numpy logical.xor on a 2D array












1















I have a 2D numpy array with only 0 and 255 values (created from a black and white image) that i'd like to XOR with another similar 2D array.



The dtype of these arrays is uint8 and their shapes are identical.



The only information and examples i've been able to find, so far, deal with 1D arrays.



Do I need to 'flatten' these 2D arrays before XOR'ing? If so, how is that done?






arrays numpy python-2.x







share|improve this question



























    1















    I have a 2D numpy array with only 0 and 255 values (created from a black and white image) that i'd like to XOR with another similar 2D array.



    The dtype of these arrays is uint8 and their shapes are identical.



    The only information and examples i've been able to find, so far, deal with 1D arrays.



    Do I need to 'flatten' these 2D arrays before XOR'ing? If so, how is that done?






    arrays numpy python-2.x







    share|improve this question

























      1












      1








      1








      I have a 2D numpy array with only 0 and 255 values (created from a black and white image) that i'd like to XOR with another similar 2D array.



      The dtype of these arrays is uint8 and their shapes are identical.



      The only information and examples i've been able to find, so far, deal with 1D arrays.



      Do I need to 'flatten' these 2D arrays before XOR'ing? If so, how is that done?






      arrays numpy python-2.x







      share|improve this question














      I have a 2D numpy array with only 0 and 255 values (created from a black and white image) that i'd like to XOR with another similar 2D array.



      The dtype of these arrays is uint8 and their shapes are identical.



      The only information and examples i've been able to find, so far, deal with 1D arrays.



      Do I need to 'flatten' these 2D arrays before XOR'ing? If so, how is that done?






      arrays numpy python-2.x




      arrays numpy python-2.x






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 28 '18 at 1:46









      yodishyodish

      329312




      329312
























          1 Answer
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          active

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          1














          numpy.logical_xor and numpy.bitwise_xor will work for 2-D arrays, as will the operators != and ^ (essentially logical and bitwise XOR, respectively).



          edit: I just noticed in your title you are looking for logical XOR, but I will leave the bitwise info there for reference in case it's helpful.



          Setup:



          a = np.random.choice([0,255], (5,5))
          
b = np.random.choice([0,255], (5,5))
          

          
>>> a
          
array([[255, 255,   0, 255, 255],
          
       [255, 255,   0, 255,   0],
          
       [255,   0,   0,   0,   0],
          
       [  0, 255, 255, 255, 255],
          
       [  0,   0, 255,   0,   0]])
          
>>> b
          
array([[  0, 255, 255, 255, 255],
          
       [255,   0,   0, 255,   0],
          
       [255,   0, 255,   0, 255],
          
       [  0,   0,   0,   0,   0],
          
       [255,   0,   0,   0, 255]])


          Logical XOR:



          >>> np.logical_xor(a,b)
          
array([[ True, False,  True, False, False],
          
       [False,  True, False, False, False],
          
       [False, False,  True, False,  True],
          
       [False,  True,  True,  True,  True],
          
       [ True, False,  True, False,  True]])
          

          
# equivalently:
          
>>> a!=b
          
array([[ True, False,  True, False, False],
          
       [False,  True, False, False, False],
          
       [False, False,  True, False,  True],
          
       [False,  True,  True,  True,  True],
          
       [ True, False,  True, False,  True]])


          Bitwise XOR:



          >>> np.bitwise_xor(a,b)
          
array([[255,   0, 255,   0,   0],
          
       [  0, 255,   0,   0,   0],
          
       [  0,   0, 255,   0, 255],
          
       [  0, 255, 255, 255, 255],
          
       [255,   0, 255,   0, 255]])
          

          
# equivalently:
          
>>> a^b
          
array([[255,   0, 255,   0,   0],
          
       [  0, 255,   0,   0,   0],
          
       [  0,   0, 255,   0, 255],
          
       [  0, 255, 255, 255, 255],
          
       [255,   0, 255,   0, 255]])





          share|improve this answer





















          • 1





            It turns out, Bitwise XOR was what I was looking for ! :) I appreciate it!

            – yodish
            Nov 28 '18 at 2:03











          • is there a numpy function that returns the intersection of two 2D arrays?

            – yodish
            Nov 28 '18 at 2:07











          • What do you mean by the intersection in this case?

