NumPy array initialization (fill with identical values) NumPy array initialization (fill with identical values) python python

NumPy array initialization (fill with identical values)


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).


Updated for Numpy 1.7.0:(Hat-tip to @Rolf Bartstra.)

a=np.empty(n); a.fill(5) is fastest.

In descending speed order:

%timeit a=np.empty(10000); a.fill(5)100000 loops, best of 3: 5.85 us per loop%timeit a=np.empty(10000); a[:]=5 100000 loops, best of 3: 7.15 us per loop%timeit a=np.ones(10000)*510000 loops, best of 3: 22.9 us per loop%timeit a=np.repeat(5,(10000))10000 loops, best of 3: 81.7 us per loop%timeit a=np.tile(5,[10000])10000 loops, best of 3: 82.9 us per loop


I believe fill is the fastest way to do this.

a = np.empty(10)a.fill(7)

You should also always avoid iterating like you are doing in your example. A simple a[:] = v will accomplish what your iteration does using numpy broadcasting.


matomo