Delete list of elements from a list

Question

Suppose I have two lists listBig and listSmall and I want to delete all elements of listSmall from listBig. Assume that all elements of listSmall are present in listBig and that if there are multiple instances of an element of listSmall in listBig, they must not be deleted, meaning deleting one is the key.

Is there a better way to achieve this than:

listBig = [1,2,3,4,5,6,7,8,4,7]
listSmall = [4,7]
deleted = 0
inner = 0
for outer in range(0,len(listSmall)):
    while not deleted == len(listSmall):
        if listSmall[outer]==listBig[inner]:
            listBig.remove(listBig[inner])
            inner=0
            deleted+=1
            break
        inner+=1
print listBig

OUTPUT: [1, 2, 3, 5, 6, 8, 4, 7]

modified the lists to show that if there are elements in listSmall which occir more than once in listBig, multiple occurrences MUST NOT BE DELETED. Thanks. Sorry for creating some confusion. — statBeginner
– statBeginner, Commented Nov 25, 2014 at 21:48
Do you have ordering constraint? Why do you remove one of the duplicated element in listBig?@user1989682 — ChesterL
– ChesterL, Commented Nov 25, 2014 at 21:54
The big list was created by adding elements from multiple small lists. I want to "undo" the effect of one of the small lists. The duplicates were added by other small lists. So they must stay. — statBeginner
– statBeginner, Commented Nov 25, 2014 at 22:07

Hasan Ramezani · Accepted Answer · 2014-11-25 21:35:13Z

2

try:

listBig = [item for item in listBig if item not in listSmall]

answered Nov 25, 2014 at 21:35

Hasan Ramezani

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1 Comment

ChesterL Over a year ago

if any list has duplicated elements, set(list_nname) would only keep one left

Cory Kramer · Accepted Answer · 2014-11-25 21:34:50Z

1

I would convert your listSmall to a set (for faster membership lookup), then you can use a simple list comprehension

>>> setSmall = set(listSmall)
>>> listBig = [i for i in listBig if i not in setSmall]
>>> listBig
[1, 2, 3, 5, 6, 8]

answered Nov 25, 2014 at 21:34

Cory Kramer

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5 Comments

ChesterL Over a year ago

if you change listSmall to set() might be better if listSmall is a long list and there is no duplicated element in listSmall. Because in works for set() in constant time, while it does linear time in list @Cyber @user1989682

Cory Kramer Over a year ago

@ChesterL I'm not sure I understand your point. As you said the in operation is faster for set than for list. For any decent sized listSmall, it would be beneficial to make a set.

statBeginner Over a year ago

Thanks. I am sorry for creating some confusion. I realised that in my case, if there are multiple occurrences of an element of listSmall in listBig, they MUST NOT BE deleted. I will edit the question.

ChesterL Over a year ago

@Cyber As long as there is no repeated element in listSmall, otherwise, set() will keep only one left

statBeginner Over a year ago

Please check the modified question.

G.Sheppard · Accepted Answer · 2014-11-25 22:02:44Z

0

If you don't want to remove the duplicates, list.remove(x) only removes the first element it finds in the list.

    listBig = [1,2,3,4,5,6,7,8,4,7]
    listSmall = [4,7]
    for item in listSmall: listBig.remove(item)

answered Nov 25, 2014 at 22:02

G.Sheppard

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Comments

Shaney96 · Accepted Answer · 2014-11-25 21:43:17Z

0

You can use list comprehension to compare every element in both lists and check whether there are any repeated elements. If so, the repeated element will be deleted from listBig

>>>'listBig[:] = [element for element in listBig if element not in listSmall] >>>print listBig

[1, 2, 3, 5, 6, 8]

answered Nov 25, 2014 at 21:43

Shaney96

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