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Looking to return all possible rolls of two dice. The return value is a list which contains tuples

I am currently trying to understand how to nest two for loops together before utilizing any comprehension loops. My mission is to return a list of every combination as a tuple, instead i am receiving a list of 6 lists, when i want a list of 36 combinations.
I have tried to iterate over the two ranges.
def build_roll_permutations():
dice = []
for i in range(1,7):
dicee = []
for j in range(1,7):
dicee.append((i,j))
dice.append(dicee)
return dice
expected results:
[(1,1),(1,2)(1,3)...etc]
my results:
[[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)], [(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)], [(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)], [(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)], [(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)], [(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]]

Just get rid of dicee and append directly to dice.
def build_roll_permutations():
dice = []
for i in range(1,7):
for j in range(1,7):
dice.append((i,j))
return dice
Note you can do this with a simple list comprehension
def build_roll_permuatations():
return [(i,j) for i in range(1,7) for j in range(1,7)]
or itertools.product (since this is, in fact, a product and not a permutation):
def build_rolls():
return list(product(range(1,7), repeat=2))

Use extend instead of append, when modifying dice:
def build_roll_permutations():
dice = []
for i in range(1, 7):
dicee = []
for j in range(1, 7):
dicee.append((i, j))
dice.extend(dicee) # line to fix
return dice
print(build_roll_permutations())
Output
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]

How can I convert this to a single line comprehension for combinations?

I've done this to list all the permutations of rolling two dice, but I need to convert it to a single line comprehension of all the combinations and I am struggling.
from itertools import product
myList = list(product(range(1,7), repeat=2))
print myList

Itertools also has a function for combinations that allow repeating elements:
import itertools
c = list(iter.combinations_with_replacement(range(1,7),2))
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 5), (5, 6), (6, 6)]

Python itertools.combinations: how to obtain the indices of the combined numbers

The result created by Python's itertools.combinations() is the combinations of numbers. For example:
a = [7, 5, 5, 4]
b = list(itertools.combinations(a, 2))
# b = [(7, 5), (7, 5), (7, 4), (5, 5), (5, 4), (5, 4)]
But I would like to also obtain the indices of the combinations such as:
index = [(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]
How can I do it?

You can use enumerate:
>>> a = [7, 5, 5, 4]
>>> list(itertools.combinations(enumerate(a), 2))
[((0, 7), (1, 5)), ((0, 7), (2, 5)), ((0, 7), (3, 4)), ((1, 5), (2, 5)), ((1, 5), (3, 4)), ((2, 5), (3, 4))]
>>> b = list((i,j) for ((i,_),(j,_)) in itertools.combinations(enumerate(a), 2))
>>> b
[(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]

You can use range to get order of indexes that combinations produce.
>>> list(itertools.combinations(range(3), 2))
[(0, 1), (0, 2), (1, 2)]
So you can use len(a):
>>> list(itertools.combinations(range(len(a)), 2))
[(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]

a = [7, 5, 5, 4]
n_combinations = 2
np.array(list(itertools.combinations(enumerate(a), n_combinations)))[...,0]
More extendable solution. Change of n_combinations does not require change of code, as it does in the example from #fredtantini. It is basically his solution expressed with numpy slicing.

2d numpy.array() comparing one string to all others, and repeated for each string

Supposed I have an input file with a list of strings (I'm using int's here for cleanliness), e.g. 1,2,3,4,5,6,...,n
I would like to generate a 2D numpy.array that looks like this:
a = [1,1], [1,2], [1,3], [1,4], ..., [1,n]
And then repeat with each consecutive string, e.g.
a = [2,1], [2,2], [2,3], [2,4], ..., [2,n], ...[n,n]
How can I go about this?

