How do I pop item off a list of tuples, based on the value of the first number in each tuple?

Question

I am trying to create a list of tuples that consists of all the variations of a set of numbers. However I want to remove any variants from the list that are the same sequence but offset by a position or two. For example: (-1,1,2) , (1,2,-1) & (2,-1,1) I would only want the first one.

Here's where I'm up to:

import itertools as it

list = [-1, 0, 1, 2]
cycles = []

list_cycle_3 = it.permutations(cycles, 3)
list_cycle_4 = it.permutations(cycles, 4)

for item in list_cycle_3:
    cycles.append(item)

for item in list_cycle_4:
    cycles.append(item)

print(cycles)

This results in:

[(-1, 0, 1), (-1, 0, 2), (-1, 1, 0), (-1, 1, 2), (-1, 2, 0), (-1, 2, 1), 
(0, -1, 1), (0, -1, 2), (0, 1, -1), (0, 1, 2), (0, 2, -1), (0, 2, 1), 
(1, -1, 0), (1, -1, 2), (1, 0, -1), (1, 0, 2), (1, 2, -1), (1, 2, 0), 
(2, -1, 0), (2, -1, 1), (2, 0, -1), (2, 0, 1), (2, 1, -1), (2, 1, 0), 
(-1, 0, 1, 2), (-1, 0, 2, 1), (-1, 1, 0, 2), (-1, 1, 2, 0), (-1, 2, 0, 1), (-1, 2, 1, 0), (0, -1, 1, 2), (0, -1, 2, 1), (0, 1, -1, 2), (0, 1, 2, -1), 
(0, 2, -1, 1), (0, 2, 1, -1), (1, -1, 0, 2), (1, -1, 2, 0), (1, 0, -1, 2), 
(1, 0, 2, -1), (1, 2, -1, 0), (1, 2, 0, -1), (2, -1, 0, 1), (2, -1, 1, 0), 
(2, 0, -1, 1), (2, 0, 1, -1), (2, 1, -1, 0), (2, 1, 0, -1)]

So what do I do next to filter the results so I only have the results I want, which are:

[(-1, 0, 1), (-1, 0, 2), (-1, 1, 0), (-1, 1, 2), (-1, 2, 0), (-1, 2, 1), 
(0, 1, 2), (0, 2, 1), (-1, 0, 1, 2), (-1, 0, 2, 1), (-1, 1, 0, 2), 
(-1, 1, 2, 0), (-1, 2, 0, 1), (-1, 2, 1, 0)]

If it helps a simple difference between the lists are that the list I want is all the tuples starting with -1, and the tuples where there is no -1 starting with 0

Answer 1

    l=[(-1, 0, 1), (-1, 0, 2), (-1, 1, 0), (-1, 1, 2), (-1, 2, 0), (-1, 2, 1),                                                                                           
    (0, -1, 1), (0, -1, 2), (0, 1, -1), (0, 1, 2), (0, 2, -1), (0, 2, 1),
    (1, -1, 0), (1, -1, 2), (1, 0, -1), (1, 0, 2), (1, 2, -1), (1, 2, 0),
    (2, -1, 0), (2, -1, 1), (2, 0, -1), (2, 0, 1), (2, 1, -1), (2, 1, 0),
    (-1, 0, 1, 2), (-1, 0, 2, 1), (-1, 1, 0, 2), (-1, 1, 2, 0), (-1, 2, 0, 1), (-1, 2, 1, 0), (0, -1, 1, 2), (0, -1, 2, 1), (0, 1, -1, 2), (0, 1, 2, -1),
    (0, 2, -1, 1), (0, 2, 1, -1), (1, -1, 0, 2), (1, -1, 2, 0), (1, 0, -1, 2),
    (1, 0, 2, -1), (1, 2, -1, 0), (1, 2, 0, -1), (2, -1, 0, 1), (2, -1, 1, 0),
    (2, 0, -1, 1), (2, 0, 1, -1), (2, 1, -1, 0), (2, 1, 0, -1)]

   l2=[]

   for i in l:
       if i[0] == -1 :
           l2.append(i)

   print(l2)

this code works

output

[(-1, 0, 1), (-1, 0, 2), (-1, 1, 0), (-1, 1, 2), (-1, 2, 0), (-1, 2, 1), (-1, 0, 1, 2), (-1, 0, 2, 1), (-1, 1, 0, 2), (-1, 1, 2, 0), (-1, 2, 0, 1), (-1, 2, 1, 0)]

Answer 2

Start with a list that does not contain the number that you want to filter against:

For example, you'll only need the ones that starts with 0. Then your list is

l = [-1, 1, 2]

Find all the two element permutatations and filter them as you like. Ant then add 0 as the first element by mapping your result set.

Example:

In [2]: from itertools import permutations
In [3]: l = [-1, 1, 2]
In [4]: p = permutations(l, 2)
In [5]: [(0, *t) for t in p]
Out[5]: [(0, -1, 1), (0, -1, 2), (0, 1, -1), (0, 1, 2), (0, 2, -1), (0, 2, 1)]

You could do a similar trick for, say, the ones that does not have -1 in them and start with 0 which is omit -1 and 0 from your list and then add 0 as the first element into your result sets items.

And do not override reserved keywords in your code:

list =  [...] # do not do that