Related
If I have a MxN numpy array denoted arr, I wish to index over all elements and adjust the values like so
for m in range(arr.shape[0]):
for n in range(arr.shape[1]):
arr[m, n] += x**2 * np.cos(m) * np.sin(n)
Where x is a random float.
Is there a way to broadcast this over the entire array without needing to loop? Thus, speeding up the run time.
You are just adding zeros, because sin(2*pi*k) = 0 for integer k.
However, if you want to vectorize this, the function np.meshgrid could help you.
Check the following example, where I removed the 2 pi in the trigonometric functions to add something unequal zero.
x = 2
arr = np.arange(12, dtype=float).reshape(4, 3)
n, m = np.meshgrid(np.arange(arr.shape[1]), np.arange(arr.shape[0]), sparse=True)
arr += x**2 * np.cos(m) * np.sin(n)
arr
Edit: use the sparse argument to reduce memory consumption.
You can use nested generators of two-dimensional arrays:
import numpy as np
from random import random
x = random()
n, m = 10,20
arr = [[x**2 * np.cos(2*np.pi*j) * np.sin(2*np.pi*i) for j in range(m)] for i in range(n)]
In [156]: arr = np.ones((2, 3))
Replace the range with arange:
In [157]: m, n = np.arange(arr.shape[0]), np.arange(arr.shape[1])
And change the first array to (2,1) shape. A (2,1) array broadcasts with a (3,) to produce a (2,3) result.
In [158]: A = 0.23**2 * np.cos(m[:, None]) * np.sin(n)
In [159]: A
Out[159]:
array([[0. , 0.04451382, 0.04810183],
[0. , 0.02405092, 0.02598953]])
In [160]: arr + A
Out[160]:
array([[1. , 1.04451382, 1.04810183],
[1. , 1.02405092, 1.02598953]])
The meshgrid suggested in the accepted answer does the same thing:
In [161]: np.meshgrid(m, n, sparse=True, indexing="ij")
Out[161]:
[array([[0],
[1]]),
array([[0, 1, 2]])]
This broadcasting may be clearer with:
In [162]: m, n
Out[162]: (array([0, 1]), array([0, 1, 2]))
In [163]: m[:, None] * 10 + n
Out[163]:
array([[ 0, 1, 2],
[10, 11, 12]])
I want to calculate the following:
but I have no idea how to do this in python, I do not want to implement this manually but use a predefined function for this, something from numpy for example.
But numpy seems to ignore that x.T should be transposed.
Code:
import numpy as np
x = np.array([1, 5])
print(np.dot(x, x.T)) # = 26, This is not the matrix it should be!
While your vectors are defined as 1-d arrays, you can use np.outer:
np.outer(x, x.T)
> array([[ 1, 5],
> [ 5, 25]])
Alternatively, you could also define your vectors as matrices and use normal matrix multiplication:
x = np.array([[1], [5]])
x # x.T
> array([[ 1, 5],
> [ 5, 25]])
You can do:
x = np.array([[1], [5]])
print(np.dot(x, x.T))
Your original x is of shape (2,), while you need a shape of (2,1). Another way is reshaping your x:
x = np.array([1, 5]).reshape(-1,1)
print(np.dot(x, x.T))
.reshape(-1,1) reshapes your array to have 1 column and implicitely takes care of number of rows.
output:
[[ 1 5]
[ 5 25]]
np.matmul(x[:, np.newaxis], [x])
I'm trying to insert elements to an empty 2d numpy array. However, I am not getting what I want.
I tried np.hstack but it is giving me a normal array only. Then I tried using append but it is giving me an error.
