numpy.chararray.resize#

method

chararray.resize(new_shape, refcheck=True)#

Change shape and size of array in-place.

Parameters:
new_shapetuple of ints, or n ints

Shape of resized array.

refcheckbool, optional

If False, reference count will not be checked. Default is True.

Returns:
None
Raises:
ValueError

If a does not own its own data or references or views to it exist, and the data memory must be changed. PyPy only: will always raise if the data memory must be changed, since there is no reliable way to determine if references or views to it exist.

SystemError

If the order keyword argument is specified. This behaviour is a bug in NumPy.

See also

resize

Return a new array with the specified shape.

Notes

This reallocates space for the data area if necessary.

Only contiguous arrays (data elements consecutive in memory) can be resized.

The purpose of the reference count check is to make sure you do not use this array as a buffer for another Python object and then reallocate the memory. However, reference counts can increase in other ways so if you are sure that you have not shared the memory for this array with another Python object, then you may safely set refcheck to False.

Examples

Shrinking an array: array is flattened (in the order that the data are stored in memory), resized, and reshaped:

>>> a = np.array([[0, 1], [2, 3]], order='C')
>>> a.resize((2, 1))
>>> a
array([[0],
       [1]])
>>> a = np.array([[0, 1], [2, 3]], order='F')
>>> a.resize((2, 1))
>>> a
array([[0],
       [2]])

Enlarging an array: as above, but missing entries are filled with zeros:

>>> b = np.array([[0, 1], [2, 3]])
>>> b.resize(2, 3) # new_shape parameter doesn't have to be a tuple
>>> b
array([[0, 1, 2],
       [3, 0, 0]])

Referencing an array prevents resizing…

>>> c = a
>>> a.resize((1, 1))
Traceback (most recent call last):
...
ValueError: cannot resize an array that references or is referenced ...

Unless refcheck is False:

>>> a.resize((1, 1), refcheck=False)
>>> a
array([[0]])
>>> c
array([[0]])