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Dynamic expansion of scipy array

Is there a way to dynamically expand a scipy array

from scipy import sci
time = sci.zeros((n,1), 'double')

Is it possible to increase the size of the time array after that?

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2 answers

It is possible to expand arrays using the resize method, but it can be a slow operation for large arrays, so avoid it if possible * .

For instance:

 import scipy as sci n=3 time = sci.zeros((n,1), 'double') print(time) # [[ 0.] # [ 0.] # [ 0.]] time.resize((n+1,2)) print(time) # [[ 0. 0.] # [ 0. 0.] # [ 0. 0.] # [ 0. 0.]]

* Instead, find out how large you need the array from the beginning, and select this form for time only once. In general, it is faster to reconfigure than to resize.

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As a result, the time array is just a Numpy Array , you can use standard Numpy methods to control them, such as numpy # insert , which returns a modified array with new elements inserted into it. Usage examples from Numpy docs (here np short for numpy ):

 >>> a = np.array([[1, 1], [2, 2], [3, 3]]) >>> a array([[1, 1], [2, 2], [3, 3]]) >>> np.insert(a, 1, 5) array([1, 5, 1, 2, 2, 3, 3]) >>> np.insert(a, 1, 5, axis=1) array([[1, 5, 1], [2, 5, 2], [3, 5, 3]])

In addition, numpy#insert faster than numpy#resize :

 >>> timeit np.insert(time, 1, 1, 1) 100000 loops, best of 3: 16.7 us per loop >>> timeit np.resize(time, (20,1)) 10000 loops, best of 3: 27.1 us per loop
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