Error:
ValueError: all the input arrays must have same number of dimensions
randomReleaseAngle1 = np.random.uniform(20.0, 77.0, size=(5, 1))
randomVelocity1 = np.random.uniform(40.0, 60.0, size=(5, 1))
randomArray =np.concatenate((randomReleaseAngle1,randomVelocity1),axis=1)
arr1 = np.empty((2,2), float)
arr = np.array([])
for i in randomArray:
data = [[170, 68.2, i[0], i[1]]]
df = pd.DataFrame(data, columns = ['height', 'release_angle', 'velocity', 'holding_angle'])
test_y_predictions = model.predict(df)
print(test_y_predictions)
if (np.any(test_y_predictions == 1)):
arr = np.hstack((arr, np.array([i[0], i[1]])))
arr1 = np.append(arr1, np.array([i[0], i[1]]), axis=0)
print(arr)
print(arr1)
I wanted to get something like
[[1.5,2.2],
[3.3,4.3],
[7.1,7.3],
[3.3,4.3],
[3.3,4.3]]
However, I'm getting
[56.60290125 49.79106307 35.45102444 54.89380834 47.09359271 49.19881675
22.96523274 44.52753514 67.19027156 54.10421167]
The recommended list append approach:
In [39]: alist = []
In [40]: for i in range(3):
...: alist.append([i, i+10])
...:
In [41]: alist
Out[41]: [[0, 10], [1, 11], [2, 12]]
In [42]: np.array(alist)
Out[42]:
array([[ 0, 10],
[ 1, 11],
[ 2, 12]])
If we start with a empty((2,2)) array:
In [47]: arr = np.empty((2,2),int)
In [48]: arr
Out[48]:
array([[139934912589760, 139934912589784],
[139934871674928, 139934871674952]])
In [49]: np.concatenate((arr, [[1,10]],[[2,11]]), axis=0)
Out[49]:
array([[139934912589760, 139934912589784],
[139934871674928, 139934871674952],
[ 1, 10],
[ 2, 11]])
Note that empty does not mean the same thing as the list []. It's a real 2x2 array, with 'unspecified' values. And those values remain when we add other arrays to it.
I could start with an array with a 0 dimension:
In [51]: arr = np.empty((0,2),int)
In [52]: arr
Out[52]: array([], shape=(0, 2), dtype=int64)
In [53]: np.concatenate((arr, [[1,10]],[[2,11]]), axis=0)
Out[53]:
array([[ 1, 10],
[ 2, 11]])
That looks more like the list append approach. But why start with the (0,2) array in the first place?
np.concatenate takes a list of arrays (or lists that can be made into arrays). I used nested lists that make (1,2) arrays. With this I can join them on axis 0.
Each concatenate makes a new array. So if done iteratively it is more expensive than the list append.
np.append just takes 2 arrays and does a concatenate. So doesn't add much. hstack tweaks shapes and joins on the 2nd (horizontal) dimension. vstack is another variant. But they all end up using concatenate.
With the hstack method, you can just reshape after you get the final array:
arr = arr.reshape(-1, 2)
print(arr)
The other method can be more easily done in a similar way:
arr1 = np.append(arr1, np.array([i[0], i[1]]) # in the loop
arr1 = arr1.reshape(-1, 2)
print(arr1)
I read about how important it is to preallocate a numpy array. In my case I am, however, not sure how to do this. I want to preallocate an nxm matrix. That sounds simple enough
M = np.zeros((n,m))
However, what if my matrix is a matrix of matrices? So what if each of these nxm elements is actually of the form
np.array([[t], [x0,x1,x2], [y0,y1,y2]])
I know that in that case, M would have the shape (n,m,3).
As an example, later I want to have something like this
[[[[0], [0,1,2], [3,4,5]],
[[1], [10,11,12], [13,14,15]]],
[[[0], [100,101,102], [103,104,105]],
[[1], [110,111,112], [113,114,115]]]]
I tried simply doing
M = np.zeros((2,2,3))
but then
M[0,0,:] = np.array([[0], [0,1,2], [3,4,5]])
will give me an error
ValueError: setting an array element with a sequence.
Can I not preallocate this monster? Or should I approach this in a completely different way?
Thanks for your help
You have to make sure you preallocate the correct number of dimensions and elements along each dimension to use simple assignments to fill it.
