How do I delete the Nth list item from a list of lists (column delete)?

Question

How do I delete a "column" from a list of lists?
Given:

L = [
     ["a","b","C","d"],
     [ 1,  2,  3,  4 ],
     ["w","x","y","z"]
    ]

I would like to delete "column" 2 to get:

L = [
     ["a","b","d"],
     [ 1,  2,  4 ],
     ["w","x","z"]
    ]

Is there a slice or del method that will do that? Something like:

del L[:][2]

Brindled

Answer 1

You could loop.

for x in L:
    del x[2]

If you're dealing with a lot of data, you can use a library that support sophisticated slicing like that. However, a simple list of lists doesn't slice.

Hermanhermann

how do I save the resulting newlist into a new variable? – Watertight

@DJ_Stuffy_K: This technique doesn't create new lists, it modifies the existing lists. However, you can copy the lists beforehand. new_L = [list(row) for row in L]... this is just one way to do things, it will create a shallow copy of each row, so when you modify new_L the original L will be unmodified. – Hermanhermann

Answer 2

just iterate through that list and delete the index which you want to delete.

for example

for sublist in list:
    del sublist[index]

Hesperian

this solution is slow – Prudery

Could you please then let me know what is the best ever solution for this Prob :) – Hesperian

Answer 3

You can do it with a list comprehension:

>>> removed = [ l.pop(2) for l in L ]
>>> print L
[['a', 'b', 'd'], [1, 2, 4], ['w', 'x', 'z']]
>>> print removed
['d', 4, 'z']

It loops the list and pops every element in position 2.

You have got list of elements removed and the main list without these elements.

Laruelarum

Answer 4

A slightly twisted version:

index = 2  # Delete column 2 
[(x[0:index] + x[index+1:]) for x in L]

Italia

Answer 5

[(x[0], x[1], x[3]) for x in L]

It works fine.

Nosewheel

Answer 6

This is a very easy way to remove whatever column you want.

L = [
["a","b","C","d"],
[ 1,  2,  3,  4 ],
["w","x","y","z"]
]
temp = [[x[0],x[1],x[3]] for x in L] #x[column that you do not want to remove]
print temp
O/P->[['a', 'b', 'd'], [1, 2, 4], ['w', 'x', 'z']]

Incoercible

Answer 7

L = [['a', 'b', 'C', 'd'], [1, 2, 3, 4], ['w', 'x', 'y', 'z']]
_ = [i.remove(i[2]) for i in L]

Cutie

Aside from the fact that using a listcomp for its side effects is generally disfavoured, this won't work unless the element you're removing happens to be first in the list. For example, if one of the sublists is [1,2,1,3], then the remove will return [2,1,3], not [1,2,3]. – Fustigate

Answer 8

If you don't mind on creating new list then you can try the following:

filter_col = lambda lVals, iCol: [[x for i,x in enumerate(row) if i!=iCol] for row in lVals]

filter_out(L, 2)

Fillender

Answer 9

An alternative to pop():

[x.__delitem__(n) for x in L]

Here n is the index of the elements to be deleted.

Foulup