Removing an element of a list

Question

I have a list of lists containing dictionaries:

[[{'event_id': 1, 'order_id': 1, 'item_id': 1, 'count': 1, 'return_count': 0, 'status': 'OK'}, 
  {'event_id': 2, 'order_id': 1, 'item_id': 1, 'count': 1, 'return_count': 0, 'status': 'OK'}],
 [{'order_id': 2, 'item_id': 1, 'event_id': 1, 'count': 3, 'return_count': 1, 'status': 'OK'}, 
  {'order_id': 2, 'event_id': 2, 'item_id': 1, 'count': 3, 'return_count': 1, 'status': 'OK'},
  {'order_id': 2, 'event_id': 1, 'item_id': 2, 'count': 4, 'return_count': 2, 'status': 'OK'}]]

For each item in the given order I only need those dictionaries whose event_id is max. So I wrote the following code:

for el in lst:
    for element in el:
        if element['event_id'] != max(x['event_id'] for x in el if element['item_id'] == x['item_id']):
            el.remove(element)

lst is the initial list. For some reason, after running the code lst remains unchanged.

changing the size of the list while iterating over it is problematic in many programming languages — Thomas Weller, Sep 02 '21 at 20:04

score 1 · Answer 1 · answered Sep 02 '21 at 20:13

This isn't in one line, but it does return dictionaries with the max event id

dictlist = [
    [{'event_id': 1, 'order_id': 1, 'item_id': 1, 'count': 1, 'return_count': 0, 'status': 'OK'}, 
     {'event_id': 2, 'order_id': 1, 'item_id': 1, 'count': 1, 'return_count': 0, 'status': 'OK'}],
    [{'order_id': 2, 'item_id': 1, 'event_id': 1, 'count': 3, 'return_count': 1, 'status': 'OK'}, 
     {'order_id': 2, 'event_id': 2, 'item_id': 1, 'count': 3, 'return_count': 1, 'status': 'OK'},
     {'order_id': 2, 'event_id': 1, 'item_id': 2, 'count': 4, 'return_count': 2, 'status': 'OK'}]]

max = 0
parsed = []
for item in dictlist:
    for i in item:
        if i['event_id'] > max:
            max = i['event_id']
for item in dictlist:
    for dic in item:
        if dic['event_id'] == max:
            parsed.append(dic)

Nothing wrong with multi-line code; many programmers find it easier to understand, maintain and modify than one-liners/comprehensions. — DisappointedByUnaccountableMod, Sep 02 '21 at 20:19

not_speshal · Answer 2 · 2021-09-02T20:22:33.637

0

sort on "event_id" and keep only the max (last element):

result = [sorted(l, key=lambda x: x["event_id"])[-1] for l in lst]

If you want to keep all dictionaries with the max "event_id":

lsts = [[x for x in l if x["event_id"]==max(l, key=lambda x: x["event_id"])["event_id"]] for l in lst]
result = [item for sublist in lsts for item in lsts]

edited Sep 02 '21 at 20:22

answered Sep 02 '21 at 20:08

not_speshal

22,093
2
15
30

this would only keep one element with event id, in the example data there are multiple elements that want to be kept – Tadhg McDonald-Jensen Sep 02 '21 at 20:13
In the example there is only one `event_id` for each `order_id` which corresponds to the max. But I see how there could be multiple. Edited. – not_speshal Sep 02 '21 at 20:15
for the first one - can be shorter if you'll remove the "reverse" keyword argument and pick the last element – absoup Sep 02 '21 at 20:20
@absofacto - Brilliant suggestion! :) – not_speshal Sep 02 '21 at 20:22
Perhaps I didn't explain the problem in the right way. If the item_id is different within same order_id, then these two should not be compared. – Mansur Sep 02 '21 at 20:29

score 0 · Accepted Answer · answered Sep 02 '21 at 20:21

0

You're trying to remove an element from a list you're iterating over, that won't work. And it's really hard to understand what you're trying to do; my suggestion would be to do something like this:

newlst = []
for el in lst:
    max_event_id = max(element['event_id'] for element in el)
    max_event_element = next(element for element in el if element['event_id'] == max_event_id)
    newlst.append(max_event_element)

The expected result ends up in the newlst variable.

answered Sep 02 '21 at 20:21

Óscar López

232,561
37
312
386

I understood my mistake. Thank you. The problem is I also need to take into account item_id. If the item_id is different for two dictionaries then these dictionaries must not be comapred – Mansur Sep 02 '21 at 20:33

Removing an element of a list

3 Answers3