Normally: a * b == np.multiply(a, b)
but in this case :
a=np.matrix(([1,2,3],[1,2,3],[1,2,3]))
b = np.array(([1,2,3]))
print( a.dot(b))
print(np.multiply(a,b))
print(a * b)
I have a problem:
[[14 14 14]]
[[1 4 9]
[1 4 9]
[1 4 9]]
Traceback (most recent call last):
File "C:\Users\FAROUQ\.spyder-py3\untitled0.py", line 28, in <module>
print(a * b)
ValueError: shapes (3,3) and (1,3) not aligned: 3 (dim 1) != 1 (dim 0)
Please, could you explain to me why ?
np.dot() product is the dot product or scaler product of vectors/matrixes. If you don't know what dot product is I recommend you to learn about it. (Linear Algebra / Calculus).
Algebraically, the dot product is the sum of the products of the corresponding entries of the two sequences of numbers.
For example:
a = [3, 2, 4]
b = [5, 7, 6]
np.dot(a, b) = 3 * 5 + 5 * 7 + 4 * 6 = 15 + 14 + 24 = 53
* in NumPy is the element-wise product of two arrays.
For example:
a = np.array([3, 2, 4])
b = np.array([5, 7, 6])
a * b = [3 * 5, 2 * 7, 4 * 6] = [15, 14, 24]
You should replace np.matrix by np.array to get the same result!
a = np.array(([1,2,3],[1,2,3],[1,2,3]))
b = np.array(([1,2,3]))
print( a.dot(b))
print(np.multiply(a,b))
print(a * b)
Additional Information
1. element-wise product: a*b or element-wise matrix multiplication np.multiply(a,b) are the same!
2. dot product : np.dot(a,b) or a.dot(b)
Examples:
Related
Equivalent transform for vectorized solution
For a given symmetric 4x4 matrix Q and a 3x4 matrix P the 3x3 matrix C is obtained through
C=P # Q # P.T
It can be shown that the output C will be symmetric again. The same problem can be formulated using only the unique elements in Q and C exploiting their symmetry. To do so, the matrices are vectorized as seen below.
I want to construct a matrix B that maps the vectorized matrices onto each other like so:
c = B # q
B must be a 6x10 and should be constructable from P only. How can I get B from P?
I tried this, but it doesnt seem to work. Maybe someone has experienced a similar problem?
import numpy as np
def vectorize(A, ord='c'):
"""
Symmetric matrix to vector e.g:
[[1, 2, 3],
[2, 4, 5],
[3, 5, 6]] -> [1, 2, 3, 4, 5, 6] (c-order, row-col)
-> [1, 2, 4, 3, 5, 6] (f-order, col-row)
"""
# upper triangle mask
m = np.triu(np.ones_like(A, dtype=bool)).flatten(order=ord)
return A.flatten(order=ord)[m]
def B(P):
B = np.zeros((6, 10))
counter = 0
# the i,j entry in C depends on the i, j columns in P
for i in range(3):
for j in range(i, 3):
coeffs = np.outer(P[i], P[j])
B[counter] = vectorize(coeffs)
counter += 1
return B
if __name__ == '__main__':
# original transform
P = np.arange(12).reshape((3, 4))
# calculated transform for vectorized matrix
_B = B(P)
# some random symmetric matrix
Q = np.array([[1, 2, 3, 4],
[2, 5, 6, 7],
[3, 6, 8, 9],
[4, 7, 9, 10]])
# if B is an equivilant transform to P, these should be similar
C = P # Q # P.T
c = _B # vectorize(Q)
print(f"q: {vectorize(Q)}\n"
f"C: {vectorize(C)}\n"
f"c: {c}")
Output:
q: [ 1 2 3 4 5 6 7 8 9 10]
C: [ 301 949 2973 1597 4997 8397]
c: [ 214 542 870 1946 3154 5438] <-- not the same
import numpy as np
def vec_from_mat(A, order='c'):
"""
packs the unique elements of symmetric matrix A into a vector
:param A: symmetric matrix
:return:
"""
return A[np.triu_indices(A.shape[0])].flatten(order=order)
def B_from_P(P):
"""
returns a 6x10 matrix that maps the 10 unique elements of a symmetric 4x4 matrix Q on the 6 unique elements of a
