The recommended way to do this is to preallocate before the loop and use slicing and indexing to insert
my_array = numpy.zeros(1,1000)
for i in xrange(1000):
#for 1D array
my_array[i] = functionToGetValue(i)
#OR to fill an entire row
my_array[i:] = functionToGetValue(i)
#or to fill an entire column
my_array[:,i] = functionToGetValue(i)
numpy does provide an array.resize()
method, but this will be far slower due to the cost of reallocating memory inside a loop. If you must have flexibility, then I'm afraid the only way is to create an array
from a list
.
EDIT: If you are worried that you're allocating too much memory for your data, I'd use the method above to over-allocate and then when the loop is done, lop off the unused bits of the array using array.resize()
. This will be far, far faster than constantly reallocating the array inside the loop.
EDIT: In response to @user248237's comment, assuming you know any one dimension of the array (for simplicity's sake):
my_array = numpy.array(10000, SOMECONSTANT)
for i in xrange(someVariable):
if i >= my_array.shape[0]:
my_array.resize((my_array.shape[0]*2, SOMECONSTANT))
my_array[i:] = someFunction()
#lop off extra bits with resize() here
The general principle is "allocate more than you think you'll need, and if things change, resize the array as few times as possible". Doubling the size could be thought of as excessive, but in fact this is the method used by several data structures in several standard libraries in other languages (java.util.Vector
does this by default for example. I think several implementations of std::vector
in C++ do this as well).