, 0, 0],
[ 0, 10, 0],
[ 0, 0, 11]]])
X[0,np.arange(3), np.arange(3)] indexes the diagonal on the first plane. np.arange(4)[:,None] is a (4,1) array, which broadcasts with a (3,) to index a (4,3) block, matching the size of x.
Related
I'm working with a dataset that stores an array of unit-vectors as arrays of the vectors' components.
How would I use vectorised code / broadcasting to write clean and compact code to give the cross product of the vectors element-wise?
For example, here's a brute force method for looping through the length of the arrays, picking out the coordinates, re-composing the two vectors, then calculating the cross product.
x = [0,0,1,1]
y = [0,1,0,1]
z = [1,0,0,1]
v1 = np.array([x,y,z])
x = [1,1,0,1]
y = [1,0,1,1]
z = [0,1,1,1]
v2 = np.array([x,y,z])
result = []
for i in range(0, len(x)):
a = [v1[0][i], v1[1][i], v1[2][i]]
b = [v2[0][i], v2[1][i], v2[2][i]]
result.append(np.cross(a,b))
result
>>>
[
array([-1, 1, 0]),
array([ 1, 0, -1]),
array([ 0, -1, 1]),
array([ 0, 0, 0])
]
I've tried to understand this question and answer to generalise it, but failed:
- Element wise cross product of vectors contained in 2 arrays with Python
np.cross can work with 2D arrays too, you just need to specify the right axes:
np.cross(v1,v2, axisa=0, axisb=0)
array([[-1, 1, 0],
[ 1, 0, -1],
[ 0, -1, 1],
[ 0, 0, 0]])
Suppose there are two lists with the same dimension, let's say [0,1,2,2] and [0,1,2,2]. How should I compare two elements at every position and return a matrix which, in this case, is [[1,0,0,0],[0,1,0,0],[0,0,1,1],[0,0,1,1]]?
i.e. f(x,y)=1 if x=y else 0, and x,y are come from the above two lists respectively.
Already tried the for-loop block with python, which is rather slow when dealing with large lists.
Is there any efficient way of handling this problem with numpy instead of using codes below:
for i in list_1:
for j in list_2:
1 if i==j else 0
You can use np.equal to compare the elements and use np.where to convert it into 1 or 0.:
import numpy as np
a = np.array([0,1,2,2])
b = np.array([0,1,2,2])
z = np.where(np.equal(a, b[:,np.newaxis]), 1, 0)
Output:
array([[1, 0, 0, 0],
[0, 1, 0, 0],
[0, 0, 1, 1],
[0, 0, 1, 1]])
Or as #Onyambu suggested, you can do
z = (a==b[:,None]).astype(int)
Try the below code, I hope this would help.
print([1 if i==j else 0 for i,j in zip([0,1,2,2],[0,1,0,2])])
Ouput :
[1, 1, 0, 1]
So given a 2d numpy array consisting of ones and zeros, I want to find every index where it is a value of one and where either to its top, left, right, or bottom consists of a zero. For example in this array
0 0 0 0 0
0 0 1 0 0
0 1 1 1 0
0 0 1 0 0
0 0 0 0 0
I only want coordinates for (1,2), (2,1), (2,3) and (3,2) but not for (2,2).
I have created code that works and creates two lists of coordinates, similar to the numpy nonzero method, however I don't think its very "pythonic" and I was hoping there was a better and more efficient way to solve this problem. (*Note this only works on arrays padded by zeros)
from numpy import nonzero
...
array= ... # A numpy array consistent of zeros and ones
non_zeros_pairs=nonzero(array)
coordinate_pairs=[[],[]]
for x, y in zip(temp[0],temp[1]):
if array[x][y+1]==0 or array[x][y-1]==0 or array[x+1][y]==0 or array[x-1][y]==0:
coordinate_pairs[0].append(x)
coordinate_pairs[1].append(y)
...
If there exist methods in numpy that can handle this for me, that would be awesome. If this question has already been asked/answered on stackoverflow before, I will gladly remove this, I just struggled to find anything. Thank You.
Setup
import scipy.signal
import numpy as np
a = np.array([[0, 0, 0, 0, 0],
[0, 0, 1, 0, 0],
[0, 1, 1, 1, 0],
[0, 0, 1, 0, 0],
[0, 0, 0, 0, 0]])
Create a window which matches the four directions from each value, and convolve. Then, you can check if elements are 1, and if their convolution is less than 4, since a value ==4 means that the value was surrounded by 1s
window = np.array([[0, 1, 0],
[1, 0, 1],
[0, 1, 0]])
m = scipy.signal.convolve2d(a, window, mode='same', fillvalue=1)
v = np.where(a & (m < 4))
list(zip(*v))
[(1, 2), (2, 1), (2, 3), (3, 2)]
trying to come up with a method to perform load combinations and transient load patterning for structural/civil engineering applications.
