I have created a list
a = [[3, 4], [5], [6, 7, 8]]
I want to delete 3 from this list. What is the command for this?
9 Answers
lots of possible ways
>>> mylist = [[3,4],[5],[6,7,8]]
>>> mylist[0] = [4]
>>> mylist
[[4], [5], [6, 7, 8]]
>>> mylist = [[3,4],[5],[6,7,8]]
>>> del mylist[0][0]
>>> mylist
[[4], [5], [6, 7, 8]]
>>> mylist = [[3,4],[5],[6,7,8]]
>>> mylist[0].remove(3)
>>> mylist
[[4], [5], [6, 7, 8]]
Take your pick :)
1 Comment
Peter Hansen
+1 for teaching the man to fish. (I also reformatted and made it clear that
del is a statement, not a function.)Easy, you can try this
del a[0][0]
Comments
Assuming, you want to delete all 3s from a list of lists:
>>> lst = [[3,4],[5],[6,7,8]]
>>> [[i for i in el if i != 3] for el in lst]
[[4], [5], [6, 7, 8]]
6 Comments
Bryan Oakley
I upvoted since the answer is useful, but after more than a year of python programming I still don't see the advantage to list comprehensions like this over something with a couple more lines of code that is perhaps a bit less obtuse. ghostdog74's answer is almost instantly recognizable whereas I always have to pause and mentally parse list comprehensions. <shrug>
SilentGhost
@Bryan: how is this obtuse? ghostdog74's answer is incorrect, as it only removes a single 3 it's also deeply nested, which ghostdog74 is trying to avoid by flattening the remove and making it even less readable or pythonic.
Bryan Oakley
I say it's obtuse because unless you have used a lot of list comprehensions, you have to mentally parse through what is happening. All developers can pretty much instantly recognize and understand nested loops much faster than they can parse something like "i for i in el". Maybe I'm just an old dog that can't learn new tricks, but I prefer explict loops over one-line list comprehensions.
Bryan Oakley
It's hard to say if ghostdog74's answer is incorrect or not. The original problem says "I want to delete 3". Not all 3's. One could interpret it as "I want to remove the first element of the first nested list" or "I want to remove the 3 from the first nested list", or a couple other ways. <shrug>
SilentGhost
@Bryan: and ghostdog74's version is doing neither.
|
a[0].remove(3)
(had to add more text so it was long enough)
Comments
Using this:
del a[0][0]
For a better understanding of lists, dictionaries, etc., I suggest you should read Dive Into Python You'll find Chapter 3 very useful.
Comments
if you don't know where "3" is,
>>> for n,i in enumerate(list):
... if 3 in i: list[n].remove(3)
...
>>> list
[[4], [5], [6, 7, 8]]
>>>
Comments
First of all, be careful because you are shadowing the built in name "list". This
a_list = [[3,4],[5],[6,7,3, 3, 8]]
def clear_list_from_item(a_list, item):
try:
while True: a_list.remove(item)
except ValueError:
return a_list
a_list = [clear_list_from_item(x, 3) for x in a_list]
This will modify your original list in place.
Comments
If I understand your question you have list in a list and you want to delete first element from first list. So use:
del a[0][0]
Comments
two easy ways to remove from a list, if you know the element of concern is at position 0:
a[0].pop (0)
del a[0][0]
listis a bad name for your object since it shadows a built-in.