Why Does Copied Object Avoid Modification in Python?

Question

I have the following data and following function that works to sort the products to the front of the list that match the specified category.

products =[{'t1': 'bbq',    'category': 2}, {'name': 't5',  'category': 3}, {'name': 't6',   'category': 3}, {'name': 't2', 'category': 2}, {'name': 't3',  'category': 2}, {'name': 't4',  'category': 1}, {'name': 't7',  'category': 1}]
category = 2

I found I have to run this function twice in order to actually get this accomplished, why? How do I need to modify the function to just have to run this once? I am trying to modify products inplace via the category sorted as specified.

def sort_by_cat(products,category):
  for p in products:
    print(p)
    if p['category']==category:
      continue
    else:
      products.append(products.pop(products.index(p)))
  return

sort_by_cat(products,category)
sort_by_cat(products,category)

Desired output:

[{'name': 't1', 'category': 2},
 {'name': 't2', 'category': 2},
 {'name': 't3', 'category': 2},
 {'name': 't4', 'category': 1},
 {'name': 't5', 'category': 3},
 {'name': 't6', 'category': 3},
 {'name': 't7', 'category': 1}]

Because you `.pop` from products while you iterate over it. Everything shifts down one index, and the subsequent iterator ends up skipping an element. Classic error. — juanpa.arrivillaga, Apr 07 '23 at 00:13
The better strategy is simply not to rely on modifying the list in-place. I would just make one pass separating into matching and non-matching, then return `nonmatching + matching` — juanpa.arrivillaga, Apr 07 '23 at 00:17
`products[:] = nonmatching + matching`, but why must this be done in-place? — juanpa.arrivillaga, Apr 07 '23 at 07:02

Why Does Copied Object Avoid Modification in Python?

0 Answers0