The criterion to satisfy for providing the new shape is that 'The new shape should be compatible with the original shape'
numpy allow us to give one of new shape parameter as -1 (eg: (2,-1) or (-1,3) but not (-1, -1)). It simply means that it is an unknown dimension and we want numpy to figure it out. And numpy will figure this by looking at the 'length of the array and remaining dimensions' and making sure it satisfies the above mentioned criteria
Now see the example.
z = np.array([[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]])
z.shape
(3, 4)
Now trying to reshape with (-1) . Result new shape is (12,) and is compatible with original shape (3,4)
z.reshape(-1)
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12])
Now trying to reshape with (-1, 1) . We have provided column as 1 but rows as unknown . So we get result new shape as (12, 1).again compatible with original shape(3,4)
z.reshape(-1,1)
array([[ 1],
[ 2],
[ 3],
[ 4],
[ 5],
[ 6],
[ 7],
[ 8],
[ 9],
[10],
[11],
[12]])
The above is consistent with numpy advice/error message, to use reshape(-1,1) for a single feature; i.e. single column
Reshape your data using
array.reshape(-1, 1)if your data has a single feature
New shape as (-1, 2). row unknown, column 2. we get result new shape as (6, 2)
z.reshape(-1, 2)
array([[ 1, 2],
[ 3, 4],
[ 5, 6],
[ 7, 8],
[ 9, 10],
[11, 12]])
Now trying to keep column as unknown. New shape as (1,-1). i.e, row is 1, column unknown. we get result new shape as (1, 12)
z.reshape(1,-1)
array([[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]])
The above is consistent with numpy advice/error message, to use reshape(1,-1) for a single sample; i.e. single row
Reshape your data using
array.reshape(1, -1)if it contains a single sample
New shape (2, -1). Row 2, column unknown. we get result new shape as (2,6)
z.reshape(2, -1)
array([[ 1, 2, 3, 4, 5, 6],
[ 7, 8, 9, 10, 11, 12]])
New shape as (3, -1). Row 3, column unknown. we get result new shape as (3,4)
z.reshape(3, -1)
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
And finally, if we try to provide both dimension as unknown i.e new shape as (-1,-1). It will throw an error
z.reshape(-1, -1)
ValueError: can only specify one unknown dimension
Say we have a 3 dimensional array of dimensions 2 x 10 x 10:
r = numpy.random.rand(2, 10, 10)
Now we want to reshape to 5 X 5 x 8:
numpy.reshape(r, shape=(5, 5, 8))
will do the job.
Note that, once you fix first dim = 5 and second dim = 5, you don't need to determine third dimension. To assist your laziness, Numpy gives the option of using -1:
numpy.reshape(r, shape=(5, 5, -1))
will give you an array of shape = (5, 5, 8).
Likewise,
numpy.reshape(r, shape=(50, -1))
will give you an array of shape = (50, 4)
You can read more at http://anie.me/numpy-reshape-transpose-theano-dimshuffle/
Reshape doesn't change the data as mentioned here. Resize changes the data as can be seen here.
Here are some examples:
>>> numpy.random.rand(2,3)
array([[ 0.6832785 , 0.23452056, 0.25131171],
[ 0.81549186, 0.64789272, 0.48778127]])
>>> ar = numpy.random.rand(2,3)
>>> ar.reshape(1,6)
array([[ 0.43968751, 0.95057451, 0.54744355, 0.33887095, 0.95809916,
0.88722904]])
>>> ar
array([[ 0.43968751, 0.95057451, 0.54744355],
[ 0.33887095, 0.95809916, 0.88722904]])
After reshape the array didn't change, but only output a temporary array reshape.
>>> ar.resize(1,6)
>>> ar
array([[ 0.43968751, 0.95057451, 0.54744355, 0.33887095, 0.95809916,
0.88722904]])
After resize the array changed its shape.
One major difference is reshape() does not change your data, but resize() does change it. resize() first accommodates all the values in the original array. After that, if extra space is there (or size of new array is greater than original array), it adds its own values. As @David mentioned in comments, what values resize() adds depends on how that is called.
You can call reshape() and resize() function in the following two ways.
numpy.resize()
ndarray.resize() - where ndarray is an n dimensional array you are resizing.
You can similarly call reshape also as numpy.reshape() and ndarray.reshape(). But here they are almost the same except the syntax.
One point to notice is that, reshape() will always try to return a view wherever possible, otherwise it would return a copy. Also, it can't tell what will be returned when, but you can make your code to raise error whenever the data is copied.
For resize() function, numpy.resize() returns a new copy of the array whereas ndarray.resize() does it in-place. But they don't go to the view thing.
Now coming to the point that what the values of extra elements should be. From the documentation, it says
If the new array is larger than the original array, then the new array is filled with repeated copies of a. Note that this behavior is different from a.resize(new_shape) which fills with zeros instead of repeated copies of a.
So for ndarray.resize() it is the value 0, but for numpy.resize() it is the values of the array itself (of course, whatever can fit in the new size). The below code snippet will make it clear.
In [40]: arr = np.array([1, 2, 3, 4])
In [41]: np.resize(arr, (2,5))
Out[41]:
array([[1, 2, 3, 4, 1],
[2, 3, 4, 1, 2]])
In [42]: arr.resize((2,5))
In [43]: arr
Out[43]:
array([[1, 2, 3, 4, 0],
[0, 0, 0, 0, 0]])
You can also see that ndarray.resize() returns None and does the resizing in-place.
More simpler way with python syntax sugar is to use
b.reshape(-1,1)
where the system will automatically compute the correct shape instead "-1"
ndarray.reshape() returns a new view, or a copy (depends on the new shape). It does not modify the array in place.
b.reshape((10, 1))
as such is effectively no-operation, since the created view/copy is not assigned to anything. The "fix" is simple:
b_new = b.reshape((10, 1))
The amount of memory used should not differ at all between the 2 shapes. Numpy arrays use the concept of strides and so the dimensions (10,) and (10, 1) can both use the same buffer; the amounts to jump to next row and column just change.