initialize a numpy array

Question

numpy.zeros

Return a new array of given shape and type, filled with zeros.

or

numpy.ones

Return a new array of given shape and type, filled with ones.

or

numpy.empty

Return a new array of given shape and type, without initializing entries.

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However, the mentality in which we construct an array by appending elements to a list is not much used in numpy, because it's less efficient (numpy datatypes are much closer to the underlying C arrays). Instead, you should preallocate the array to the size that you need it to be, and then fill in the rows. You can use numpy.append if you must, though.

The way I usually do that is by creating a regular list, then append my stuff into it, and finally transform the list to a numpy array as follows :

import numpy as np
big_array = [] #  empty regular list
for i in range(5):
    arr = i*np.ones((2,4)) # for instance
    big_array.append(arr)
big_np_array = np.array(big_array)  # transformed to a numpy array

of course your final object takes twice the space in the memory at the creation step, but appending on python list is very fast, and creation using np.array() also.

Introduced in numpy 1.8:

numpy.full

Return a new array of given shape and type, filled with fill_value.

Examples:

>>> import numpy as np
>>> np.full((2, 2), np.inf)
array([[ inf,  inf],
       [ inf,  inf]])
>>> np.full((2, 2), 10)
array([[10, 10],
       [10, 10]])

Array analogue for the python's

a = []
for i in range(5):
    a.append(i)

is:

import numpy as np

a = np.empty((0))
for i in range(5):
    a = np.append(a, i)

To initialize a numpy array with a specific matrix:

import numpy as np

mat = np.array([[1, 1, 0, 0, 0],
                [0, 1, 0, 0, 1],
                [1, 0, 0, 1, 1],
                [0, 0, 0, 0, 0],
                [1, 0, 1, 0, 1]])

print mat.shape
print mat

output:

(5, 5)
[[1 1 0 0 0]
 [0 1 0 0 1]
 [1 0 0 1 1]
 [0 0 0 0 0]
 [1 0 1 0 1]]

numpy.fromiter() is what you are looking for:

big_array = numpy.fromiter(xrange(5), dtype="int")

It also works with generator expressions, e.g.:

big_array = numpy.fromiter( (i*(i+1)/2 for i in xrange(5)), dtype="int" )

If you know the length of the array in advance, you can specify it with an optional 'count' argument.

You do want to avoid explicit loops as much as possible when doing array computing, as that reduces the speed gain from that form of computing. There are multiple ways to initialize a numpy array. If you want it filled with zeros, do as katrielalex said:

big_array = numpy.zeros((10,4))

EDIT: What sort of sequence is it you're making? You should check out the different numpy functions that create arrays, like numpy.linspace(start, stop, size) (equally spaced number), or numpy.arange(start, stop, inc). Where possible, these functions will make arrays substantially faster than doing the same work in explicit loops