Check out the itertools library. The product function seems to be what you are after
In [19]: list( product([1, 2, 3, 4, 5], repeat=2) )
Out[19]:
[(1, 1),
(1, 2),
(1, 3),
(1, 4),
(1, 5),
(2, 1),
(2, 2),
(2, 3),
(2, 4),
(2, 5),
(3, 1),
(3, 2),
(3, 3),
(3, 4),
(3, 5),
(4, 1),
(4, 2),
(4, 3),
(4, 4),
(4, 5),
(5, 1),
(5, 2),
(5, 3),
(5, 4),
(5, 5)]

Generating all unique pair permutations

I need to generate all possible pairings, but with the constraint that a particular pairing only occurs once in the results. So for example:
import itertools
for perm in itertools.permutations(range(9)):
print zip(perm[::2], perm[1::2])
generates all possible two-paired permutations; here's a small subset of the output:
...
[(8, 4), (7, 6), (5, 3), (0, 2)]
[(8, 4), (7, 6), (5, 3), (1, 0)]
[(8, 4), (7, 6), (5, 3), (1, 2)]
[(8, 4), (7, 6), (5, 3), (2, 0)]
[(8, 4), (7, 6), (5, 3), (2, 1)]
[(8, 5), (0, 1), (2, 3), (4, 6)]
[(8, 5), (0, 1), (2, 3), (4, 7)]
[(8, 5), (0, 1), (2, 3), (6, 4)]
[(8, 5), (0, 1), (2, 3), (6, 7)]
[(8, 5), (0, 1), (2, 3), (7, 4)]
[(8, 5), (0, 1), (2, 3), (7, 6)]
[(8, 5), (0, 1), (2, 4), (3, 6)]
[(8, 5), (0, 1), (2, 4), (3, 7)]
[(8, 5), (0, 1), (2, 4), (6, 3)]
...
How do I further filter it so that I only ever see (8,4) once (throughout all of the filtered permutations), and (8,5) only once, and (0,1) only once, and (4,7) only once, etc.?
Basically I want the permutations such that each two-element pairing happens only once.
I'll bet there's an additional itertool that would solve this but I'm not expert enough to know what it is.
Update: Gareth Rees is correct -- I was completely unaware that I was trying to solve the round-robin problem. I have an additional constraint which is that what I'm doing is grouping people for pair-programming exercises. Thus, if I have an odd number of people, I need to create a group of three to include an odd person for each exercise. My current thinking is to (1) make an even number of people by adding in an invisible person. Then, after the pairing, find the person paired with the invisible person and randomly place them into an existing group to form a team of three. However, I wonder if there isn't already an algorithm or adjustment to round-robin that does this in a better way.
Update 2: Theodros' solution produces exactly the right result without the inelegant futzing about I describe above. Everyone's been amazingly helpful.

I'd like to share a different implementation of round-robin scheduling that makes use of the deque-data structure from the Standard Library:
from collections import deque
def round_robin_even(d, n):
for i in range(n - 1):
yield [[d[j], d[-j-1]] for j in range(n/2)]
d[0], d[-1] = d[-1], d[0]
d.rotate()
def round_robin_odd(d, n):
for i in range(n):
yield [[d[j], d[-j-1]] for j in range(n/2)]
d.rotate()
def round_robin(n):
d = deque(range(n))
if n % 2 == 0:
return list(round_robin_even(d, n))
else:
return list(round_robin_odd(d, n))
print round_robin(5)
[[[0, 4], [1, 3]],
[[4, 3], [0, 2]],
[[3, 2], [4, 1]],
[[2, 1], [3, 0]],
[[1, 0], [2, 4]]]
print round_robin(2)
[[[0, 1]]]
It puts the objects(ints here) in the deque. Then it rotates and builds consecutive pairs taking from both ends towards the middle. One can imagine this as folding the deque in the middle back on itself. To make it clear:
Case uneven elements:
round 1 round 2 # pairs are those numbers that sit
---------- --------- # on top of each other
0 1 2 3 4 8 0 1 2 3
8 7 6 5 7 6 5 4
In case of even elements an additional step is required.
(I missed the first time cause I only checked the uneven case. This yielded a horribly wrong algorithm... which shows me how important it is to check edge cases when implementing an algorithm...)
This special step is that I swap the two leftmost elements (which are the first and last elements of the deque) before each rotation -- this means the 0 stays all the time upper left.
Case even elements:
round 1 round 2 # pairs are those numbers that sit
---------- --------- # on top of each other
0 1 2 3 0 7 1 2
7 6 5 4 6 5 4 3
What haunts me about this version is the amount of code duplication, but I couldn't find a way to improve while keeping it as readable. Here's my first implementation, which is less readable IMO:
def round_robin(n):
is_even = (n % 2 == 0)
schedule = []
d = deque(range(n))
for i in range(2 * ((n - 1) / 2) + 1):
schedule.append(
[[d[j], d[-j-1]] for j in range(n/2)])
if is_even:
d[0], d[-1] = d[-1], d[0]
d.rotate()
return schedule
Update to account for the updated question:
To allow in the uneven case for groups of three you just need to change round_robin_odd(d, n):
def round_robin_odd(d, n):
for i in range(n):
h = [[d[j], d[-j-1]] for j in range(n/2)]
h[-1].append(d[n/2])
yield h
d.rotate()
This gives:
print round_robin(5)
[[[0, 4], [1, 3, 2]],
[[4, 3], [0, 2, 1]],
[[3, 2], [4, 1, 0]],
[[2, 1], [3, 0, 4]],
[[1, 0], [2, 4, 3]]]

