The Python Oracle

2-D arrays with numpy arange

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Chapters
00:00 2-D Arrays With Numpy Arange
01:15 Answer 1 Score 0
01:29 Answer 2 Score 3
01:57 Answer 3 Score 5
02:45 Accepted Answer Score 4
03:07 Thank you

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Full question
https://stackoverflow.com/questions/4072...

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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

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Tags
#python #arrays #numpy #multidimensionalarray

#avk47


ANSWER 1

Score 5


To do the first one with numpy:

>>> a = np.arange(11)
>>> a[:,None]+a
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10],  
  [ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11],  
  [ 2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12],  
  [ 3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13],  
  [ 4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14],  
  [ 5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15],  
  [ 6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16],  
  [ 7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17],  
  [ 8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18],  
  [ 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19],  
  [10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]])

For the second array, @Divakar has a good approach. Maybe a bit simpler syntax to do this:

>>> (a%a[:,None])==0
array([[ True,  True,  True,  True,  True,  True,  True,  True,  True, True,  True],
   [ True,  True,  True,  True,  True,  True,  True,  True,  True, True,  True],
   [ True, False,  True, False,  True, False,  True, False,  True, False,  True],
   [ True, False, False,  True, False, False,  True, False, False, True, False],
   [ True, False, False, False,  True, False, False, False,  True, False, False],
   [ True, False, False, False, False,  True, False, False, False, False,  True],
   [ True, False, False, False, False, False,  True, False, False, False, False],
   [ True, False, False, False, False, False, False,  True, False, False, False],
   [ True, False, False, False, False, False, False, False,  True, False, False],
   [ True, False, False, False, False, False, False, False, False, True, False],
   [ True, False, False, False, False, False, False, False, False, False,  True]], dtype=bool)


ACCEPTED ANSWER

Score 4


Per your first question:

np.add(*np.indices((nrow, ncol)))

For nrow=5, ncol=6 you get

array([[0, 1, 2, 3, 4, 5],
       [1, 2, 3, 4, 5, 6],
       [2, 3, 4, 5, 6, 7],
       [3, 4, 5, 6, 7, 8],
       [4, 5, 6, 7, 8, 9]])

This method doesn't use the numpy.arange function, though I find it more readable. Moreover, it supports cases when nrow != ncol.



ANSWER 3

Score 3


For the second array, here's one approach with broadcasting -

a = np.arange(10)
out = (np.mod(a,a[:,None])==0) & (a[:,None]!=0)

Sample run -

In [511]: a = np.arange(10)

In [512]: print (np.mod(a,a[:,None])==0) & (a[:,None]!=0)
[[False False False False False False False False False False]
 [ True  True  True  True  True  True  True  True  True  True]
 [ True False  True False  True False  True False  True False]
 [ True False False  True False False  True False False  True]
 [ True False False False  True False False False  True False]
 [ True False False False False  True False False False False]
 [ True False False False False False  True False False False]
 [ True False False False False False False  True False False]
 [ True False False False False False False False  True False]
 [ True False False False False False False False False  True]]


ANSWER 4

Score 0


You can try this for the first array you want:

np.array([range(i,i+10) for i in range(10)])

and for the second array:

np.array([[i>0 and j%i==0 for j in range(10)] for i in range(10)])