How to loop through a list of objects without changing the objects themselves?

Question

I have a list of boxes defined by their coordinates and geometry as follows:

box = [x y w h]

I would like to loop through these boxes as shown below - problem is I need to change the boxes' representation in the function 'findMatchingbox) - so I boxes it to [xmin ymin xmax ymax] in this function. Problem is, when I give 'i' into the function in permanently changes the boxes in my list, I tried first letting a temp_i = i in the function, and then performing the necessary steps but to no avail.

I'm guessing it must be because python only keeps one copy of each box in memory, how would I send the specific box (i) into the function, extract the necessary information from it after transforming it, without changing the actual box? Can you make a copy of it?

for i in bboxes:
    # Determine if detection belongs to an existing object
    print('1:\t',i)
    boxIDx = self.findMatchingBox(i)
    print('2:\t')

Output:

1:   [464, 282, 48, 48]
2:   [464, 282, 512, 330]

I think it might be helpful to show a minimal example of the code in findMatchingBox — M.T
– M.T, Commented Mar 2, 2016 at 9:14
I think I actually figured it out: I have to make a temporary copy of i and do the calculations on that - I did this using temp_i = list(i). It doesnt work to simply say temp_i = i. — Peter Turner
– Peter Turner, Commented Mar 2, 2016 at 9:16

GLaDOS · Accepted Answer · 2016-03-02 09:18:27Z

1

In python you have mutable and immutable objects.

a = [1,2,3]
b = a
b[0] = 0
print(a)
[0,2,3]

If you want to change the values without changing the list everywhere, you need to copy it.

a = [1,2,3]
b = a.copy()
b[0] = 0
print(a)

[1,2,3]

print(b)

[0,2,3]

answered Mar 2, 2016 at 9:18

GLaDOS

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Comments

Martin Bonner supports Monica · Accepted Answer · 2016-03-02 09:25:04Z

1

You need to clone the list before you modify it. So, if you want to change just this bit of code:

for i in bboxes:
    # Determine if detection belongs to an existing object
    print('1:\t', i)
    box = i(:)
    boxIDx = self.findMatchingBox(box)
    print('2:\t', i)

However better would be to make 'findMatchingBox' clone its argument if it's going to modify it:

def findMatchinBox(self, box)
    box = box(:)
    ....

(it's always a bad idea to modify your arguments without cloning them, unless the modification is the point of the function.)

The reason that just saying:

temp_i = i

doesn't work, is that list objects in python are reference objects. (Think of them a bit like pointers in C if that helps).

It's rather unfortunate that python provides (so far) four ways to clone the list:

temp_i = list(i)
temp_i = i.copy()
temp_i = i.deepcopy()
temp_i = i[:]

Note that with this particular example, copy and deepcopy behave the same - that wouldn't be the case if the list contained (eg) dicts rather than integers.

Personally I think the slice notation is most pythonic. Others may disagree.

edited Mar 2, 2016 at 9:25

answered Mar 2, 2016 at 9:16

Martin Bonner supports Monica

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1 Comment

danidee Over a year ago

you should also add deepcopy as copy won't work for nested lists because of shallow copying

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