Python: Increment pointer/view to numpy array

Question

If I was writing C++ I could do the following:

double foo [9] = {0, 1, 2, 3, 4, 5, 6, 7, 8}; 
double* bar = foo;
cout << bar[0] << ", " << bar[1] << ", " << bar[2] << "\n";
bar += 3;
cout << bar[0] << ", " << bar[1] << ", " << bar[2] << "\n";
bar += 3;
cout << bar[0] << ", " << bar[1] << ", " << bar[2] << "\n";

This would print:

0, 1, 2
3, 4, 5
6, 7, 8

I now want to do the same in numpy:

foo = np.arange((9))
bar = foo[:3]
print(bar)
???
print(bar)
???
print(bar)

But I don't know what to put at the ??? The result would hopefully be:

[0 1 2]
[3 4 5]
[6 7 8]

In other words, I'm looking for a way to change where the view bar is pointing at foo, using the existing reference bar.

That is I would like to be able to write the ??? as bar = f(bar), where f is some function defined by the user.

Edit:

This is how I'm doing it currently:

def f(idx=[0]): #mutable default argument
    idx[0] += 3
    return foo[idx[0]-3:idx[0]]
for key in key_list:
    a_dict[key] = f()
    b_dict[key] = f()

for view in range(num_views):
    a_list.append(f())
    b_list.append(f())

It needs to be in a vector because scipy.optimize.least_squares needs all values stored in the same vector.

Why do you want to do this? Python and C++ are very different languages. You should learn to use Python on it's own terms, and not to use things like pointer arithmetic, which python doesn't really support (Python doesn't even have pointers). — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Dec 20, 2018 at 19:15
To make my code easier to maintain and read. In my specific case I need to add many different vectors into a vector, and split them out again, and this would make it very easy to read. — Atnas
– Atnas, Commented Dec 20, 2018 at 19:22
@Atnas if you actually show us what you are trying to do we could probably tell you the numpy-friendly way to do it. This question is a classic case of the X-Y problem. So, instead of asking "how do you do pointer arithmetic in Numpy/Python", before people go down some hacky road of "well, maybe if you import ctypes ..." or "well, using Cython..." you could just tell us what you are trying to achieve — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Dec 20, 2018 at 20:19

Tarifazo · Accepted Answer · 2018-12-21 12:24:37Z

1

A pythonic way is to use indeces:

foo = np.arange(9)
bar = np.arange(3)
print(foo[bar])
>> array([0, 1, 2])
bar += 3
print(foo[bar])
>> array([3, 4, 5])
bar += 3
print(foo[bar])
>> array([6, 7, 8])

A less pythonic way (but more similar to what you are coding) is to use an iterator:

foo = np.arange(9)
bar = foo.reshape(3,3).__iter__()   # or use bar = iter(foo.reshape(3,3))
bar.__next__()   # or use next(bar)
>> array([0, 1, 2])
bar.__next__()
>> array([3, 4, 5])
bar.__next__()
>> array([6, 7, 8])

edited Dec 21, 2018 at 12:24

answered Dec 20, 2018 at 19:43

Tarifazo

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9 Comments

Atnas Over a year ago

Isn't it a bit more pythonic to use the iter() and next() functions?

juanpa.arrivillaga Over a year ago

@Atnas yes, it is. Or to just use a for-loop and not call either of those functions explicitly.

juanpa.arrivillaga Over a year ago

Also, iterators are extremely pythonic.

juanpa.arrivillaga Over a year ago

so, just for sub in foo.reshape(3, -1): print(sub)

Tarifazo Over a year ago

@juanpa.arrivillaga, OP did not ask for a for loop, but for an explicit iterator object.

|

sundance · Accepted Answer · 2018-12-20 19:13:10Z

0

foo = np.arange(9)

bar = foo[:3]
print(bar)

bar = foo[3:6]
print(bar)

bar = foo[6:]
print(bar)

answered Dec 20, 2018 at 19:13

sundance

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1 Comment

juanpa.arrivillaga Over a year ago

this is the right way to do this, but this isn't what the OP is asking for.

ilias iliadis · Accepted Answer · 2018-12-20 20:29:11Z

0

foo = np.arange((9))
loopcounter = 3
startindex = 0
stopindex = startindex + loopcounter
print(foo[startindex:stopindex])
startindex = stopindex
stopindex = startindex + loopcounter
print(foo[startindex:stopindex])
startindex = stopindex
stopindex = startindex + loopcounter
print(foo[startindex:stopindex])

...

answered Dec 20, 2018 at 20:29

ilias iliadis

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