I have a set of lists and a list with all permutations of the set of lists.
mylist = []
mylist.append(['a', 'b', 'c', 'd', 'e'])
mylist.append(['1', '2', '3', '4', '5'])
mylist.append(['f', 'g', 'h', 'i', 'j'])
# Print all permutations
print(list(itertools.product(*mylist)))
Output
[('a', '1', 'f'), ('a', '1', 'g'), ('a', '1', 'h'), ('a', '1', 'i'), ('a', '1', 'j'), ('a', '2', 'f'), ('a', '2', 'g'), ('a', '2', 'h'), ('a', '2', 'i'), ('a', '2', 'j'), ('a', '3', 'f'), ('a', '3', 'g'), ('a', '3', 'h'), ('a', '3', 'i'), ('a', '3', 'j'), ('a', '4', 'f'), ('a', '4', 'g'), ('a', '4', 'h'), ('a', '4', 'i'), ('a', '4', 'j'), ('a', '5', 'f'), ('a', '5', 'g'), ('a', '5', 'h'), ('a', '5', 'i'), ('a', '5', 'j'), ('b', '1', 'f'), ('b', '1', 'g'), ('b', '1', 'h'), ('b', '1', 'i'), ('b', '1', 'j'), ('b', '2', 'f'), ('b', '2', 'g'), ('b', '2', 'h'), ('b', '2', 'i'), ('b', '2', 'j'), ('b', '3', 'f'), ('b', '3', 'g'), ('b', '3', 'h'), ('b', '3', 'i'), ('b', '3', 'j'), ('b', '4', 'f'), ('b', '4', 'g'), ('b', '4', 'h'), ('b', '4', 'i'), ('b', '4', 'j'), ('b', '5', 'f'), ('b', '5', 'g'), ('b', '5', 'h'), ('b', '5', 'i'), ('b', '5', 'j'), ('c', '1', 'f'), ('c', '1', 'g'), ('c', '1', 'h'), ('c', '1', 'i'), ('c', '1', 'j'), ('c', '2', 'f'), ('c', '2', 'g'), ('c', '2', 'h'), ('c', '2', 'i'), ('c', '2', 'j'), ('c', '3', 'f'), ('c', '3', 'g'), ('c', '3', 'h'), ('c', '3', 'i'), ('c', '3', 'j'), ('c', '4', 'f'), ('c', '4', 'g'), ('c', '4', 'h'), ('c', '4', 'i'), ('c', '4', 'j'), ('c', '5', 'f'), ('c', '5', 'g'), ('c', '5', 'h'), ('c', '5', 'i'), ('c', '5', 'j'), ('d', '1', 'f'), ('d', '1', 'g'), ('d', '1', 'h'), ('d', '1', 'i'), ('d', '1', 'j'), ('d', '2', 'f'), ('d', '2', 'g'), ('d', '2', 'h'), ('d', '2', 'i'), ('d', '2', 'j'), ('d', '3', 'f'), ('d', '3', 'g'), ('d', '3', 'h'), ('d', '3', 'i'), ('d', '3', 'j'), ('d', '4', 'f'), ('d', '4', 'g'), ('d', '4', 'h'), ('d', '4', 'i'), ('d', '4', 'j'), ('d', '5', 'f'), ('d', '5', 'g'), ('d', '5', 'h'), ('d', '5', 'i'), ('d', '5', 'j'), ('e', '1', 'f'), ('e', '1', 'g'), ('e', '1', 'h'), ('e', '1', 'i'), ('e', '1', 'j'), ('e', '2', 'f'), ('e', '2', 'g'), ('e', '2', 'h'), ('e', '2', 'i'), ('e', '2', 'j'), ('e', '3', 'f'), ('e', '3', 'g'), ('e', '3', 'h'), ('e', '3', 'i'), ('e', '3', 'j'), ('e', '4', 'f'), ('e', '4', 'g'), ('e', '4', 'h'), ('e', '4', 'i'), ('e', '4', 'j'), ('e', '5', 'f'), ('e', '5', 'g'), ('e', '5', 'h'), ('e', '5', 'i'), ('e', '5', 'j')]
From above list I want to extract 10 items under the following conditions:
The a should appear 6 times, the b 3 times and the c 1 time.
