Python: list of lists

Question

Running the code

listoflists = []
list = []
for i in range(0,10):
    list.append(i)
    if len(list)>3:
        list.remove(list[0])
        listoflists.         


        

Answer 1

You're also not going to get the output you're hoping for as long as you append to listoflists only inside the if-clause.

Try something like this instead:

import copy

listoflists = []
list = []
for i in range(0,10):
    list.append(i)
    if len(list)>3:
        list.remove(list[0])
    listoflists.append((copy.copy(list), copy.copy(list[0])))
print(listoflists)

Answer 2

First, I strongly recommend that you rename your variable list to something else. list is the name of the built-in list constructor, and you're hiding its normal function. I will rename list to a in the following.

Python names are references that are bound to objects. That means that unless you create more than one list, whenever you use a it's referring to the same actual list object as last time. So when you call

listoflists.append((a, a[0]))

you can later change a and it changes what the first element of that tuple points to. This does not happen with a[0] because the object (which is an integer) pointed to by a[0] doesn't change (although a[0] points to different objects over the run of your code).

You can create a copy of the whole list a using the list constructor:

listoflists.append((list(a), a[0]))

Or, you can use the slice notation to make a copy:

listoflists.append((a[:], a[0]))

Answer 3

When you run the code

listoflists.append((list, list[0]))

You are not (as I think you expect) adding a copy of list to the end of listoflists. What you are doing is adding a reference to list to the end of listoflists. Thus, every time you update list, it updates every reference to list, which in this case, is every item in listoflists

What you could do instead is something like this:

listoflists = []
for i in range(1, 10):
    listoflists.append((range(i), 0))

Answer 4

Time traveller here

List_of_list =[([z for z in range(x-2,x+1) if z >= 0],y) for y in range(10) for x in range(10)]

This should do the trick. And the output is this:

[([0], 0), ([0, 1], 0), ([0, 1, 2], 0), ([1, 2, 3], 0), ([2, 3, 4], 0),  ([3, 4, 5], 0), ([4, 5, 6], 0), ([5, 6, 7], 0), ([6, 7, 8], 0), ([7, 8, 9], 0), ([0], 1), ([0, 1], 1), ([0, 1, 2], 1), ([1, 2, 3], 1), ([2, 3, 4], 1), ([3, 4, 5], 1), ([4, 5, 6], 1), ([5, 6, 7], 1), ([6, 7, 8], 1), ([7, 8, 9], 1), ([0], 2), ([0, 1], 2), ([0, 1, 2], 2), ([1, 2, 3], 2), ([2, 3, 4], 2), ([3, 4, 5], 2), ([4, 5, 6], 2), ([5, 6, 7], 2), ([6, 7, 8], 2), ([7, 8, 9], 2), ([0], 3), ([0, 1], 3), ([0, 1, 2], 3), ([1, 2, 3], 3), ([2, 3, 4], 3), ([3, 4, 5], 3), ([4, 5, 6], 3), ([5, 6, 7], 3), ([6, 7, 8], 3), ([7, 8, 9], 3), ([0], 4), ([0, 1], 4), ([0, 1, 2], 4), ([1, 2, 3], 4), ([2, 3, 4], 4), ([3, 4, 5], 4), ([4, 5, 6], 4), ([5, 6, 7], 4), ([6, 7, 8], 4), ([7, 8, 9], 4), ([0], 5), ([0, 1], 5), ([0, 1, 2], 5), ([1, 2, 3], 5), ([2, 3, 4], 5), ([3, 4, 5], 5), ([4, 5, 6], 5), ([5, 6, 7], 5), ([6, 7, 8], 5), ([7, 8, 9], 5), ([0], 6), ([0, 1], 6), ([0, 1, 2], 6), ([1, 2, 3], 6), ([2, 3, 4], 6), ([3, 4, 5], 6), ([4, 5, 6], 6), ([5, 6, 7], 6), ([6, 7, 8], 6), ([7, 8, 9], 6), ([0], 7), ([0, 1], 7), ([0, 1, 2], 7), ([1, 2, 3], 7), ([2, 3, 4], 7), ([3, 4, 5], 7), ([4, 5, 6], 7), ([5, 6, 7], 7), ([6, 7, 8], 7), ([7, 8, 9], 7), ([0], 8), ([0, 1], 8), ([0, 1, 2], 8), ([1, 2, 3], 8), ([2, 3, 4], 8), ([3, 4, 5], 8), ([4, 5, 6], 8), ([5, 6, 7], 8), ([6, 7, 8], 8), ([7, 8, 9], 8), ([0], 9), ([0, 1], 9), ([0, 1, 2], 9), ([1, 2, 3], 9), ([2, 3, 4], 9), ([3, 4, 5], 9), ([4, 5, 6], 9), ([5, 6, 7], 9), ([6, 7, 8], 9), ([7, 8, 9], 9)]    

This is done by list comprehension(which makes looping elements in a list via one line code possible). The logic behind this one-line code is the following:

(1) for x in range(10) and for y in range(10) are employed for two independent loops inside a list

(2) (a list, y) is the general term of the loop, which is why it is placed before two for's in (1)

(3) the length of the list in (2) cannot exceed 3, and the list depends on x, so

[z for z in range(x-2,x+1)] 

is used

(4) because z starts from zero but range(x-2,x+1) starts from -2 which isn't what we want, so a conditional statement if z >= 0 is placed at the end of the list in (2)

[z for z in range(x-2,x+1) if z >= 0] 

Answer 5

I came here because I'm new with python and lazy so I was searching an example to create a list of 2 lists, after a while a realized the topic here could be wrong... this is a code to create a list of lists:

listoflists = []
for i in range(0,2):
    sublist = []
    for j in range(0,10)
        sublist.append((i,j))
    listoflists.append(sublist)
print listoflists

this the output [ [(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9)], [(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9)] ]

The problem with your code seems to be you are creating a tuple with your list and you get the reference to the list instead of a copy. That I guess should fall under a tuple topic...

Answer 6

The list variable (which I would recommend to rename to something more sensible) is a reference to a list object, which can be changed.

On the line

listoflists.append((list, list[0]))

You actually are only adding a reference to the object reference by the list variable. You've got multiple possibilities to create a copy of the list, so listoflists contains the values as you seem to expect:

Use the copy library

import copy
listoflists.append((copy.copy(list), list[0]))

use the slice notation

listoflists.append((list[:], list[0]))