python comparing two matrices

Question

In the below shown matrices i want to match the first element in both the matrices. If the first element is equal then i need match the second element from both the matrices and so on.. if the elements are same then print "same" else print "not same"....

My question is how this in the optimal way also for m*n where m=n always

 for i in a1:
     for j in a2:
        if i!=j:
           break
         else:
           //compare the next corresponding columns and print "same" or "not same"


 a1=[1,44,55],[2,33,66],[3,77,91]  

 a2=[1,44,55],[2,45,66],[3,77,91]    

 OR 

 a1=[1,44,55]
    [2,33,66]
    [3,77,91]  

 a2=[1,44,55]
    [2,45,66]
    [3,77,91]

I wanted to draw that in a matrix form so i had shown as indicated — Rajeev
– Rajeev, Commented Jun 1, 2012 at 13:33

Roy Hyunjin Han · Accepted Answer · 2012-08-02 17:48:33Z

16

You may run into some issues due to floating point rounding errors.

>>> import numpy as np
>>> x = np.array(1.1)
>>> print x * x
1.21
>>> x * x == 1.21
False
>>> x * x
1.2100000000000002
>>> np.allclose(x * x, 1.21)
True

To compare whether two numerical matrices are equal, it is recommended that you use np.allclose()

>>> import numpy as np
>>> x = np.array([[1.1, 1.3], [1.3, 1.1]])
>>> y = np.array([[1.21, 1.69], [1.69, 1.21]])
>>> x * x
array([[ 1.21,  1.69],
       [ 1.69,  1.21]])
>>> x * x == y
array([[False, False],
       [False, False]], dtype=bool)
>>> np.allclose(x * x, y)
True

answered Aug 2, 2012 at 17:48

Roy Hyunjin Han

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Comments

Benjamin Wohlwend · Accepted Answer · 2012-06-01 16:46:35Z

11

What's wrong with a1 == a2?

In [1]: a1=[[1,44,55],
   ...:     [2,33,66],
   ...:     [3,77,91]]

In [2]: a2=[[1,44,55],
   ...:     [2,45,66], # <- second element differs
   ...:     [3,77,91]]

In [3]: a1 == a2
Out[3]: False

In [4]: a1=[[1,44,55],
   ...:     [2,33,66],
   ...:     [3,77,91]]

In [5]: a2=[[1,44,55],
   ...:     [2,33,66],
   ...:     [3,77,91]]

In [6]: a1 == a2
Out[6]: True

edited Jun 1, 2012 at 16:46

answered Jun 1, 2012 at 13:41

Benjamin Wohlwend

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2 Comments

UselesssCat Over a year ago

So a == b compares list recursively?

Layne B Over a year ago

I get a ValueError when I do this

starwarswii · Accepted Answer · 2023-05-07 03:32:56Z

6

In case you are using numpy arrays, use the .all() and .any().

x = np.array([[1, 2, 3], [4, 5, 6]])
y = np.array([[1, 2, 3], [4, 5, 6]])
(x == y).all()
>> True

x = np.array([[1, 2, 3], [4, 5, 6]])
y = np.array([[11, 21, 31], [41, 4, 61]])
(x == y).any()
>> True # since x[1][0] == y[1][1]

edited May 7, 2023 at 3:32

starwarswii

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answered Jan 27, 2020 at 3:40

coda

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Comments

KurzedMetal · Accepted Answer · 2012-06-01 13:45:10Z

1

If you wanna do operations over matrix, numpy is the best library you could use

In [11]: a = numpy.matrix([[1,44,55],
    ...:                   [2,33,66],
    ...:                   [3,77,91]])

In [12]: b = numpy.matrix([[1,44,55],
    ...:                   [2,45,66],
    ...:                   [3,77,91]])

In [13]: a == b
Out[13]: 
matrix([[ True,  True,  True],
        [ True, False,  True],
        [ True,  True,  True]], dtype=bool)

answered Jun 1, 2012 at 13:45

KurzedMetal

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Comments

rajitbanerjee · Accepted Answer · 2020-12-11 13:18:45Z

1

Below is solution with lists without using numpy.

def isIdentical(a: list, b: list) -> bool:
    rows, cols = len(a), len(a[0])
    return all([a[i][j] == b[i][j] for j in range(cols) for i in range(rows)])

answered Dec 11, 2020 at 13:18

rajitbanerjee

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Comments

Fatima Ansarizadeh · Accepted Answer · 2022-05-27 00:57:24Z

0

from scipy import sparse
I = np.arange(0, 3).repeat(3)     
J =np.arange(9) 
V = np.arange(9) 
W = sparse.coo_matrix((V, (I, J)), shape=(9,9))
print(np.array_equiv(W.todense(),W.todense().T)) #False Shape consistent (broadcastable)
print(np.array_equal(W.todense(),W.todense().T)) #False (exact shape)
print(W.todense().all()==W.todense().T.all()) #True
print(W.todense==W.todense().T)
print(W.todense()[:,1], W.todense()[1,:])

Why using print(W.todense().all()==W.todense().T.all()) #True returns True, while W is not symmetric?

answered May 27, 2022 at 0:57

Fatima Ansarizadeh

1

1 Comment

Community Over a year ago

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