3x3 matrix C to linearize the equation C=PTQP to c=Bv
:param P: 3x4 matrix
:return: B with shape (6, 10)
"""
n, m = P.shape
b1, b2 = (n * (n + 1) // 2), (m * (m + 1) // 2)
B = np.zeros((b1, b2))
for a, (i, j) in enumerate(zip(*np.triu_indices(n))):
coeffs = np.outer(P[i], P[j])
# collect coefficients from lower and upper triangle of symmetric matrix
B[a] = vec_from_mat(coeffs) + vec_from_mat(np.triu(coeffs.T, k=1))
return B
I have 3 arrays down below a and b combine to make a_and_b. a is multiplied by a_multiplier and b gets multiplied by b_multiplier. I am trying to make a ternary statement where it checks the indexes of a_and_b matches with either a or b then it does the necessary multiplications with a_multiplier or b_multiplier. How would I be able to modify a_and_b ternary function so that the Expected Output works?
import numpy as np
a_multiplier = 3
b_multiplier = 5
a = np.array([5,32,1,4])
b = np.array([1,5,11,3])
a_and_b = np.array([5,1,32,5,1,11,4,3])
a_and_b = a_and_b[::2] * 3,a_and_b[1::2] * 5 if a_and_b[::2] == a
else a_and_b[::2] * 5,a_and_b[1::2] * 3
Expected Output:
[15, 5, 96, 25, 3, 55, 12, 15]
IIUC, you could do directly:
import numpy as np
a_multiplier = 3
b_multiplier = 5
a = np.array([5, 32, 1, 4])
b = np.array([1, 5, 11, 3])
res = np.vstack([a * a_multiplier, b * b_multiplier]).flatten("F")
print(res)
Output
[15 5 96 25 3 55 12 15]
I am an undergraduate student who loves programming. I encountered a problem today and I don't know how to solve this problem.
I looked for "Python - string to matrix representation" (Python - string to matrix representation) for help, but I am still confused about this problem.
The problem is in the following:
Given a string of whitespace separated numbers, create an nxn matrix (a 2d list where with the same number of columns as rows)and return it. The string will contain a perfect square number of integers. The int() and split() functions may be useful.
Example:
Input: '1 2 3 4 5 6 7 8 9'
Output: [[1,2,3],[4,5,6],[7,8,9]]
Example 2:
Input: '1'
Output: [[1]]
My answer:
import numpy as np
def string_to_matrix(str_in):
str_in_split = str_in.split()
answer = []
for element in str_in_split:
newarray = []
for number in element.split():
newarray.append(int(number))
answer.append(newarray)
print (answer)
The test results are in the following:
Traceback (most recent call last):
File "/grade/run/test.py", line 20, in test_whitespace
self.assertEqual(string_to_matrix('1 2 3 4'), [[1,2],[3,4]])
AssertionError: None != [[1, 2], [3, 4]]
Stdout:
[[4]]
as well as
Traceback (most recent call last):
File "/grade/run/test.py", line 15, in test_small
self.assertEqual(string_to_matrix('1 2 3 4'), [[1,2],[3,4]])
AssertionError: None != [[1, 2], [3, 4]]
Stdout:
[[4]]
as well as
Traceback (most recent call last):
File "/grade/run/test.py", line 10, in test_one
self.assertEqual(string_to_matrix('1'), [[1]])
AssertionError: None != [[1]]
Stdout:
[[1]]
as well as
Traceback (most recent call last):
File "/grade/run/test.py", line 25, in test_larger
self.assertEqual(string_to_matrix('4 3 2 1 8 7 6 5 12 11 10 9 16 15 14 13'), [[4,3,2,1], [8,7,6,5], [12,11,10,9], [16,15,14,13]])
AssertionError: None != [[4, 3, 2, 1], [8, 7, 6, 5], [12, 11, 10, 9], [16, 15, 14, 13]]
Stdout:
[[13]]
I am still confused how to solve this problem. Thank you very much for your help!