without patterning it's fairly simple:
list of load results = [[d],[t1],...,[ti]], where [ti] = transient load result as a numpy array = A
list of combos = [[1,0,....,0],[0,1,....,1], [dfi, tf1,.....,tfi]] , where tfi = code load factor for transient load = B
in python this works as numpy.dot(A,B)
so my issue arises where:
`list of load results = [[d],[t1],.....[ti]]`, where [t1] = [[t11]......[t1i]] for i pattern possibilities and [t1i] = numpy array
so I have a nested array within another array and want to multiply by a matrix of load combinations. Is there a way to implement this in one matrix operation, I can come up with a method by looping the pattern possibilities then a dot product with the load combos, but this is computationally expensive. Any thoughts?
Thanks
for an example not considering patterning see: https://github.com/buddyd16/Structural-Engineering/blob/master/Analysis/load_combo_test.py
essential I need a method that gives similar results assuming that for loads = np.array([[D],[Ex],[Ey],[F],[H],[L],[Lr],[R],[S],[Wx],[Wy]]) --> [L],[Lr],[R],[S] are actually nested arrays ie if D = 1x500 array/vector, L, Lr, R, or S could = 100x500 array.
my simple solution is:
combined_pattern = []
for pattern in load_patterns:
loads = np.array([[D],[Ex],[Ey],[F],[H],[L[pattern]],[Lr[pattern]],[R[pattern]],[S[pattern]],[Wx],[Wy]])
combined_pattern.append(np.dot(basic_factors, loads))
Simpler Example:
import numpy as np
#Simple
A = np.array([1,0,0])
B = np.array([0,1,0])
C = np.array([0,0,1])
Loads = np.array([A,B,C])
Factors = np.array([[1,1,1],[0.5,0.5,0.5],[0.25,0.25,0.25]])
result = np.dot(Factors, Loads)
# Looking for a faster way to accomplish the below operation
# this works but will be slow for large data sets
# bi can be up to 1x5000 in size and i can be up to 500
A = np.array([1,0,0])
b1 = np.array([1,0,0])
b2 = np.array([0,1,0])
b3 = np.array([0,0,1])
B = np.array([b1,b2,b3])
C = np.array([0,0,1])
result_list = []
for load in B:
Loads = np.array([A,load,C])
Factors = np.array([[1,1,1],[0.5,0.5,0.5],[0.25,0.25,0.25]])
result = np.dot(Factors, Loads)
result_list.append(result)
edit: Had Factors and Loads reversed in the np.dot().
In your simple example, the array shapes are:
In [2]: A.shape
Out[2]: (3,)
In [3]: Loads.shape
Out[3]: (3, 3)
In [4]: Factors.shape
Out[4]: (3, 3)
In [5]: result.shape
Out[5]: (3, 3)
The rule in dot is that the last dimension of Loads pairs with the 2nd to the last of Factors
result = np.dot(Loads,Factors)
(3,3) dot (3,3) => (3,3) # 3's in common
(m,n) dot (n,l) => (m,l) # n's in common
In the iteration, A,load and C are all (3,) and Loads is (3,3).
result_list is a list of 3 (3,3) arrays, and np.array(result_list) would be (3,3,3).
Let's make a 3d array of all the Loads:
In [16]: Bloads = np.array([np.array([A,load,C]) for load in B])
In [17]: Bloads.shape
Out[17]: (3, 3, 3)
In [18]: Bloads
Out[18]:
array([[[1, 0, 0],
[1, 0, 0],
[0, 0, 1]],
[[1, 0, 0],
[0, 1, 0],
[0, 0, 1]],
[[1, 0, 0],
[0, 0, 1],
[0, 0, 1]]])
I can easily do a dot of this Bloads and Factors with einsum:
In [19]: np.einsum('lkm,mn->lkn', Bloads, Factors)
Out[19]:
array([[[1. , 1. , 1. ],
[1. , 1. , 1. ],
[0.25, 0.25, 0.25]],
[[1. , 1. , 1. ],
[0.5 , 0.5 , 0.5 ],
[0.25, 0.25, 0.25]],
[[1. , 1. , 1. ],
[0.25, 0.25, 0.25],
[0.25, 0.25, 0.25]]])
einsum isn't the only way, but it's the easiest way (for me) to keep track of dimensions.
It's even easier to keep dimensions straight if they differ. Here they are all 3, so it's hard to keep them separate. But if B was (5,4) and Factors (4,2), then Bloads would be (5,3,4), and the einsum result (5,3,2) (the size 4 dropping out in the dot).
Constructing Bloads without a loop is a bit trickier, since the rows of B are interleaved with A and C.
