How does numpy reshape works?
Question:
I have data in a numpy array:
a = np.arange(100)
a = a.reshape((20,5))
When I type
a[:10]
it returns
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29],
[30, 31, 32, 33, 34],
[35, 36, 37, 38, 39],
[40, 41, 42, 43, 44],
[45, 46, 47, 48, 49]])
Now i decided to reshape the array into 3d array.
b = a.reshape((5,4,5))
array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]],
[[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29],
[30, 31, 32, 33, 34],
[35, 36, 37, 38, 39]],
[[40, 41, 42, 43, 44],
[45, 46, 47, 48, 49],
[50, 51, 52, 53, 54],
[55, 56, 57, 58, 59]],
[[60, 61, 62, 63, 64],
[65, 66, 67, 68, 69],
[70, 71, 72, 73, 74],
[75, 76, 77, 78, 79]],
[[80, 81, 82, 83, 84],
[85, 86, 87, 88, 89],
[90, 91, 92, 93, 94],
[95, 96, 97, 98, 99]]])
How do I slice b to that I obtain the values like a[:10]?
I tried
b[:10,0,:5]
array([[ 0, 1, 2, 3, 4],
[10, 11, 12, 13, 14],
[20, 21, 22, 23, 24],
[30, 31, 32, 33, 34],
[40, 41, 42, 43, 44],
[50, 51, 52, 53, 54],
[60, 61, 62, 63, 64],
[70, 71, 72, 73, 74],
[80, 81, 82, 83, 84],
[90, 91, 92, 93, 94]])
But its not correct.
Thank you in advance!
Answers:
When you use b = a.reshape((5,4,5)) you just create a different view on the same data used by the array a. (ie changes to the elements of a will appear in b). reshape() does not copy data in this case, so it is a very fast operation. Slicing b and slicing a accesses the same memory, so there shouldn’t be any need for a different syntax for the b array (just use a[:10]). If you have created a copy of the data, perhaps with np.resize(), and discarded a, just reshape b: b.reshape((20,5))[:10].
By reshaping (20,5) to (5,4,5), there’s no way you can pull out the 1st half of the values. You can’t split those 5 rows into 2 even groups:
In [9]: b[:2]
Out[9]:
array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]],
[[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29],
[30, 31, 32, 33, 34],
[35, 36, 37, 38, 39]]])
In [10]: b[:3]
Out[10]:
array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]],
[[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29],
[30, 31, 32, 33, 34],
[35, 36, 37, 38, 39]],
[[40, 41, 42, 43, 44],
[45, 46, 47, 48, 49],
[50, 51, 52, 53, 54],
[55, 56, 57, 58, 59]]])
The last row of a[:10] is in the middle of b[3,:,:].
Note that b[:2] is (2,4,5), 8 rows of a, grouped into 2 sets of 4.
Now if you’d done c=a.reshape(4,5,5), then c[:2] would have those same 10 rows – in 2 sets of 5. And c[:2].reshape(10,-1) will look just like a[:10].
There could be a programmatic way to get what you want, but not a python slice.
It is important to understand what every component in the shape tells us about the arrangement. I like to think in terms of vectors.
Let’s talk about the shape (20, 5) – this would mean, I have 20 vectors where every vector has 5 elements.
For the shape (5, 4, 5) – this would mean, I have 5 vectors, where each vector again has 4 vectors where every vector within has 5 elements.
This might sound complicated but with some deliberation, this could be understood.
Coming to your question, by a[:10] you want to retrieve the first 10 rows where each row should be a vector containing 5 elements but using a shape of (5, 4, 5).
This is only possible if you retrieve the first 4 vectors from 1st vector of the leftmost dimension (5), next 4 vectors from the next vector and next 2 from the 3rd.
Python slicing might not be the best tool to achieve this.
I have data in a numpy array:
a = np.arange(100)
a = a.reshape((20,5))
When I type
a[:10]
it returns
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29],
[30, 31, 32, 33, 34],
[35, 36, 37, 38, 39],
[40, 41, 42, 43, 44],
[45, 46, 47, 48, 49]])
Now i decided to reshape the array into 3d array.
b = a.reshape((5,4,5))
array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]],
[[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29],
[30, 31, 32, 33, 34],
[35, 36, 37, 38, 39]],
[[40, 41, 42, 43, 44],
[45, 46, 47, 48, 49],
[50, 51, 52, 53, 54],
[55, 56, 57, 58, 59]],
[[60, 61, 62, 63, 64],
[65, 66, 67, 68, 69],
[70, 71, 72, 73, 74],
[75, 76, 77, 78, 79]],
[[80, 81, 82, 83, 84],
[85, 86, 87, 88, 89],
[90, 91, 92, 93, 94],
[95, 96, 97, 98, 99]]])
How do I slice b to that I obtain the values like a[:10]?
I tried
b[:10,0,:5]
array([[ 0, 1, 2, 3, 4],
[10, 11, 12, 13, 14],
[20, 21, 22, 23, 24],
[30, 31, 32, 33, 34],
[40, 41, 42, 43, 44],
[50, 51, 52, 53, 54],
[60, 61, 62, 63, 64],
[70, 71, 72, 73, 74],
[80, 81, 82, 83, 84],
[90, 91, 92, 93, 94]])
But its not correct.
Thank you in advance!
When you use b = a.reshape((5,4,5)) you just create a different view on the same data used by the array a. (ie changes to the elements of a will appear in b). reshape() does not copy data in this case, so it is a very fast operation. Slicing b and slicing a accesses the same memory, so there shouldn’t be any need for a different syntax for the b array (just use a[:10]). If you have created a copy of the data, perhaps with np.resize(), and discarded a, just reshape b: b.reshape((20,5))[:10].
By reshaping (20,5) to (5,4,5), there’s no way you can pull out the 1st half of the values. You can’t split those 5 rows into 2 even groups:
In [9]: b[:2]
Out[9]:
array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]],
[[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29],
[30, 31, 32, 33, 34],
[35, 36, 37, 38, 39]]])
In [10]: b[:3]
Out[10]:
array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]],
[[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29],
[30, 31, 32, 33, 34],
[35, 36, 37, 38, 39]],
[[40, 41, 42, 43, 44],
[45, 46, 47, 48, 49],
[50, 51, 52, 53, 54],
[55, 56, 57, 58, 59]]])
The last row of a[:10] is in the middle of b[3,:,:].
Note that b[:2] is (2,4,5), 8 rows of a, grouped into 2 sets of 4.
Now if you’d done c=a.reshape(4,5,5), then c[:2] would have those same 10 rows – in 2 sets of 5. And c[:2].reshape(10,-1) will look just like a[:10].
There could be a programmatic way to get what you want, but not a python slice.
It is important to understand what every component in the shape tells us about the arrangement. I like to think in terms of vectors.
Let’s talk about the shape (20, 5) – this would mean, I have 20 vectors where every vector has 5 elements.
For the shape (5, 4, 5) – this would mean, I have 5 vectors, where each vector again has 4 vectors where every vector within has 5 elements.
This might sound complicated but with some deliberation, this could be understood.
Coming to your question, by a[:10] you want to retrieve the first 10 rows where each row should be a vector containing 5 elements but using a shape of (5, 4, 5).
This is only possible if you retrieve the first 4 vectors from 1st vector of the leftmost dimension (5), next 4 vectors from the next vector and next 2 from the 3rd.
Python slicing might not be the best tool to achieve this.