The 5 should appear 5 times (the rest is random)
The f should appear 2 times and the i should appear 4 times (the rest is random)
Your list of permutations actually contains only combinations. To impose frequencies to your selection of 10, you can pre-fill parts of the combinations with the required values and complete the rest with random values from the remaining elements of the corresponding list. Then shuffle the parts before assembling then into the 10 combinations items:
from random import choices,sample
part0 = sample(['a']*6 + ['b']*3 + choices("cde",k=1) ,10)
part1 = sample(['5']*5 + choices("1234",k=5) ,10)
part2 = sample(['f']*2 + ['i']*4 + choices("ghj",k=4) ,10)
result = list(zip(part0,part1,part2))
Output:
print(*result,sep="\n")
('a', '5', 'i')
('a', '3', 'j')
('b', '5', 'i')
('a', '4', 'g')
('b', '5', 'h')
('b', '5', 'g')
('a', '4', 'f')
('a', '3', 'i')
('a', '1', 'i')
('d', '5', 'f')
\ \ \____ 4 times 'i', 2 times 'f'
\ \________ 5 times '5'
\____________ 6 times 'a', 3 times 'b'
Note that this may produce duplicate combinations. To work around that, you can place the 4 lines in a loop that regenerates the combinations until they are all distinct (with your conditions and a selection of 10 this results in 3.75 attempts on average):
For example:
result = set()
while len(result) != 10:
part0 = sample(['a']*6 + ['b']*3 + choices("cde",k=1) ,10)
part1 = sample(['5']*5 + choices("1234",k=5) ,10)
part2 = sample(['f']*2 + ['i']*4 + choices("ghj",k=4) ,10)
result = set(zip(part0,part1,part2))
If you really want permutations, then you can randomize the position of items produced by zip:
...
result = set(map(lambda c:tuple(sample(c,3)),zip(part0,part1,part2)))
which will then give actual permutations in the result (becoming less likely to have duplicates and only needs 1.22 attempts on average):
('5', 'b', 'i')
('2', 'j', 'a')
('b', 'j', '5')
('g', '2', 'a')
('2', 'f', 'a')
('i', '5', 'a')
('g', '5', 'b')
('i', '1', 'a')
('d', '5', 'f')
('i', '4', 'a')
Related
I need to initialize a multiIndex DataFrame from given data.
id = ['a','b','c'] ;
days = [2,5,4], which means, each id has its corresponding duration of days, i.e. 'a' has day 1,2; 'b' has day 1,..,5; and 'c' has day 1,...4. In another words, the days varies for each id.
And within each day, there are 4 periods, prd = [0,1,2,3].
What I expect to have, is a MultiIndex of DataFrame for each id, at each day and each period.
MultiIndex([('a',1,0),
('a',1,1),
('a',1,2),
('a',1,3),
('a',2,0),
('a',2,1),
('a',2,2),
('a',2,3),
('b',1,0),
('b',1,1),
('b',1,2),
...
('b',5,1),
('b',5,2),
('b',5,3),
('c',1,0),
('c',1,1),
('c',1,2),
...
('c',4,1),
('c',4,2),
('c',4,3),
],
names=['id','day','prd']
)
I tried to handle in python:
Because the days are different for different id, I generate two complete lists of id and days, by loop and list comprehension, and then zip them together to get the tuple pairs. And then I use itertools.product() to combine with period. But what I get is like
[(('a',1),0),
(('a',1),1),
(('a',1),2),....]
If I use pd.MultiIndex.from_product(), I got similar results, that first two index are in a group, third one separated.
Since product won't help either way, the old fashion is to also stretch prd into long and complete list to match the other two fellas , and zip them at once.
I really want to know if there is a better way of generate index from beginning, better than such a long way of loops, list comprehension, zip and product to combine them together. Is there anything in Pandas can handle this case, other than native python data structures?
Many thanks!