In [38]: np.stack((A[None,:].repeat(3,0),B,C[None,:].repeat(3,0)),1)
Out[38]:
array([[[1, 0, 0],
[1, 0, 0],
[0, 0, 1]],
[[1, 0, 0],
[0, 1, 0],
[0, 0, 1]],
[[1, 0, 0],
[0, 0, 1],
[0, 0, 1]]])
To understand this test the subexpressions, e.g. A[None,:], the repeat etc.
Equivalently:
np.array((A[None,:].repeat(3,0),B,C[None,:].repeat(3,0))).transpose(1,0,2)
Let's say I have a numpy array
a b c
A = i j k
u v w
I want to compare the value central element with some of its eight neighbor elements (along the axis or along the diagonal). Is there any faster way except the nested for loop (it's too slow for big matrix)?
To be more specific, what I want to do is compare value of element with it's neighbors and assign new values.
For example:
if (j == 1):
if (j>i) & (j>k):
j = 999
else:
j = 0
if (j == 2):
if (j>c) & (j>u):
j = 999
else:
j = 0
...
something like this.
Your operation contains lots of conditionals, so the most efficient way to do it in the general case (any kind of conditionals, any kind of operations) is using loops. This could be done efficiently using numba or cython. In special cases, you can implement it using higher level functions in numpy/scipy. I'll show a solution for the specific example you gave, and hopefully you can generalize from there.
Start with some fake data:
A = np.asarray([
[1, 1, 1, 2, 0],
[1, 0, 2, 2, 2],
[0, 2, 0, 1, 0],
[1, 2, 2, 1, 0],
[2, 1, 1, 1, 2]
])
We'll find locations in A where various conditions apply.
1a) The value is 1
1b) The value is greater than its horizontal neighbors
2a) The value is 2
2b) The value is greater than its diagonal neighbors
Find locations in A where the specified values occur:
cond1a = A == 1
cond2a = A == 2
This gives matrices of boolean values, of the same size as A. The value is true where the condition holds, otherwise false.
Find locations in A where each element has the specified relationships to its neighbors:
# condition 1b: value greater than horizontal neighbors
f1 = np.asarray([[1, 0, 1]])
cond1b = A > scipy.ndimage.maximum_filter(
A, footprint=f1, mode='constant', cval=-np.inf)
# condition 2b: value greater than diagonal neighbors
f2 = np.asarray([
[0, 0, 1],
[0, 0, 0],
[1, 0, 0]
])
cond2b = A > scipy.ndimage.maximum_filter(
A, footprint=f2, mode='constant', cval=-np.inf)
As before, this gives matrices of boolean values indicating where the conditions are true. This code uses scipy.ndimage.maximum_filter(). This function iteratively shifts a 'footprint' to be centered over each element of A. The returned value for that position is the maximum of all elements for which the footprint is 1. The mode argument specifies how to treat implicit values outside boundaries of the matrix, where the footprint falls off the edge. Here, we treat them as negative infinity, which is the same as ignoring them (since we're using the max operation).
Set values of the result according to the conditions. The value is 999 if conditions 1a and 1b are both true, or if conditions 2a and 2b are both true. Else, the value is 0.
result = np.zeros(A.shape)
result[(cond1a & cond1b) | (cond2a & cond2b)] = 999
The result is:
[
[ 0, 0, 0, 0, 0],
[999, 0, 0, 999, 999],
[ 0, 0, 0, 999, 0],
[ 0, 0, 999, 0, 0],
[ 0, 0, 0, 0, 999]
]
You can generalize this approach to other patterns of neighbors by changing the filter footprint. You can generalize to other operations (minimum, median, percentiles, etc.) using other kinds of filters (see scipy.ndimage). For operations that can be expressed as weighted sums, use 2d cross correlation.
This approach should be much faster than looping in python. But, it does perform unnecessary computations (for example, it's only necessary to compute the max when the value is 1 or 2, but we're doing it for all elements). Looping manually would let you avoid these computations. Looping in python would probably be much slower than the code here. But, implementing it in numba or cython would probably be faster because these tools generate compiled code.
I used numpy's:
concatenate to pad with zeroes
dstack and roll to align correctly
Apply custom_roll twice along different dimensions and subtract original.
import numpy as np
def custom_roll(a, axis=0):
n = 3
a = a.T if axis==1 else a
pad = np.zeros((n-1, a.shape[1]))
a = np.concatenate([a, pad], axis=0)
ad = np.dstack([np.roll(a, i, axis=0) for i in range(n)])
a = ad.sum(2)[1:-1, :]
a = a.T if axis==1 else a
return a
Consider the following ndarray:
A = np.arange(25).reshape(5, 5)
A
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]])
sum_of_eight_around_me = custom_roll(custom_roll(A), axis=1) - A
sum_of_eight_around_me
array([[ 12., 20., 25., 30., 20.],
[ 28., 48., 56., 64., 42.],
[ 53., 88., 96., 104., 67.],
[ 78., 128., 136., 144., 92.],
[ 52., 90., 95., 100., 60.]])