NumPy 1.8 introduced np.full(), which is a more direct method than empty() followed by fill() for creating an array filled with a certain value:

>>> np.full((3, 5), 7)
array([[ 7.,  7.,  7.,  7.,  7.],
       [ 7.,  7.,  7.,  7.,  7.],
       [ 7.,  7.,  7.,  7.,  7.]])

>>> np.full((3, 5), 7, dtype=int)
array([[7, 7, 7, 7, 7],
       [7, 7, 7, 7, 7],
       [7, 7, 7, 7, 7]])

This is arguably the way of creating an array filled with certain values, because it explicitly describes what is being achieved (and it can in principle be very efficient since it performs a very specific task).

Use numpy.full():

import numpy as np

np.full(
  shape=10,
  fill_value=3,
  dtype=np.int
)
    
> array([3, 3, 3, 3, 3, 3, 3, 3, 3, 3])

There are lots of ways to do this. The first one-liner that occurred to me is tile:

>>> numpy.tile(2, 25)
array([2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 
       2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 
       2, 2, 2, 2, 2])

You can tile a value in any shape:

>>> numpy.tile(2, (5, 5))
array([[2, 2, 2, 2, 2],
       [2, 2, 2, 2, 2],
       [2, 2, 2, 2, 2],
       [2, 2, 2, 2, 2],
       [2, 2, 2, 2, 2]])

However, as a number of answers below indicate, this isn't the fastest method. It's designed for tiling arrays of any size, not just single values, so if you really just want to fill an array with a single value, then it's much faster to allocate the array first, and then use slice assignment:

>>> a = numpy.empty((5, 5), dtype=int)
>>> a[:] = 2
>>> a
array([[2, 2, 2, 2, 2],
       [2, 2, 2, 2, 2],
       [2, 2, 2, 2, 2],
       [2, 2, 2, 2, 2],
       [2, 2, 2, 2, 2]])

According to a few tests I did, there aren't any faster approaches. However, two of the approaches mentioned in answers below are equally fast: ndarray.fill and numpy.full.

These tests were all done in ipython, using Python 3.6.1 on a newish mac running OS 10.12.6. Definitions:

def fill_tile(value, shape):
    return numpy.tile(value, shape)

def fill_assign(value, shape, dtype):
    new = numpy.empty(shape, dtype=dtype)
    new[:] = value
    return new

def fill_fill(value, shape, dtype):
    new = numpy.empty(shape, dtype=dtype)
    new.fill(value)
    return new

def fill_full(value, shape, dtype):
    return numpy.full(shape, value, dtype=dtype)

def fill_plus(value, shape, dtype):
    new = numpy.zeros(shape, dtype=dtype)
    new += value
    return new

def fill_plus_oneline(value, shape, dtype):
    return numpy.zeros(shape, dtype=dtype) + value

for f in [fill_assign, fill_fill, fill_full, fill_plus, fill_plus_oneline]:
    assert (fill_tile(2, (500, 500)) == f(2, (500, 500), int)).all()

tile is indeed quite slow:

In [3]: %timeit fill_tile(2, (500, 500))
947 µs ± 10.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Slice assignment ties with ndarray.fill and numpy.full for first place:

In [4]: %timeit fill_assign(2, (500, 500), int)
102 µs ± 1.37 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [5]: %timeit fill_fill(2, (500, 500), int)
102 µs ± 1.99 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [6]: %timeit fill_full(2, (500, 500), int)
102 µs ± 1.47 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In-place broadcasted addition is only slightly slower:

In [7]: %timeit fill_plus(2, (500, 500), int)
179 µs ± 3.7 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

And non-in-place broadcasted addition is only slightly slower than that:

In [8]: %timeit fill_plus_oneline(2, (500, 500), int)
213 µs ± 4.74 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

You can use np.full_like:

B = np.full_like(A, 1)

This will create an array with the same properties as A and will fill it with 1.

In case you want to fill it with 1 there is a also a convenience function: np.ones_like

B = np.ones_like(A)

Your example does not work because B.fill does not return anything. It works "in-place". So you fill your B but you immediatly overwrite your variable B with the None return of fill. It would work if you use it like this:

A=np.array([[2,2], [2,2]])
B=np.copy(A)
B.fill(1)