Assuming you don't want numpy and want to use a list of lists:
def string_to_matrix(str_in):
nums = str_in.split()
n = int(len(nums) ** 0.5)
return list(map(list, zip(*[map(int, nums)] * n)))
nums = str_in.split() splits by any whitespace, n is the side length of the result, map(int, nums) converts the numbers to integers (from strings), zip(*[map(int, nums)] * n) groups the numbers in groups of n, list(map(list, zip(*[map(int, nums)] * n))) converts the tuples produced by zip into lists.
Assuming you want to make this dynamic.
str_in = '1 2 3 4 5 6 7 8 9'
a = str_in.split(" ")
r_shape = int(math.sqrt(len(a)))
np.array([int(x) for x in a]).reshape(r_shape, r_shape)
Use split, create the 1D numpy array, then use reshape:
>>> s = '1 2 3 4 5 6 7 8 9'
>>> np.array([s.split(), dtype=int).reshape(3,3)
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
If you don't know the size of the array, but you know it's square (same width / height), then you can use math.sqrt to get the inputs for reshape:
>>> import math
>>> s = '1 2 3 4 5 6 7 8 9'
>>> arr = np.array(s.split(), dtype=int)
>>> size = int(math.sqrt(len(arr)))
>>> arr.reshape(size, size)
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
Given that you will always get perfect square number of ints:
import numpy as np
input_strings = '1 2 3 4 5 6 7 8 9'
arr = np.array(input_strings.split(), dtype=int)
n = int(len(arr) ** 0.5)
arr.reshape(n, n)
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
Note: in your case, str.split is better off without explicit sep in order to work fine with multiple whitespaces in between the digits.
import numpy as np
def string_to_matrix(str_in):
str_in_split = str_in.split()
numbers = list(map(int, str_in_split))
size = r_shape = int(np.sqrt(len(numbers)))
return np.array(numbers).reshape(r_shape, r_shape)
This is why you always got: AssertionError: None != ...
assertEqual(A, string_to_matrix("...")) verifies if A is equals to the value returned by string_to_matrix. In your code you don't return anything so it is None
The other issue is how you splitted the string, the easier options is to split everything and convert to number, and then reshape to sqrt(number of elements). This assumes that input length can be splited to form a nxn matrix
import math
string = "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16"
stringItems = string.split(" ")
numberOfItems = len(stringItems)
if math.sqrt(numberOfItems) - int(math.sqrt(numberOfItems)) == 0:
width = int(math.sqrt(numberOfItems))
array = []
finalArray = []
for i in range (0, width):
for j in range (0, width):
array.insert(j, stringItems[0])
stringItems.pop(0)
finalArray.insert(i, array)
array = []
print finalArray
else:
print "I require a string with length equal to a square number please"
Hey so I'm working on this code for a material analysis. I have a matrix generated for each layer of the material and I want to save each of these matrices as their own element. The way I was doing this was by saving it to a dictionary. I then form one matrix by summing all the values of the dictionary. Now I do this for three different conditions which leaves me with 3 matrices: A, B, and D. I want to make a matrix of all of these so that it looks like:
| A B |
| B D |
However I can't get it to print properly as it always says matrix: then one of the matrices such as A. It prints the second matrix, B, on the third line where A ended instead of being next to A. I also need to perform future operations on this massive matrix so I'm wondering what the best way to go about that would be. This is a part of my code:
Qbars = {}
for i in plies:
Qbar11 = Q11 * math.cos(float(thetas[j]))**4 + Q22 *math.sin(float(thetas[j]))**4 + \
2 * (Q12 + 2 * Q66) * math.sin(float(thetas[j]))**2 * math.cos(float(thetas[j]))**2
Qbar22 = Q11 * math.sin(float(thetas[j]))**4 + Q22 *math.cos(float(thetas[j]))**4 + \
2 * (Q12 + 2 * Q66) * math.sin(float(thetas[j]))**2 * math.cos(float(thetas[j]))**2
Qbar12 = (Q11 + Q22 - 4 * Q66) * math.sin(float(thetas[j]))**2 * \
math.cos(float(thetas[j]))**2 + Q12 * (math.cos(float(thetas[j]))**4 + \
math.sin(float(thetas[j]))**4)
Qbar66 = (Q11 + Q22 - 2 * Q12 - 2 * Q66) * math.sin(float(thetas[j]))**2 * \
math.cos(float(thetas[j])) **2 + Q66 * (math.sin(float(thetas[j]))**4 + \
math.cos(float(thetas[j]))**4)
Qbar16 = (Q11 - Q12 - 2 * Q66) * math.cos(float(thetas[j]))**3 * \
math.sin(float(thetas[j])) - (Q22 - Q12 - 2 * Q66) * math.cos(float(thetas[j])) * \
math.sin(float(thetas[j]))**3
Qbar26 = (Q11 - Q12 - 2 * Q66) * math.cos(float(thetas[j])) * \
math.sin(float(thetas[j]))**3 - (Q22 - Q12 - 2 * Q66) * \
math.cos(float(thetas[j]))**3 * math.sin(float(thetas[j]))
Qbar = np.matrix ([[Qbar11, Qbar12, Qbar16], [Qbar12, Qbar22, Qbar26], \
[Qbar16, Qbar26, Qbar66]])
Qbars[i] = Qbar
if len(thetas) == 1:
j = 0
else:
j = j + 1
k=0
Alist = {}
for i in plies:
Alist[i] = Qbars[i].dot(h[k])
if len(h) == 1:
k = 0
else:
k = k + 1
A = sum(Alist.values())
ABD = ([A, B],[B, D])
print ABD
One of the next operations I intend to perform would be to multiply the matrix by a 6x1 array that would look like such:
| Nx | | A A A B B B |
| Ny | | A A A B B B |
| Nxy| | A A A B B B |
------ * ----------------
| Mx | | B B B D D D |
| My | | B B B D D D |
| Mxy| | B B B D D D |
What would be the best way to go about doing this?
EDIT: I made this shorter code to reproduce what I'm dealing with, I couldn't think of how to make it even smaller.
import os
import numpy as np
import math
os.system('cls')
ang = raw_input("ENTER 0 (SPACE) 45 ")
thetas = [int(i) for i in ang.split()]
x = 40
h = [3, 5]
y = [1,2]
j = 0
Qbars = {}
for i in y:
theta = [thetas[j] * math.pi / 180]
Q = math.sin(float(thetas[j]))
Qbar = np.matrix ([[Q, Q, Q], [Q, Q, Q], [Q, Q, Q]])
Qbars[i] = Qbar
if len(thetas) == 1:
j = 0
else:
j = j + 1
print Qbars
k=0
Alist = {}
for i in y:
Alist[i] = Qbars[i].dot(h[k])
if len(h) == 1:
k = 0
else:
k = k + 1
A = sum(Alist.values())
AAAA = ([A, A], [A, A])
print AAAA
test = raw_input("Press ENTER to close")
As others have noted, the matrix class is pretty much deprecated by now. They are more limited than ndarrays, with very little additional functionality. The main reason why people prefer to use numpy matrices is that linear algebra (in particular, matrix multiplication) works more naturally for matrices.
