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Numpy array assignment with copy
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Closed 6 months ago.
I'm trying to make a function where I can give it a matrix and the constants and solving it using crammers rule. This will soon be used with complex numbers.
import numpy as np
import sympy as smp
def crammer(matrix, constants):
D = np.linalg.det(matrix)
dets = []
dets.append(D)
for i in range(0, len(constants[0]), 1):
Dv = matrix
Dv[:, i:i+1] = constants.T
print(Dv)
Dd = np.linalg.det(Dv)
dets.append(Dd)
return dets
Mat = np.array([[2, 1,-1],
[3, 2, 2],
[4,-2, 3]])
Con = np.array([[1,13,9]])
print(crammer(Mat, Con))
I get this as the result:
[33.000000000000014, 33.000000000000014, 0.0, 0.0]
The first two are right the Determinate is D:33 and Dx:33 but Dy and Dz should be Dy:66 and Dz: 99.
following Crammers Rule it should be:
[[ 1 1 -1]
[13 2 2]
[ 9 -2 3]]
[[ 2 1 -1]
[ 3 13 2]
[ 4 9 3]]
[[ 2 2 1]
[ 3 2 13]
[ 4 -2 9]]
when I print Dv at the beginning of the for loop I get the following:
[[ 1 1 -1]
[13 2 2]
[ 9 -2 3]]
[[ 1 1 -1]
[13 13 2]
[ 9 9 3]]
[[ 1 1 1]
[13 13 13]
[ 9 9 9]]
I tried printing the matrix at the top of the for loop as well and I get the same problem. As I can tell my for loop is is changing my original matrix and I don't understand why.
Unsure what the algorithm itself is trying to accomplish, but from a Python perspective, by assigning Dv = matrix, you're creating a shallow copy by reference that then gets modified by the rest of the code in the loop.
I replaced this assignment with Dv = deepcopy(matrix) (use from copy import deepcopy) to create a deep copy, fresh from the original matrix.
That produced this result, which seems in line with what you predicted:
[33.000000000000014, 33.000000000000014, 66.00000000000003, 99.00000000000007]
copy documentation
Related
I am trying to find a way to run a function through a whole ndarray using the indexes of every value as parameters. A regular loop is quite slow and I can't find a way to make it work using numpy built-in functions. The next example code summarizes what i'm trying to do. Thanks in advance.
import numpy as np
arr = np.arange(10).reshape(2, 5)
print(arr)
def example_func(row, col, value):
return(row + col + value)
for row in range(arr.shape[0]):
for col in range(arr.shape[1]):
arr[row, col] = example_func(row, col, arr[row, col])
print(arr)
[[0 1 2 3 4]
[5 6 7 8 9]]
[[ 0 2 4 6 8]
[ 6 8 10 12 14]]
What you try to do can be done with meshgrid.
Return coordinate matrices from coordinate vectors.
n_rows, n_cols = arr.shape
col_matrix, row_matrix = np.meshgrid(np.arange(n_cols), np.arange(n_rows))
result = arr + col_matrix + row_matrix
print(result)
This returns
[[ 0 2 4 6 8]
[ 6 8 10 12 14]]
Given a numpy ndarray A, return its rank array.
Input : [[ 9 4 15 0 18]
[16 19 8 10 1]]
Return value: [[4 2 6 0 8]
[7 9 3 5 1]]
**but I didnt solve actually I solve but my solution is wrong
how can I solve ? please help meemphasized text**
import numpy as np
array=np.array([[9,4,15,0,18],[16,19,8,10,1]])
array_1=np.array([9,4,15,0,18])
array_2=np.array([16,19,8,10,1])
temp1 = array_1.argsort()
temp2 = array_2.argsort()
ranks1 = np.arange(len(array_1))[temp1.argsort()]
rankss=ranks1.argsort()
ranks2 = np.arange(len(array_2))[temp2.argsort()]
print(ranks1*array.ndim)
print(rankss)
If you pass axis=None to argsort then the source array
is flattened at the first step (and only then arg-sorted).