Create the combinations using list comprehension with zip:
id = ['a','b','c']
prd = [0,1,2,3]
days = [2,5,4]
result = [(idx, i, p) for d, idx in zip(days, id) for i in range(1, d+1) for p in prd]
print (pd.MultiIndex.from_tuples(result))
MultiIndex([('a', 1, 0),
('a', 1, 1),
('a', 1, 2),
('a', 1, 3),
('a', 2, 0),
('a', 2, 1),
('a', 2, 2),
('a', 2, 3),
('b', 1, 0),
('b', 1, 1),
('b', 1, 2),
('b', 1, 3),
('b', 2, 0),
('b', 2, 1),
('b', 2, 2),
('b', 2, 3),
('b', 3, 0),
('b', 3, 1),
('b', 3, 2),
('b', 3, 3),
('b', 4, 0),
('b', 4, 1),
('b', 4, 2),
('b', 4, 3),
('b', 5, 0),
('b', 5, 1),
('b', 5, 2),
('b', 5, 3),
('c', 1, 0),
('c', 1, 1),
('c', 1, 2),
('c', 1, 3),
('c', 2, 0),
('c', 2, 1),
('c', 2, 2),
('c', 2, 3),
('c', 3, 0),
('c', 3, 1),
('c', 3, 2),
('c', 3, 3),
('c', 4, 0),
('c', 4, 1),
('c', 4, 2),
('c', 4, 3)],
)
You can use np.repeat and np.tile here. You can use this when dealing with large id, prd, days.
You need to repeat id values w.r.t days and each element len(prd) times, we can use np.multiply here.
You need to tile prd values by the total number of days, we can use np.sum here.
Use pd.MultiIndex.from_arrays to build desired output.
id = ['a','b','c']
prd = [0,1,2,3]
days = [2,5,4]
x = np.repeat(id,np.multiply(days, len(prd)))
y = np.concatenate([np.arange(1, i+1).repeat(len(prd)) for i in days])
z = np.tile(prd,np.sum(days))
pd.MultiIndex.from_arrays([x,y,z])
# Equivalent to
# pd.MultiIndex.from_tuples(np.c_[x,y,z].tolist())
# x y z
# | | |
# V V V
MultiIndex([('a', '1', '0'),
('a', '1', '1'),
('a', '1', '2'),
('a', '1', '3'),
('a', '2', '0'),
('a', '2', '1'),
('a', '2', '2'),
('a', '2', '3'),
('b', '1', '0'),
('b', '1', '1'),
('b', '1', '2'),
('b', '1', '3'),
('b', '2', '0'),
('b', '2', '1'),
('b', '2', '2'),
('b', '2', '3'),
('b', '3', '0'),
('b', '3', '1'),
('b', '3', '2'),
('b', '3', '3'),
('b', '4', '0'),
('b', '4', '1'),
('b', '4', '2'),
('b', '4', '3'),
('b', '5', '0'),
('b', '5', '1'),
('b', '5', '2'),
('b', '5', '3'),
('c', '1', '0'),
('c', '1', '1'),
('c', '1', '2'),
('c', '1', '3'),
('c', '2', '0'),
('c', '2', '1'),
('c', '2', '2'),
('c', '2', '3'),
('c', '3', '0'),
('c', '3', '1'),
('c', '3', '2'),
('c', '3', '3'),
('c', '4', '0'),
('c', '4', '1'),
('c', '4', '2'),
('c', '4', '3')],
)
Let's say I have two lists;
a = ['A', 'B', 'C', 'D']
b = ['1', '2', '3', '4']
I know I can get every permutation of these two lists like so:
for r in itertools.product(a, b): print (r[0] + r[1])
But what I'm looking for is every pairwise combination stored in a tuple. So, for example, some combinations would be:
[(A, 1), (B, 2), (C, 3), (D, 4)]
[(A, 1), (B, 3), (C, 2), (D, 4)]
[(A, 1), (B, 4), (C, 3), (D, 2)]
[(A, 1), (B, 3), (C, 4), (D, 2)]
[(A, 1), (B, 2), (C, 4), (D, 3)]
So it would iterate through every possible combination so that no letter has the same number value. I'm at a loss for an efficient way to do this (particularly since I need to scale this to three lists in my actual example)
We can make permutations of one of the lists (for example here the latter one), and then zip each permutation together with the first list, like:
from functools import partial
from itertools import permutations
def pairwise_comb(xs, ys):
return map(partial(zip, xs), permutations(ys))
or in case all subelements should be lists (here these are still iterables that can take a specific shape when you "materialize" these):
from functools import partial
from itertools import permutations
def pairwise_comb(xs, ys):
return map(list, map(partial(zip, xs), permutations(ys)))
For the given sample input, we obtain:
>>> for el in pairwise_comb(a, b):
... print(list(el))
...