            – sacuL
            Nov 28 '18 at 2:08











          • Returning an array that contains 255 values only where they exist (in the same location) in both array1 and array2.

            – yodish
            Nov 28 '18 at 2:12








          • 1





            in your case you could do something like ((a == 255) & (b == 255)) * 255

            – sacuL
            Nov 28 '18 at 2:42











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1














          numpy.logical_xor and numpy.bitwise_xor will work for 2-D arrays, as will the operators != and ^ (essentially logical and bitwise XOR, respectively).



          edit: I just noticed in your title you are looking for logical XOR, but I will leave the bitwise info there for reference in case it's helpful.



          Setup:



          a = np.random.choice([0,255], (5,5))
          
b = np.random.choice([0,255], (5,5))
          

          
>>> a
          
array([[255, 255,   0, 255, 255],
          
       [255, 255,   0, 255,   0],
          
       [255,   0,   0,   0,   0],
          
       [  0, 255, 255, 255, 255],
          
       [  0,   0, 255,   0,   0]])
          
>>> b
          
array([[  0, 255, 255, 255, 255],
          
       [255,   0,   0, 255,   0],
          
       [255,   0, 255,   0, 255],
          
       [  0,   0,   0,   0,   0],
          
       [255,   0,   0,   0, 255]])


          Logical XOR:



          >>> np.logical_xor(a,b)
          
array([[ True, False,  True, False, False],
          
       [False,  True, False, False, False],
          
       [False, False,  True, False,  True],
          
       [False,  True,  True,  True,  True],
          
       [ True, False,  True, False,  True]])
          

          
# equivalently:
          
>>> a!=b
          
array([[ True, False,  True, False, False],
          
       [False,  True, False, False, False],
          
       [False, False,  True, False,  True],
          
       [False,  True,  True,  True,  True],
          
       [ True, False,  True, False,  True]])


          Bitwise XOR:



          >>> np.bitwise_xor(a,b)
          
array([[255,   0, 255,   0,   0],
          
       [  0, 255,   0,   0,   0],
          
       [  0,   0, 255,   0, 255],
          
       [  0, 255, 255, 255, 255],
          
       [255,   0, 255,   0, 255]])
          

          
# equivalently:
          
>>> a^b
          
array([[255,   0, 255,   0,   0],
          
       [  0, 255,   0,   0,   0],
          
       [  0,   0, 255,   0, 255],
          
       [  0, 255, 255, 255, 255],
          
       [255,   0, 255,   0, 255]])





          share|improve this answer





















          • 1





            It turns out, Bitwise XOR was what I was looking for ! :) I appreciate it!

            – yodish
            Nov 28 '18 at 2:03











          • is there a numpy function that returns the intersection of two 2D arrays?

            – yodish
            Nov 28 '18 at 2:07











          • What do you mean by the intersection in this case?

            – sacuL
            Nov 28 '18 at 2:08











          • Returning an array that contains 255 values only where they exist (in the same location) in both array1 and array2.

            – yodish
            Nov 28 '18 at 2:12








          • 1





            in your case you could do something like ((a == 255) & (b == 255)) * 255

            – sacuL
            Nov 28 '18 at 2:42
















          1














          numpy.logical_xor and numpy.bitwise_xor will work for 2-D arrays, as will the operators != and ^ (essentially logical and bitwise XOR, respectively).



          edit: I just noticed in your title you are looking for logical XOR, but I will leave the bitwise info there for reference in case it's helpful.



          Setup:



          a = np.random.choice([0,255], (5,5))
          
b = np.random.choice([0,255], (5,5))
          

          
>>> a
          
array([[255, 255,   0, 255, 255],
          
       [255, 255,   0, 255,   0],
          
       [255,   0,   0,   0,   0],
          
       [  0, 255, 255, 255, 255],
          
       [  0,   0, 255,   0,   0]])
          
>>> b
          
array([[  0, 255, 255, 255, 255],
          
       [255,   0,   0, 255,   0],
          
       [255,   0, 255,   0, 255],
          
       [  0,   0,   0,   0,   0],
          
       [255,   0,   0,   0, 255]])


          Logical XOR:



          >>> np.logical_xor(a,b)
          
array([[ True, False,  True, False, False],
          
       [False,  True, False, False, False],
          
       [False, False,  True, False,  True],
          
       [False,  True,  True,  True,  True],
          
       [ True, False,  True, False,  True]])
          

