Removing a list in a list by only 1 of its Items[Python]

Question

I have a system that takes in loans which contain a unique ID, a name and a book title. Layout is like this;

[[0, Name, title],[2, Name, title],[1, Name, title]]

Now i've been trying to find a way to remove one of the lists in the list, but only by ID. I want to be able to just give my program the 2, and it removes the entire list with that 2 in it. Note; the list with ID 2 is not always on index number 2, so listname.pop(ID) is not the solution either. It has to be some solution using listname.remove(), but with remove you have to give the entire list as an argument. Or am I missing something?

Just rebuild the list without the sublist that contains the selected value at index 0. What have you tried so far? — roganjosh, Jun 08 '19 at 11:07

score 0 · Accepted Answer · answered Jun 08 '19 at 11:16

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Using list-comprehension

l=[[0, 'Name', 'title'],[2, 'Name', 'title'],[1, 'Name', 'title']] 
print([i for i in l if i[0]!=2])
#[[0, 'Name', 'title'], [1, 'Name', 'title']]

answered Jun 08 '19 at 11:16

Smart Manoj

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I didn't think of something this simple because of all the stuff I had around my code, thanks for this simple solution. – d n Jun 08 '19 at 11:35

score 0 · Answer 2 · answered Jun 08 '19 at 11:16

0

you can use this list comprehension

new_list = [l for l in x if l[0] != 2]

This has the 2 hard-coded, but you can replace it with a variable name

index = 2
new_list = [l for l in x if l[0] != index]

answered Jun 08 '19 at 11:16

drxl

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Removing a list in a list by only 1 of its Items[Python]

2 Answers2