Indexing python array with a python array with redundant elements

Question

I'm experiencing a problem with array indexing. Suppose you have an array a and another array b you want to use to use as index for a in order to assign some values to the position pointed by b elements.

a=numpy.zeros(5)
print a

[ 0.  0.  0.  0.  0.]

Now I would like to increase the second element twice

b=numpy.array([1,1])
a[b]+=1.
print a

[ 0.  1.  0.  0.  0.]

while I expected to have

[ 0.  2.  0.  0.  0.]

There are no problems if the array b has no redundancies (all values of its elements are different). Has somebody got a solution for such a problem which avoids using for loops? Is it a bug in numpy? Thanks in advance

Ferdinand Beyer · Accepted Answer · 2011-08-02 15:03:40Z

7

When you use an integer array for indexing another array, NumPy cannot create an adequate view, since the resulting array may not be representable with strides. Therefore, it will return a copy:

>>> a = np.zeros(5)
>>> b = np.array([1, 1])
>>> c = a[b]
>>> c
array([ 0.,  0.])
>>> c.base is a
False

When using this index with in-place operations like +=, NumPy will interpret it differently than you expect. Instead of "Walk the index array and perform the operation on each element in turn", it will first select all values that are indexed by b (in this case, just one element with index 1), then perform the operation on these elements once.

answered Aug 2, 2011 at 15:03

Ferdinand Beyer

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1 Comment

fightthewar Over a year ago

Thanks for the clear answer. In your opinion, is there a clever way round this problem (i.e using arrays or dictionaries maybe)? I'm working with very big arrays so I would try anything to avoid loops...

HYRY · Accepted Answer · 2011-08-03 03:54:57Z

2

or you can use bincount():

a=numpy.zeros(5)
idx = numpy.bincount([0,0,0,1,1,3,3])
a[:len(idx)]+=idx

answered Aug 3, 2011 at 3:54

HYRY

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Comments

eumiro · Accepted Answer · 2011-08-02 15:07:18Z

1

You can try:

a += numpy.histogram(b, numpy.arange(len(a)+1))[0]

This will return a = array([ 0., 2., 0., 0., 0.])

answered Aug 2, 2011 at 15:07

eumiro

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