How to remove items from list conditionally

Question

I have a list:

bigdumblist = [
    (0, 0, {'product_id': 2, 'product_uom_qty': 90}),
    (0, 0, {'product_id': 3, 'product_uom_qty': 5}),
    (0, 0, {'product_id': 5, 'product_uom_qty': 69})
]

I want to remove all items from the list where 'product_id' is not 2 or 3, like so:

[
   (0, 0, {'product_id': 2, 'product_uom_qty': 90}),
   (0, 0, {'product_id': 3, 'product_uom_qty': 5})
]

What I have tried:

def not_in(item):
    if item["product_id"] is not 2 or 3:
        bigdumblist.remove((0, 0, {'product_id': 5, 'product_uom_qty': 69}))

for _, _, item in bigdumblist:
    not_in(item)
    break

print(bigdumblist)

Which works but obviously including (0, 0, {'product_id': 5, 'product_uom_qty': 69}) is not a solution. How can I properly remove specific items in the list?

You shouldn't modify a list while you're iterating over it. – Barmar Aug 17 '20 at 21:23 — Barmar, Aug 17 '20 at 21:23
`if item["product_id"] not in (2, 3):` – Barmar Aug 17 '20 at 21:23 — Barmar, Aug 17 '20 at 21:23
Not opposed to making a new list – apexprogramming Aug 17 '20 at 21:25 — apexprogramming, Aug 17 '20 at 21:25

score 3 · Accepted Answer · answered Aug 17 '20 at 21:25

3

You could use a list comprehension like so:

[x for x in bigdumblist if x[2]['product_id'] in [2,3]]

answered Aug 17 '20 at 21:25

Franz

357
3
11

Barmar · Answer 2 · 2020-08-17T21:28:57.817

You shouldn't modify a list while you're iterating over it. You could make a copy of the list first, but it's simpler to use a list comprehension.

Since the list comprehension specifies which elements to keep rather than delete, the condition must be inverted.

bigdumblist = [item for item in bigdumblist if item[2]['product_id'] in (2, 3)]

Note that

if item["product_id"] is not 2 or 3:

is parsed as if it were

if (item["product_id"] is not 2) or 3:

Since 3 is always truthy, this condition will always succeed. Logical operators like and and or are not automatically distributed over relational operators.

score 0 · Answer 3 · answered Aug 17 '20 at 21:26

0

Try a conditional list comprehension like this....

[x for x in bigdumblist if x[2]['product_id'] in [2,3]]

answered Aug 17 '20 at 21:26

caldweln

126
3

Gustavo Kawamoto · Answer 4 · 2020-08-17T21:29:59.217

0

You should really use list comprehension for that kind of filter, something like this:

bigdumblist = [
    (0, 0, {'product_id': 2, 'product_uom_qty': 90}),
    (0, 0, {'product_id': 3, 'product_uom_qty': 5}),
    (0, 0, {'product_id': 5, 'product_uom_qty': 69})
]

newlist = [i for i in bigdumblist if i[2]['product_id'] not in [2,3]]

edited Aug 17 '20 at 21:29

answered Aug 17 '20 at 21:26

Gustavo Kawamoto

2,665
18
27

score 0 · Answer 5 · edited Aug 17 '20 at 21:36

0

List comprehension is one way to do this.

You probably want something similar to:

filtered_list = [item for item in original_list if item["product_id"] in (2, 3)]

Another option is filter:

def valid_products(item):
    return item["product_id"] in (2,3)

filtered_list = list(filter(valid_products, original_list))

edited Aug 17 '20 at 21:36

mkrieger1

19,194
5
54
65

answered Aug 17 '20 at 21:30

edgars

1,038
1
7
17

score -1 · Answer 6 · answered Aug 17 '20 at 21:28

-1

items = [item for item in bigdumblist if item item[2]['product_id'] in (2,3)]

answered Aug 17 '20 at 21:28

AnGG

679
3
9

How to remove items from list conditionally

6 Answers6