Handling of duplicate indices in NumPy assignments

Question

I am setting the values of multiple elements in a 2D array, however my data sometimes contains multiple values for a given index.

It seems that the "later" value is always assigned (see examples below) but is this behaviour guaranteed or is there a chance I will get inconsistent results? How do I know that I can interpret "later" in the way that I would like in a vectorized assignment?

i.e. in my first example will a definitely always contain 4 and in the second example would it ever print values[0]?

Very simple example:

import numpy as np
indices = np.zeros(5,dtype=np.int)
a[indices] = np.arange(5)
a # array([4])

Another example

import numpy as np

grid = np.zeros((1000, 800))

# generate indices and values
xs = np.random.randint(0, grid.shape[0], 100)
ys = np.random.randint(0, grid.shape[1], 100)
values = np.random.rand(100)

# make sure we have a duplicate index
print values[0], values[5]
xs[0] = xs[5]
ys[0] = ys[5]

grid[xs, ys] = values

print "output value is", grid[xs[0], ys[0]]
# always prints value of values[5]

Answer 1

In NumPy 1.9 and later this will in general not be well defined.

The current implementation iterates over all (broadcasted) fancy indexes (and the assignment array) at the same time using separate iterators, and these iterators all use C-order. In other words, currently, yes you can. Since you maybe want to know it more exact. If you compare mapping.c in NumPy, which handles these things, you will see that it uses PyArray_ITER_NEXT, which is documented to be in C-order.

For the future I would paint the picture differently. I think it would be good to iterate all indices + the assignment array together using the newer iterator. If this is done, then the order could be kept open for the iterator to decide the fastest way. If you keep it open to the iterator, it is hard to say what would happen, but you cannot be certain that your example works (probably the 1-d case you still can, but...).

So, as far as I can tell it works currently, but it is undocumented (for all I know) so if you actually think that this should be ensured, you would need to lobby for it and best write some tests to make sure it can be guaranteed. Because at least am tempted to say: if it makes things faster, there is no reason to ensure C-order, but of course maybe there is a good reason hidden somewhere...

The real question here is: Why do you want that anyway? ;)

Answer 2

I know this has been satisfactorily answered, but I wanted to mention that it is documented as being the "last value" (perhaps informally) in the Tentative Numpy Tutorial under Indexing with Arrays of Indices:

However, when the list of indices contains repetitions, the assignment is done several times, leaving behind the last value:

>>> a = arange(5)
>>> a[[0,0,2]]=[1,2,3]  
>>> a
array([2, 1, 3, 3, 4])  

This is reasonable enough, but watch out if you want to use Python's += construct, as it may not do what you expect:

>>> a = arange(5) 
>>> a[[0,0,2]]+=1  
>>> a
array([1, 1, 3, 3, 4])  

Even though 0 occurs twice in the list of indices, the 0th element is only incremented once. This is because Python requires a+=1 to be equivalent to a=a+1.

Answer 3

I'm not answering you question directly, I'd just like to make a point that even if you can rely on that behavior being consistent, you better not.

Consider:

a = np.zeros(4)
x = np.arange(4)
indices = np.zeros(4,dtype=np.int)
a[indices] += x

At this point, is it reasonable to assume that a.sum() is a's previous sum + x.sum()?

assert a.sum() == x.sum()
--> AssertionError 

a
= array([ 3.,  0.,  0.,  0.])

In your case, when assigning to an array using duplicate indices, the result is intuitive: assignment to the same index takes place multiple times, thus only the last assignment "sticks" (it overwrites previous ones).

But this is not the case in this example. It is no longer intuitive. If it were, the in-place addition would have taken place multiple times, because addition is cumulative in its nature.

So, put another way, you are risking getting caught in this trap:

• you start working with duplicate indices
• you see all is well, behavior is exactly as you expect
• you stop paying attention to the crucial fact that your operations involve duplicate indices. After all, it makes no difference, does it?
• you start using the same indices in different contexts, e.g. as above
• deep regrets :)

So, quoting @seberg:

The real question here is: Why do you want that anyway? ;)

Answer 4

I found a way with with numpy to do this operation this is clearly not optimal but its faster than looping (with a python for loop)

with: numpy.bincount

size = 5
a = np.arange(size)
index = [0,0,2]
values = [1,2,3]
a[index] += values
a
[2 1 5 3 4]

witch is not correct but:

size = 5
a = np.arange(size)
index = [0,0,2]
values = [1,2,3]
result = np.bincount(index, values, size)
a += result
a
[3 1 5 3 4]

wich is good !

Answer 5

Time changes. An updated answer should be given.

size = 5
a = np.arange(size)
index = [0,0,2]
values = [1,2,3]
np.add.at(a,[0,0,2],values)
a