          
# equivalently:
          
>>> a!=b
          
array([[ True, False,  True, False, False],
          
       [False,  True, False, False, False],
          
       [False, False,  True, False,  True],
          
       [False,  True,  True,  True,  True],
          
       [ True, False,  True, False,  True]])


          Bitwise XOR:



          >>> np.bitwise_xor(a,b)
          
array([[255,   0, 255,   0,   0],
          
       [  0, 255,   0,   0,   0],
          
       [  0,   0, 255,   0, 255],
          
       [  0, 255, 255, 255, 255],
          
       [255,   0, 255,   0, 255]])
          

          
# equivalently:
          
>>> a^b
          
array([[255,   0, 255,   0,   0],
          
       [  0, 255,   0,   0,   0],
          
       [  0,   0, 255,   0, 255],
          
       [  0, 255, 255, 255, 255],
          
       [255,   0, 255,   0, 255]])





          share|improve this answer





















          • 1





            It turns out, Bitwise XOR was what I was looking for ! :) I appreciate it!

            – yodish
            Nov 28 '18 at 2:03











          • is there a numpy function that returns the intersection of two 2D arrays?

            – yodish
            Nov 28 '18 at 2:07











          • What do you mean by the intersection in this case?

            – sacuL
            Nov 28 '18 at 2:08











          • Returning an array that contains 255 values only where they exist (in the same location) in both array1 and array2.

            – yodish
            Nov 28 '18 at 2:12








          • 1





            in your case you could do something like ((a == 255) & (b == 255)) * 255

            – sacuL
            Nov 28 '18 at 2:42














          1












          1








          1







          numpy.logical_xor and numpy.bitwise_xor will work for 2-D arrays, as will the operators != and ^ (essentially logical and bitwise XOR, respectively).



          edit: I just noticed in your title you are looking for logical XOR, but I will leave the bitwise info there for reference in case it's helpful.



          Setup:



          a = np.random.choice([0,255], (5,5))
          
b = np.random.choice([0,255], (5,5))
          

          
>>> a
          
array([[255, 255,   0, 255, 255],
          
       [255, 255,   0, 255,   0],
          
       [255,   0,   0,   0,   0],
          
       [  0, 255, 255, 255, 255],
          
       [  0,   0, 255,   0,   0]])
          
>>> b
          
array([[  0, 255, 255, 255, 255],
          
       [255,   0,   0, 255,   0],
          
       [255,   0, 255,   0, 255],
          
       [  0,   0,   0,   0,   0],
          
       [255,   0,   0,   0, 255]])


          Logical XOR:



          >>> np.logical_xor(a,b)
          
array([[ True, False,  True, False, False],
          
       [False,  True, False, False, False],
          
       [False, False,  True, False,  True],
          
       [False,  True,  True,  True,  True],
          
       [ True, False,  True, False,  True]])
          

          
# equivalently:
          
>>> a!=b
          
array([[ True, False,  True, False, False],
          
       [False,  True, False, False, False],
          
       [False, False,  True, False,  True],
          
       [False,  True,  True,  True,  True],
          
       [ True, False,  True, False,  True]])


          Bitwise XOR:



          >>> np.bitwise_xor(a,b)
          
array([[255,   0, 255,   0,   0],
          
       [  0, 255,   0,   0,   0],
          
       [  0,   0, 255,   0, 255],
          
       [  0, 255, 255, 255, 255],
          
       [255,   0, 255,   0, 255]])
          

          
# equivalently:
          
>>> a^b
          
array([[255,   0, 255,   0,   0],
          
       [  0, 255,   0,   0,   0],
          
       [  0,   0, 255,   0, 255],
          
       [  0, 255, 255, 255, 255],
          
       [255,   0, 255,   0, 255]])





          share|improve this answer















          numpy.logical_xor and numpy.bitwise_xor will work for 2-D arrays, as will the operators != and ^ (essentially logical and bitwise XOR, respectively).



          edit: I just noticed in your title you are looking for logical XOR, but I will leave the bitwise info there for reference in case it's helpful.