So probably the shortest code is:
result = arr.argsort(axis=None).reshape(arr.shape)
No need to explicitely flatten the source array and only
a single call to argsort.
I'm new to numpy and am trying to do some slicing and indexing with arrays. My goal is to take an array, and use slicing and indexing to square the last column, and then subtract the first column from that result. I then want to put the new column back into the old array.
I've been able to figure out how to slice and index the column to get the result I want for the last column. My problem however is that when I try to put it back into my original array, I get the wrong output (as seen below).
theNumbers = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]])
sliceColumnOne = theNumbers[:,0]
sliceColumnThree = theNumbers[:,3]**2
editColumnThree = sliceColumnThree - sliceColumnOne
newArray = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12],[editColumnThree]])
print("nums:\n{}".format(newArray))
I want the output to be
[[ 1 2 3 15]
[ 5 6 7 59]
[ 9 10 11 135]
[ 13 14 15 243]]
However mine becomes:
[list([1, 2, 3, 4]) list([5, 6, 7, 8]) list([9, 10, 11, 12])
list([array([ 15, 59, 135, 243])])]
Any suggestions on how to fix this?
Just assign the last numpy array row to the new one "theNumbers[3] = editColumnThree"
Code:
import numpy as np
theNumbers = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]])
sliceColumnOne = theNumbers[:,0]
sliceColumnThree = theNumbers[:,3]**2
editColumnThree = sliceColumnThree - sliceColumnOne
theNumbers[3] = editColumnThree
print("nums:\n{}".format(theNumbers))
Output:
[[ 1 2 3 4]
[ 5 6 7 8]
[ 9 10 11 12]
[ 15 59 135 243]]
newArray = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12],[editColumnThree]])
print("nums:\n{}".format(newArray))
this way, editColumnThree is the last row, not column. You can use
newArray = theNumbers.copy() # if a copy is needed
newArray[:,-1] = editColumnThree # replace last (-1) column
If you just want to stack the vectors on top of eachother, use vstack:
import numpy as np
theNumbers = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]])
newNumbers = np.vstack(theNumbers)
print(newNumbers)
>>>[[ 1 2 3 4]
[ 5 6 7 8]
[ 9 10 11 12]
[13 14 15 16]]
But the issue here isn't just that you need to stack these numbers, you are mixing up columns and rows. You are changing a row instead of a column. To change the column, update the last element in each row:
import numpy as np
theNumbers = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]])
LastColumn = theNumbers[:,3]**2
FirstColumn = theNumbers[:,0]
editColumnThree = LastColumn - FirstColumn
for i in range(4):
theNumbers[i,3] = editColumnThree [i]
print(theNumbers)
>>>[[ 1 2 3 15]
[ 5 6 7 59]
[ 9 10 11 135]
[ 13 14 15 243]]
I have just started using numpy and I am getting confused about how to use arrays. I have seen several Stack Overflow answers on numpy arrays but they all deal with how to get the desired result (I know how to do this, I just don't know why I need to do it this way). The consensus that I've seen is that arrays are better than matrices because they are a more basic class and less restrictive. I understand you can transpose an array which to me means there is a distinction between a row and a column, but the multiplication rules all produce the wrong outputs (compared to what I am expecting).
