Delete an object from a list of objects in python

Question

I have a list of objects.

My list format:

my_list = [
    {
        "y": "9",
        "x": "num.nine"
    },
    {
        "y": "8",
        "x": "eight"
    },
    {
        "y": "7",
        "x": "num.seven"
    },
    {
        "y": "6",
        "x": "six"
    },
    {
        "y": "5",
        "x": "num.five"
    }
]

I want to delete the objects where my_list['x'] is not like num.

This is what I did:

for val in my_list:
    if('num' not in val['x']):
        del val['x']
        del val['y']

pprint(my_list)

The result of that code is:

[
 {'x': 'num.nine', 'y': '9'},
 {},
 {'x': 'num.seven', 'y': '7'},
 {},
 {'x': 'num.five', 'y': '5'}
]

How do I delete the whole object?

My expected result is:

[
 {'x': 'num.nine', 'y': '9'},
 {'x': 'num.seven', 'y': '7'},
 {'x': 'num.five', 'y': '5'}
]

my_list = [obj for obj in my_list if obj['x'].startswith('num.')] — Olvin Roght
– Olvin Roght, Commented Jul 17, 2020 at 20:20
list comprehension only returning entries you want, or for loop making a COPY and using my_list.delete, deleting the entries you dont want — Derek Eden
– Derek Eden, Commented Jul 17, 2020 at 21:02

Ehtesham Siddiqui · Accepted Answer · 2020-07-18 02:54:41Z

2

Why delete? Alternatively

my_list = [val for val in my_list if 'num' in val['x']]

You can use filter function as well. (Note that filter in python2 returns a list, and in python3 it returns a iterator)

PS. I am not sure whether the filter condition you have written is exactly what you need, yet I have used the same condition. Most likely you want to include val when val['x'] is starting with "num."

edited Jul 18, 2020 at 2:54

answered Jul 17, 2020 at 20:24

Ehtesham Siddiqui

3223 silver badges6 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

juanpa.arrivillaga Over a year ago

Terminology note, filter returns an iterator, not a generator.

Parth Shah · Accepted Answer · 2020-07-17 20:24:30Z

1

Add what you need to a new list:

new_list = []
for elem in my_list:
    if('num' in elem['x']):
        new_list.append(elem)

answered Jul 17, 2020 at 20:24

Parth Shah

1,51517 silver badges26 bronze badges

Comments

Ken Kinder · Accepted Answer · 2020-07-17 20:26:39Z

0

The most efficient thing to is probably to use the Python filter function. Consider this example:

>>> my_list = my_list=[ 
         { 
             "y": "9", 
             "x": "num.nine" 
         }, 
         { 
             "y": "8", 
             "x": "eight" 
         }, 
         { 
             "y": "7", 
             "x": "num.seven" 
         }, 
         { 
             "y": "6", 
             "x": "six" 
         }, 
         { 
             "y": "5", 
             "x": "num.five" 
         } 
    ]                                                                                                                                                                                          
>>> my_list = list(filter(lambda item: 'num' in item['x'], my_list))                                                                                                                           
>>> my_list                                                                                                                                                                                    
[{'y': '9', 'x': 'num.nine'},
 {'y': '7', 'x': 'num.seven'},
 {'y': '5', 'x': 'num.five'}]

Note that you're creating a new list with this approach, rather than modifying the old one in place. That's probably a better approach, because if you simultaneously iterate over an object while also modifying it, you'll get surprising results.

answered Jul 17, 2020 at 20:26

Ken Kinder

13.2k6 gold badges54 silver badges72 bronze badges

12 Comments

Mark Over a year ago

In my experience filter() is measurably slower than a list comprehension.

Olvin Roght Over a year ago

@MarkMeyer, depends on size of initial list. In one of questions I've seen detailed test, but not sure if I can find link now. Upd. Tested with timeit and my simple test shows that list comprehension is slightly slower (I used list with 200 dictionaries for test).

Mark Over a year ago

I would be interested to see that @OlvinRoght — with the kind of data and filter in this question, the comprehension is considerably faster whether the list has 5 items or 100000.

Ken Kinder Over a year ago

The reason a list comprehension may not be more efficient is that if the list is especially large, and you don't cast the output of filter() to a list, you can iterate sequentially over the return value of filter() and it won't have to hold a new list in memory or build it ahead of time.

Olvin Roght Over a year ago

@KenKinder, I mean that alternatively you can use (obj for obj in my_list if obj['x'].startswith('num.')) which also won't create list in memory.

|

eyalfu · Accepted Answer · 2020-07-17 20:28:28Z

0

you can do it using list comprehension, this would be the answer:

new_list = [obj for obj in my_list if 'num' in obj['x']]

edited Jul 17, 2020 at 20:28

answered Jul 17, 2020 at 20:25

eyalfu

364 bronze badges

Collectives™ on Stack Overflow

Delete an object from a list of objects in python

4 Answers 4

1 Comment

Comments

12 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

Comments

12 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related