Time complexity of numpy.transpose

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What is the time-complexity of np.transpose?

In my opinion, it loops two for loops inside so, that means it should have O(n2) complexity but can someone confirm on that? Also, is there any way so, that I can reduce the time-complexity of matrix transpose

python-3.x numpy python
3 Answers

In memory the matrices are represented as blocks of contiguous memory, that is as if it were a one-dimensional array. N-dimensions are an abstraction that we humans use to make the problem more understandable. For numpy, transposing the matrix is ​​simply an axis change, but the memory is not changed.

So the temporal complexity is O(1) because to transpose an array, numpy just swaps the shape and stride information for each axis.

No data needs to be copied for this to happen. Numpy can simply change how it looks at the underlying memory to construct the new array.

If you want to deepen the topic you can see this beautiful answer with illustrations

user_93024880

It is O(1), because it does not copy data at all. Just modifies the shape and strides.

>>> A = np.random.rand(3,4)
>>> A.flags
  C_CONTIGUOUS : True
  F_CONTIGUOUS : False
  OWNDATA : True
  WRITEABLE : True
  ALIGNED : True
  WRITEBACKIFCOPY : False
  UPDATEIFCOPY : False
>>> np.transpose(A).flags
  C_CONTIGUOUS : False
  F_CONTIGUOUS : True
  OWNDATA : False
  WRITEABLE : True
  ALIGNED : True
  WRITEBACKIFCOPY : False
  UPDATEIFCOPY : False

Note that C_CONTIGUOUS, F_CONTIGUOUS were swapped (i.e. major iteration order changes), and the transposed array has OWNDATA false (i.e. it is just a view into the original array's data).

Related tip: when you have a view like this, to find the array owning the data you can check the base attribute

>>> np.transpose(A).base is A
True
user_6740390

Yes adding to that we can also transpose matrix by using zip 

list(zip(*matrix))
user_110037250
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