Deleting a desired list in a 2d list

Question

Say I have a 2d list like this:

list = [[1, 2, 3], ["a", 5, 6], [7, 8, 9], [10, 11, 12]]

and I want to delete a list that starts with a. Here is what I tried:

for i in range(0, len(list)):
     if list[i][0] == "a":
          list.pop(i)

Here, I am trying to delete a list that starts with a. However, it gives me an error saying

IndexError: list index out of range

I am not sure what the problem is here. It's weird because when I try print(list[i]), it prints the matching line, which is `["a", 2, 3].

I am expecting list = [[1, 2, 3], [7, 8, 9], [10, 11, 12]] as my final list.

The problem is that you modify your list while iterating over it. Since you delete items from it, then at some point your list is shorter than what `len(list)` used to be. — cglacet, Mar 20 '19 at 22:48

gmds · Accepted Answer · 2019-03-20T22:53:23.777

8

Use a list comprehension:

>>> l = [[1, 2, 3], ["a", 5, 6], [7, 8, 9], [10, 11, 12]]
>>> new_l = [sublist for sublist in l if sublist[0] != 'a']
>>> new_l
[[1, 2, 3], [7, 8, 9], [10, 11, 12]]

In general, it is not a good idea to name your variables after inbuilts (list) since you will shadow those inbuilts.

Also, as others have noted, your original problem was mutating the list while iterating over it.

edited Mar 20 '19 at 22:53

answered Mar 20 '19 at 22:47

gmds

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score 1 · Answer 2 · edited Mar 20 '19 at 23:11

1

Use a list comprehension:

[x for x in my_list if x[0] != 'a']

edited Mar 20 '19 at 23:11

Ramy M. Mousa

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answered Mar 20 '19 at 22:50

Russ Brown

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CloudEmber · Answer 3 · 2019-06-27T17:58:59.520

What's happening is that your list length gets shorter as you delete items from it. It will still try to loop at i = 3.

You could use a while loop instead since it checks every time. Also since you're not using the popped value, use del

list = [[1, 2, 3], ["a", 5, 6], [7, 8, 9], [10, 11, 12]]
i = 0
while i < len(list):
    if list[i][0] == "a":
        del list[i]
    else:
        i += 1

Related: Difference between del, remove and pop on lists

Devesh Kumar Singh · Answer 4 · 2019-04-30T16:07:36.513

1

You can also use filter for this!

li = [[1, 2, 3], ["a", 5, 6], [7, 8, 9], [10, 11, 12]]

#Filter on first element not being equal to a
res = filter(lambda x: x[0] != 'a', li)

#Print the result
print(list(res))
#[[1, 2, 3], [7, 8, 9], [10, 11, 12]]

edited Apr 30 '19 at 16:07

answered Apr 30 '19 at 15:50

Devesh Kumar Singh

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score 0 · Answer 5 · answered Mar 20 '19 at 23:11

How about an approach using filter? Also I renamed your list to l so you don't overwrite the built-in data type.

from operator import itemgetter

f = itemgetter(0)

def test(sublist):
    return f(sublist) != 'a'

list(filter(test, l))

[[1, 2, 3], [7, 8, 9], [10, 11, 12]]

Deleting a desired list in a 2d list

5 Answers5