CODEWITHSUNDEEP
pythonlist
Shin euichul
Shin euichul

Reputation: 63

Want to remove object itself from list in Python

I am very new to programming and not good at English. plz understand even if I make mistakes. here is my question, I want to remove object is self form list. how can I get result as like first one I tried even though I switch to asd=ad_material_name2

what I tried is 

ad_material_name=['a', 'b', 'c']
ad_material_name2=['a', 'b', 'c']

for aa in ad_material_name:  
    asd=['a', 'b', 'c']
    asd.remove(aa)
    print(asd)

it's result came out like this

['b', 'c']
['a', 'c']
['a', 'b']

so, I thought it worked. but, when I tried this

ad_material_name=['a', 'b', 'c']
ad_material_name2=['a', 'b', 'c']

for aa in ad_material_name:  
    asd=ad_material_name2
    asd.remove(aa)
    print(asd)

it came out like this.

['b', 'c']
['c']
[]

I thought from here it might be a problem of ad_material_name=['a', 'b', 'c']=ad_material_name2 and it some how effect result. so, I tried this.

ad_material_name=['a', 'b', 'c']
ad_material_name2=['a', 'b', 'c', 'b']

for aa in ad_material_name:  
    asd=ad_material_name2
    asd.remove(aa)
    print(asd)

but, it came out like this

['b', 'c', 'd']
['c', 'd']
['d']

how can I get result as like first one I tried even though I switch to asd=ad_material_name2

Upvotes: 0

Views: 98

Answers (4)

Sapphire_Brick
Sapphire_Brick

Reputation: 1672

The problem with your second code is that Python lists are mutable; they can be changed. Assigning a variable to a list, and then changing the variable- changes the list as well.
e.g:

def make123(array):
    array[2] = 3
    array[1] = 2
    array[0] = 1

e = [3, 5, 6]
make123(e)

print(e) # [1, 2, 3]

You might wanna use a tuple, then convert it to a list:

ad_material_name=['a', 'b', 'c']
ad_material_name2=('a', 'b', 'c')

for aa in ad_material_name:  
    asd=list(ad_material_name2)
    asd.remove(aa)
    print(asd)
"""
['b', 'c']
['a', 'c']
['a', 'b']
"""

Upvotes: 0

Vlad Bezden
Vlad Bezden

Reputation: 89527

You might use combinations function from itertools to achieve the same result.

from itertools import combinations

ad_material_name=['a', 'b', 'c']

for x, y in combinations(ad_material_name, r=2):
    print([x, y])

or if you need to create ad_material_name2 variable and assign permutations to it, you can do following:

from itertools import combinations

ad_material_name=['a', 'b', 'c']

ad_material_name2 = [[*comb] for comb in combinations(ad_material_name, r=2)]

print(ad_material_name2)

# output
[['a', 'b'], ['a', 'c'], ['b', 'c']]

Upvotes: 2

alan23273850
alan23273850

Reputation: 262

modify

asd=ad_material_name2

to

asd=ad_material_name2.copy()

so that the "remove" opertion will not modify the list ad_material_name2 directly.

Upvotes: 2

user12471443
user12471443

Reputation: 9

It's the difference that comes from deep copy and shallow copy. Your problem will be solved by using deepcopy.

import copy
ad_material_name=['a', 'b', 'c']
ad_material_name2=['a', 'b', 'c', 'b']

for aa in ad_material_name:
    asd=copy.deepcopy(ad_material_name2)
    asd.remove(aa)
    print(asd)

Upvotes: -1

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