Is it possible to convert a one dimensional array
a = np.array([1,2,3])
to a two dimensional array that is equivalent to
b = np.array([[1],[2],[3]])
- without creating a copy of the array and also
- with
bbeing contiguous?
-
What do you mean by "without creating a copy of the data?"Ffisegydd– Ffisegydd2014-10-23 08:38:18 +00:00Commented Oct 23, 2014 at 8:38
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@Ffisegydd I do not want the data to be copied in memory. Rather I want numpy to use the same memory but interpret it differently.Woltan– Woltan2014-10-23 11:01:03 +00:00Commented Oct 23, 2014 at 11:01
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2 Answers
You can do this using np.newaxis and the T transpose method.
import numpy as np
a = np.array([1,2,3])
a = a[np.newaxis].T
print(a)
# [[1]
# [2]
# [3]]
1 Comment
Woltan
If I run this code,
a is not contiguous and it also appears that the data is copied...Reshaping the array does not copy data (where possible*) and retains C contiguity (usually):
>>> a = np.array([1,2,3])
>>> b = a.reshape(3, 1)
>>> b
array([[1],
[2],
[3]])
>>> b.flags
C_CONTIGUOUS : True
F_CONTIGUOUS : False
OWNDATA : False
WRITEABLE : True
ALIGNED : True
UPDATEIFCOPY : False
*Edit: Reshaping often creates a view of an array, but this is not always possible (docs). There are ways to check that the data has not been copied and that a and b share the same underlying data. For example see here and here.
In the above case, trying a couple of tests showed that reshape created a view of a (no data copied):
>>> a.data == b.data
True
>>> np.may_share_memory(a, b) # function from linked answer above
True
2 Comments
Woltan
I though about this very same solution. How do you know that the data is not copied? Doesn't
id(a) != id(b) prove that the data is copied?
Alex Riley
@Woltan updated my answer slightly. Sharing data between arrays is not always easy to determine (the
id test isn't reliable because of the fundamental differences between Python and NumPy objects).