However, as far as I can tell you're using np.dot rather than the overloaded arithmetic operators of the matrix class to begin with, so you would not see any loss of functionality from using np.array instead. Furthermore, if you would switch to python 3.5 or newer, you could use the # matrix multiplication operator that would let you write things such as
Alist[i] = Qbars[i] # h[k]
In the following I'll use the ndarray class instead of the matrix class for the above reasons.
So, your question has two main parts: creating your block matrix and multiplying the result with a vector. I suggest using an up-to-date numpy version, since there's numpy.block introduced in version 1.13. This conveniently does exactly what you want it to do:
>>> import numpy as np
>>> A,B,C = (np.full((3,3),k) for k in range(3))
>>> A
array([[0, 0, 0],
[0, 0, 0],
[0, 0, 0]])
>>> B
array([[1, 1, 1],
[1, 1, 1],
[1, 1, 1]])
>>> C
array([[2, 2, 2],
[2, 2, 2],
[2, 2, 2]])
>>> np.block([[A,B],[B,C]])
array([[0, 0, 0, 1, 1, 1],
[0, 0, 0, 1, 1, 1],
[0, 0, 0, 1, 1, 1],
[1, 1, 1, 2, 2, 2],
[1, 1, 1, 2, 2, 2],
[1, 1, 1, 2, 2, 2]])
Similarly, you can concatenate your two 3-length vectors using np.concatenate or one of the stacking methods (these are available in older versions too).
Now, the problem is that you can't multiply a matrix of shape (6,1) with a matrix of shape (6,6), so the question is what you're really trying to do here. In case you want to multiply each element of your matrix with the corresponding row of your vector, you can just multiply your arrays (of class np.ndarray!) and make use of array broadcasting:
>>> Q = np.block([[A,B],[B,C]]) # (6,6)-shape array
>>> v = np.arange(6).reshape(-1,1) # (6,1)-shape array
>>> v * Q
array([[ 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 1, 1, 1],
[ 0, 0, 0, 2, 2, 2],
[ 3, 3, 3, 6, 6, 6],
[ 4, 4, 4, 8, 8, 8],
[ 5, 5, 5, 10, 10, 10]])
The other option is that you want to do matrix-vector multiplication, but then either you have to transpose your vector (in order to multiply it with the matrix from the right) or swap the order of the matrix and the vector (multiplying the vector with the matrix from the left). Example for the former:
>>> v.T # Q # python 3.5 and up
array([[12, 12, 12, 27, 27, 27]])
>>> v.T.dot(Q)
array([[12, 12, 12, 27, 27, 27]])
Another benefit of arrays (rather than matrices) is that arrays can be multidimensional. Instead of putting numpy arrays inside a dict and summing them that way, you could define a 3d array (a collection of 2d arrays along a third axis), then you could sum along the third dimension. One huge benefit of numpy is its efficient memory need and performance, and these aspects are strongest if you use numpy objects and methods all through your code. Mixing native python objects (such as dicts, zips, loops) typically hinders performance.
The two arrays:
a = numpy.array([[2,3,2],[5,6,1]])
b = numpy.array([3,5])
c = a * b
What I want is:
c = [[6,9,6],
[25,30,5]]
But, I am getting this error:
ValueError: operands could not be broadcast together with shapes (2,3) (2)
How to multiply a nD array with 1D array, where len(1D-array) == len(nD array)?
You need to convert array b to a (2, 1) shape array, use None or numpy.newaxis in the index tuple:
import numpy
a = numpy.array([[2,3,2],[5,6,1]])
b = numpy.array([3,5])
c = a * b[:, None]
Here is the document.
Another strategy is to reshape the
second array, so it has the same number of dimensions as the first array:
c = a * b.reshape((b.size, 1))
print(c)
# [[ 6 9 6]
# [25 30 5]]
Alternatively, the shape attribute of the second array can be modified in-place:
b.shape = (b.size, 1)
print(a.shape) # (2, 3)
print(b.shape) # (2, 1)
print(a * b)
# [[ 6 9 6]
# [25 30 5]]