[Numpy-discussion] iterate over multiple arrays

[Sturla Molden]

Den 12.09.2011 08:52, skrev David Froger:
> Hy everybody,
>
> I'm wondering what is the (best) way to apply the same function to multiple
> arrays.
>

I tried to experiment a bit with this. Here is from an ipython session:

Create some arrays:

In [1]: import numpy as np

In [2]: a = np.zeros(4)

In [3]: b = a+1

In [4]: c = a+2

In [5]: d = a+3




Create an array with dtype=object to store the four arrays a-d:

In [6]: e = np.zeros(4, dtype=object)

In [7]: e[:] = a,b,c,d

In [8]: e

Out[8]:

array([[ 0. 0. 0. 0.], [ 1. 1. 1. 1.], [ 2. 2. 2. 2.],

[ 3. 3. 3. 3.]], dtype=object)




Modify array e inplace:

In [9]: e += 1

In [10]: e

Out[10]:

array([[ 1. 1. 1. 1.], [ 2. 2. 2. 2.], [ 3. 3. 3. 3.],

[ 4. 4. 4. 4.]], dtype=object)




This did not modify arrays a-d though:


In [11]: a

Out[11]: array([ 0., 0., 0., 0.])




Albeit e[0] was array a prior to the iadd:

In [12]: e[:] = a,b,c,d

In [13]: e[0] is a

Out[13]: True



We can apply a function to the arrays in e, getting array([f(a), f(b), 
f(c), f(d)]):


In [14]: np.sum(e)

Out[14]: array([ 6., 6., 6., 6.])




Observe that if e was a 2D array, np.sum(e) would have returned a 
scalar, like so:

In [18]: g = np.array((a,b,c,d))

In [19]: g

Out[19]:

array([[ 0., 0., 0., 0.],

[ 1., 1., 1., 1.],

[ 2., 2., 2., 2.],

[ 3., 3., 3., 3.]])

In [20]: np.sum(g)

Out[20]: 24.0



Which means:


We can create an array of arrays, and have numpy broadcast an ufunc to 
multiple arrays.


The other obvious way is, as mentioned by others, to keep the arrays in 
a normal
Python container (e.g. list) and use a for loop or functional 
programming (map,
apply, reduce).


Sturla



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