Here is the test code I have written along with the outputs:
a = numpy.array([1,2,3,4])
print(a)
>>> [1 2 3 4]
print(a.T) # Transpose
>>> [1 2 3 4] # No apparent affect
b = numpy.array( [ [1], [2], [3], [4] ] )
print(b)
>>> [[1]
[2]
[3]
[4]] # Column (Expected)
print(b.T)
>>> [[1 2 3 4]] # Row (Expected, transpose seems to work here)
print((b.T).T)
>>> [[1]
[2]
[3]
[4]] # Column (All of these are as expected,
# unlike for declaring the array as a row vector)
# The following are element wise multiplications of a
print(a*a)
>>> [ 1 4 9 16]
print(a * a.T) # Row*Column
>>> [ 1 4 9 16] # Inner product scalar result expected
print(a.T * a) # Column*Row
>>> [ 1 4 9 16] # Outer product matrix result expected
print(b*b)
>>> [[1]
[4]
[9]
[16]] # Expected result, element wise multiplication in a column
print(b * b.T) # Column * Row (Outer product)
>>> [[ 1 2 3 4]
[ 2 4 6 8]
[ 3 6 9 12]
[ 4 8 12 16]] # Expected matrix result
print(b.T * (b.T)) # Column * Column (Doesn't make much sense so I expected elementwise multiplication
>>> [[ 1 4 9 16]]
print(b.T * (b.T).T) # Row * Column, inner product expected
>>> [[ 1 2 3 4]
[ 2 4 6 8]
[ 3 6 9 12]
[ 4 8 12 16]] # Outer product result
I know that I can use numpy.inner() and numpy.outer() to achieve the affect (that is not a problem), I just want to know if I need to keep track of whether my vectors are rows or columns.
I also know that I can create a 1D matrix to represent my vectors and the multiplication works as expected. I'm trying to work out the best way to store my data so that when I look at my code it is clear what is going to happen - right now the maths just looks confusing and wrong.
I only need to use 1D and 2D tensors for my application.
I'll try annotating your code
a = numpy.array([1,2,3,4])
print(a)
>>> [1 2 3 4]
print(a.T) # Transpose
>>> [1 2 3 4] # No apparent affect
a.shape will show (4,). a.T.shape is the same. It kept the same number of dimensions, and performed the only meaningful transpose - no change. Making it (4,1) would have added a dimension, and destroyed the A.T.T roundtrip.
b = numpy.array( [ [1], [2], [3], [4] ] )
print(b)
>>> [[1]
[2]
[3]
[4]] # Column (Expected)
print(b.T)
>>> [[1 2 3 4]] # Row (Expected, transpose seems to work here)
b.shape is (4,1), b.T.shape is (1,4). Note the extra set of []. If you'd created a as a = numpy.array([[1,2,3,4]]) its shape too would have been (1,4).
The easy way to make b would be b=np.array([[1,2,3,4]]).T (or b=np.array([1,2,3,4])[:,None] or b=np.array([1,2,3,4]).reshape(-1,1))
Compare this to MATLAB
octave:3> a=[1,2,3,4]
a =
1 2 3 4
octave:4> size(a)
ans =
1 4
octave:5> size(a.')
ans =
4 1
Even without the extra [] it has initialed the matrix as 2d.
numpy has a matrix class that imitates MATLAB - back in the time when MATLAB allowed only 2d.
In [75]: m=np.matrix('1 2 3 4')
In [76]: m
Out[76]: matrix([[1, 2, 3, 4]])
In [77]: m.shape
Out[77]: (1, 4)
In [78]: m=np.matrix('1 2; 3 4')
In [79]: m
Out[79]:
matrix([[1, 2],
[3, 4]])
I don't recommend using np.matrix unless it really adds something useful to your code.
Note the MATLAB talks of vectors, but they are really just their matrix with only one non-unitary dimension.
# The following are element wise multiplications of a
print(a*a)
>>> [ 1 4 9 16]
print(a * a.T) # Row*Column
>>> [ 1 4 9 16] # Inner product scalar result expected
This behavior follows from a.T == A. As you noted, * produces element by element multiplication. This is equivalent to the MATLAB .*. np.dot(a,a) gives the dot or matrix product of 2 arrays.
print(a.T * a) # Column*Row
>>> [ 1 4 9 16] # Outer product matrix result expected
No, it is still doing elementwise multiplication.
I'd use broadcasting, a[:,None]*a[None,:] to get the outer product. Octave added this in imitation of numpy; I don't know if MATLAB has it yet.