          Setup:



          a = np.random.choice([0,255], (5,5))
          
b = np.random.choice([0,255], (5,5))
          

          
>>> a
          
array([[255, 255,   0, 255, 255],
          
       [255, 255,   0, 255,   0],
          
       [255,   0,   0,   0,   0],
          
       [  0, 255, 255, 255, 255],
          
       [  0,   0, 255,   0,   0]])
          
>>> b
          
array([[  0, 255, 255, 255, 255],
          
       [255,   0,   0, 255,   0],
          
       [255,   0, 255,   0, 255],
          
       [  0,   0,   0,   0,   0],
          
       [255,   0,   0,   0, 255]])


          Logical XOR:



          >>> np.logical_xor(a,b)
          
array([[ True, False,  True, False, False],
          
       [False,  True, False, False, False],
          
       [False, False,  True, False,  True],
          
       [False,  True,  True,  True,  True],
          
       [ True, False,  True, False,  True]])
          

          
# equivalently:
          
>>> a!=b
          
array([[ True, False,  True, False, False],
          
       [False,  True, False, False, False],
          
       [False, False,  True, False,  True],
          
       [False,  True,  True,  True,  True],
          
       [ True, False,  True, False,  True]])


          Bitwise XOR:



          >>> np.bitwise_xor(a,b)
          
array([[255,   0, 255,   0,   0],
          
       [  0, 255,   0,   0,   0],
          
       [  0,   0, 255,   0, 255],
          
       [  0, 255, 255, 255, 255],
          
       [255,   0, 255,   0, 255]])
          

          
# equivalently:
          
>>> a^b
          
array([[255,   0, 255,   0,   0],
          
       [  0, 255,   0,   0,   0],
          
       [  0,   0, 255,   0, 255],
          
       [  0, 255, 255, 255, 255],
          
       [255,   0, 255,   0, 255]])






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 28 '18 at 1:54

























          answered Nov 28 '18 at 1:49









          sacuLsacuL

          30.6k41943




          30.6k41943








          • 1





            It turns out, Bitwise XOR was what I was looking for ! :) I appreciate it!

            – yodish
            Nov 28 '18 at 2:03











          • is there a numpy function that returns the intersection of two 2D arrays?

            – yodish
            Nov 28 '18 at 2:07











          • What do you mean by the intersection in this case?

            – sacuL
            Nov 28 '18 at 2:08











          • Returning an array that contains 255 values only where they exist (in the same location) in both array1 and array2.

            – yodish
            Nov 28 '18 at 2:12








          • 1





            in your case you could do something like ((a == 255) & (b == 255)) * 255

            – sacuL
            Nov 28 '18 at 2:42














          • 1





            It turns out, Bitwise XOR was what I was looking for ! :) I appreciate it!

            – yodish
            Nov 28 '18 at 2:03











          • is there a numpy function that returns the intersection of two 2D arrays?

            – yodish
            Nov 28 '18 at 2:07











          • What do you mean by the intersection in this case?

            – sacuL
            Nov 28 '18 at 2:08











          • Returning an array that contains 255 values only where they exist (in the same location) in both array1 and array2.

            – yodish
            Nov 28 '18 at 2:12








          • 1





            in your case you could do something like ((a == 255) & (b == 255)) * 255

            – sacuL
            Nov 28 '18 at 2:42








          1




          1





          It turns out, Bitwise XOR was what I was looking for ! :) I appreciate it!

          – yodish
          Nov 28 '18 at 2:03





          It turns out, Bitwise XOR was what I was looking for ! :) I appreciate it!

          – yodish
          Nov 28 '18 at 2:03













          is there a numpy function that returns the intersection of two 2D arrays?

          – yodish
          Nov 28 '18 at 2:07





          is there a numpy function that returns the intersection of two 2D arrays?

          – yodish
          Nov 28 '18 at 2:07













          What do you mean by the intersection in this case?

          – sacuL
          Nov 28 '18 at 2:08





          What do you mean by the intersection in this case?

          – sacuL
          Nov 28 '18 at 2:08













          Returning an array that contains 255 values only where they exist (in the same location) in both array1 and array2.

          – yodish
          Nov 28 '18 at 2:12







          Returning an array that contains 255 values only where they exist (in the same location) in both array1 and array2.

          – yodish
          Nov 28 '18 at 2:12






          1




          1





          in your case you could do something like ((a == 255) & (b == 255)) * 255

          – sacuL
          Nov 28 '18 at 2:42





          in your case you could do something like ((a == 255) & (b == 255)) * 255

          – sacuL
          Nov 28 '18 at 2:42




















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