For example you want to save 3 2x3 matrices:
number_of_matrices = 3
matrix_dim_1 = 2
matrix_dim_2 = 3
M = np.empty((number_of_matrices, matrix_dim_1, matrix_dim_2))
M[0] = np.array([[ 0, 1, 2], [ 3, 4, 5]])
M[1] = np.array([[100, 101, 102], [103, 104, 105]])
M[2] = np.array([[ 10, 11, 12], [ 13, 14, 15]])
M
#array([[[ 0., 1., 2.], # matrix 1
# [ 3., 4., 5.]],
#
# [[ 100., 101., 102.], # matrix 2
# [ 103., 104., 105.]],
#
# [[ 10., 11., 12.], # matrix 3
# [ 13., 14., 15.]]])
You're approach contains some problems. The array you want to save is not a valid ndimensional numpy array:
np.array([[0], [0,1,2], [3,4,5]])
# array([[0], [0, 1, 2], [3, 4, 5]], dtype=object)
# |----!!----|
# ^-------^----------^ 3 items in first dimension
# ^ 1 item in first item of 2nd dim
# ^--^--^ 3 items in second item of 2nd dim
# ^--^--^ 3 items in third item of 2nd dim
It just creates an 3 item array containing python list objects. You probably want to have an array containing numbers so you need to care about dimensions. Your np.array([[0], [0,1,2], [3,4,5]]) could be a 3x1 array or a 3x3 array, numpy doesn't know what to do in this case and saves it as objects (the array now has only 1 dimension!).
The other problem is that you want to set one element of the preallocated array with another array that contains more than one element. This is not possible (except you already have an object-array). You have two options here:
Fill as many elements in the preallocated array as are required by the array:
M[0, :, :] = np.array([[0,1,2], [3,4,5]])
# ^--------------------^--------^ First dimension has 2 items
# ^---------------^-^-^ Second dimension has 3 items
# ^------------------------^-^-^ dito
# if it's the first dimension you could also use M[0]
Create a object array and set the element (not recommended, you loose most of the advantages of numpy arrays):
M = np.empty((3), dtype='object')
M[0] = np.array([[0,1,2], [3,4,5]])
M[1] = np.array([[0,1,2], [3,4,5]])
M[2] = np.array([[0,1,2], [3,4,5]])
M
#array([array([[0, 1, 2],
# [3, 4, 5]]),
# array([[0, 1, 2],
# [3, 4, 5]]),
# array([[0, 1, 2],
# [3, 4, 5]])], dtype=object)
If you know you will only store values t, y, x for each point in n,m then it may be easier, and faster computationally, to have three numpy arrays.
So:
M_T = np.zeros((n,m))
M_Y = np.zeros((n,m))
M_X = np.zeros((n,m))
I believe you can now type 'normal' python operators to do array logic, such as:
MX = np.ones((n,m))
MY = np.ones((n,m))
MT = MX + MY
MT ** MT
_ * 7.5
By defining array-friendly functions (similarly to MATLAB) you will get a big speed increase for calculations.
Of course if you need more variables at each point then this may become unwieldy.
I have two arrays of the same dimension:
a = np.array([ 1, 1, 2, 0, 0, 1])
b = np.array([50, 51, 6, 10, 3, 2])
I want to sum the elements of b according to the indices in a.
The ith element of the matrix I want will be the sum of all values b[j] such that a[j]==i.
So the result should be a 3-dimensional array of [10 + 3, 50 + 51 + 2, 6]
Is there a numpy way to do this? I have some very large arrays that I need to sum like this over multiple dimensions, so it would NOT be convenient to to have to perform explicit loops.
numpy.bincount has a weights parameter which does just what you need:
In [36]: np.bincount(a, weights=b)
Out[36]: array([ 13., 103., 6.])
In case you are not using numpy, something as simple as :
res = [0]*len(set(a))
for i, v in enumerate(b):
res[a[i]] += v
Assuming the indices in a are always 0-based and a continuous sequence.