In the following * is always element by element multiplication. It's broadcasting that produces matrix/outer product results.
print(b*b)
>>> [[1]
[4]
[9]
[16]] # Expected result, element wise multiplication in a column
A (4,1) * (4,1)=>(4,1). Same shapes all around.
print(b * b.T) # Column * Row (Outer product)
>>> [[ 1 2 3 4]
[ 2 4 6 8]
[ 3 6 9 12]
[ 4 8 12 16]] # Expected matrix result
Here (4,1)*(1,4)=>(4,4) product. The 2 size 1 dimensions have been replicated so it becomes, effectively a (4,4)*(4,4). How would you do replicate this in MATLAB - with .*?
print(b.T * (b.T)) # Column * Column (Doesn't make much sense so I expected elementwise multiplication
>>> [[ 1 4 9 16]]
* is elementwise regardless of expectations. Think b' .* b' in MATLAB.
print(b.T * (b.T).T) # Row * Column, inner product expected
>>> [[ 1 2 3 4]
[ 2 4 6 8]
[ 3 6 9 12]
[ 4 8 12 16]] # Outer product result
Again * is elementwise; inner requires a summation in addition to multiplication. Here broadcasting again applies (1,4)*(4,1)=>(4,4).
np.dot(b,b) or np.trace(b.T*b) or np.sum(b*b) give 30.
When I worked in MATLAB I frequently checked the size, and created test matrices that would catch dimension mismatches (e.g. a 2x3 instead of a 2x2 matrix). I continue to do that in numpy.
The key things are:
numpy arrays may be 1d (or even 0d)
A (4,) array is not exactly the same as a (4,1) or (1,4)`.
* is elementwise - always.
broadcasting usually accounts for outer like behavior
"Transposing" is, from a numpy perspective, really only a meaningful concept for two-dimensional structures:
>>> import numpy
>>> arr = numpy.array([1,2,3,4])
>>> arr.shape
(4,)
>>> arr.transpose().shape
(4,)
So, if you want to transpose something, you'll have to make it two-dimensional:
>>> arr_2d = arr.reshape((4,1)) ## four rows, one column -> two-dimensional
>>> arr_2d.shape
(4, 1)
>>> arr_2d.transpose().shape
(1, 4)
Also, numpy.array(iterable, **kwargs) has a key word argument ndmin, which will, set to ndmin=2 prepend your desired shape with as many 1 as necessary:
>>> arr_ndmin = numpy.array([1,2,3,4],ndmin=2)
>>> arr_ndmin.shape
(1, 4)
Yes, they do.
Your question is already answered. Though I assume you are a Matlab user? If so, you may find this guide useful: Moving from MATLAB matrices to NumPy arrays
I have to use dynamic programming in a python script.
I defined a numpy array u with shape=(N,K).
I want to pick one element for each column, therefore generating a K-uplets.
How would you proceed to loop efficiently across all K-uplets generated this way ? A solution would be to use
import itertools
itertools.combination_with_replacement(list,K)
where list = [0..N-1], but I will need to build iteratively each of my K-uplets using the output (index) of the itertools method.
Is there a more direct way to proceed ?
Thanks
Vincent
You can build the K-uplet with arr[ind, np.arange(K)]. Of course, that's actually a NumPy ndarray, but they are easy to convert to tuplets if you really want tuplets: tuple(arr[ind, np.arange(K)]).
import numpy as np
import itertools as IT
N, K = 5,3
arr = np.arange(N*K).reshape(N,K)
print(arr)
# [[ 0 1 2]
# [ 3 4 5]
# [ 6 7 8]
# [ 9 10 11]
# [12 13 14]]
for ind in IT.combinations_with_replacement(range(N), K):
print(arr[ind, np.arange(K)])
# [0 1 2]
# [0 1 5]
# [0 1 8]
# [ 0 1 11]
# ...