Pass the list to set to get make sure each tuple only exists once.
>>> from itertools import permutations
>>> set( [ zip( perm[::2], perm[1::2] ) for perm in permutations( range( 9 ) ) ] )
set([(7, 3), (4, 7), (1, 3), (4, 8), (5, 6), (2, 8), (8, 0), (3, 2), (2, 1), (6, 2), (1, 6), (5, 1), (3, 7), (2, 5), (8, 5), (0, 3), (5, 8), (4, 0), (1, 2), (3, 8), (3, 1), (6, 7), (2, 0), (8, 1), (7, 6), (3, 0), (6, 3), (1, 5), (7, 2), (3, 6), (0, 4), (8, 6), (3, 5), (4, 1), (6, 4), (5, 4), (2, 6), (8, 2), (2, 7), (7, 1), (4, 5), (8, 3), (1, 4), (6, 0), (7, 5), (2, 3), (0, 7), (8, 7), (4, 2), (1, 0), (0, 8), (6, 5), (4, 6), (0, 1), (5, 3), (7, 0), (6, 8), (3, 4), (6, 1), (5, 7), (5, 2), (0, 2), (7, 4), (0, 6), (1, 8), (4, 3), (1, 7), (0, 5), (5, 0), (7, 8), (2, 4), (8, 4)])
From your description you want each of the 2-tuple permutations of the range( 9 ) the above should give you all of the various permutations, based on your code. But, this is pretty inefficient.
However you can further simplify your code by doing the following:
>>> list( permutations( range( 9 ), 2 ) )
[(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (1, 0), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (2, 0), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (3, 0), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (4, 0), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (4, 7), (4, 8), (5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (5, 7), (5, 8), (6, 0), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 7), (6, 8), (7, 0), (7, 1), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6), (7, 8), (8, 0), (8, 1), (8, 2), (8, 3), (8, 4), (8, 5), (8, 6), (8, 7)]
The method permutations also takes a length argument that will allow you to specify the length of the tuple returned. So, you were using the correct itertool provided function, but missed the tuple length parameter.
itertools.permutations documentation

As MatthieuW says in this answer, it looks as if you are trying to generate a schedule for a round-robin tournament. This can be easily generated using this algorithm, the main difficulty being the handling of an odd number of teams (when each team gets a bye in one round).
def round_robin_schedule(n):
"""
Generate a round-robin tournament schedule for `n` teams.
"""
m = n + n % 2 # Round up to even number.
for r in xrange(m - 1):
def pairing():
if r < n - 1:
yield r, n - 1
for i in xrange(m // 2 - 1):
p, q = (r + i + 1) % (m - 1), (m + r - i - 2) % (m - 1)
if p < n - 1 and q < n - 1:
yield p, q
yield list(pairing())
For example, with nine teams:
>>> list(round_robin_schedule(9))
[[(0, 8), (2, 7), (3, 6), (4, 5)],
[(1, 8), (2, 0), (4, 7), (5, 6)],
[(2, 8), (3, 1), (4, 0), (6, 7)],
[(3, 8), (4, 2), (5, 1), (6, 0)],
[(4, 8), (5, 3), (6, 2), (7, 1)],
[(5, 8), (6, 4), (7, 3), (0, 1)],
[(6, 8), (7, 5), (0, 3), (1, 2)],
[(7, 8), (0, 5), (1, 4), (2, 3)],
[(0, 7), (1, 6), (2, 5), (3, 4)]]