[('A', '1'), ('B', '2'), ('C', '3'), ('D', '4')]
[('A', '1'), ('B', '2'), ('C', '4'), ('D', '3')]
[('A', '1'), ('B', '3'), ('C', '2'), ('D', '4')]
[('A', '1'), ('B', '3'), ('C', '4'), ('D', '2')]
[('A', '1'), ('B', '4'), ('C', '2'), ('D', '3')]
[('A', '1'), ('B', '4'), ('C', '3'), ('D', '2')]
[('A', '2'), ('B', '1'), ('C', '3'), ('D', '4')]
[('A', '2'), ('B', '1'), ('C', '4'), ('D', '3')]
[('A', '2'), ('B', '3'), ('C', '1'), ('D', '4')]
[('A', '2'), ('B', '3'), ('C', '4'), ('D', '1')]
[('A', '2'), ('B', '4'), ('C', '1'), ('D', '3')]
[('A', '2'), ('B', '4'), ('C', '3'), ('D', '1')]
[('A', '3'), ('B', '1'), ('C', '2'), ('D', '4')]
[('A', '3'), ('B', '1'), ('C', '4'), ('D', '2')]
[('A', '3'), ('B', '2'), ('C', '1'), ('D', '4')]
[('A', '3'), ('B', '2'), ('C', '4'), ('D', '1')]
[('A', '3'), ('B', '4'), ('C', '1'), ('D', '2')]
[('A', '3'), ('B', '4'), ('C', '2'), ('D', '1')]
[('A', '4'), ('B', '1'), ('C', '2'), ('D', '3')]
[('A', '4'), ('B', '1'), ('C', '3'), ('D', '2')]
[('A', '4'), ('B', '2'), ('C', '1'), ('D', '3')]
[('A', '4'), ('B', '2'), ('C', '3'), ('D', '1')]
[('A', '4'), ('B', '3'), ('C', '1'), ('D', '2')]
[('A', '4'), ('B', '3'), ('C', '2'), ('D', '1')]
This thus results in 24 possible ways to combine this, since the order of 'A', 'B', 'C' and 'D' remains fixed, and the 4 characters can be assigned in 4! ways, or 4! = 4×3×2×1 = 24.
It may be a lot easier than you think. What about:
import itertools
a = ['A', 'B', 'C', 'D']
b = ['1', '2', '3', '4']
for aperm in itertools.permutations(a):
for bperm in itertools.permutations(b):
print(list(zip(aperm, bperm)))
First outputs:
[('A', '1'), ('B', '2'), ('C', '3'), ('D', '4')]
[('A', '1'), ('B', '2'), ('C', '4'), ('D', '3')]
[('A', '1'), ('B', '3'), ('C', '2'), ('D', '4')]
[('A', '1'), ('B', '3'), ('C', '4'), ('D', '2')]
[('A', '1'), ('B', '4'), ('C', '2'), ('D', '3')]
[('A', '1'), ('B', '4'), ('C', '3'), ('D', '2')]
[('A', '2'), ('B', '1'), ('C', '3'), ('D', '4')]
[('A', '2'), ('B', '1'), ('C', '4'), ('D', '3')]
[('A', '2'), ('B', '3'), ('C', '1'), ('D', '4')]
...
(There are 576 lines printed for these two 4-element lists)
Edit: If you want to generalize this to more iterables, you could do something like:
import itertools
a = ['A', 'B', 'C', 'D']
b = ['1', '2', '3', '4']
gens = [itertools.permutations(lst) for lst in (a,b)]
for perms in itertools.product(*gens):
print(list(zip(*perms)))
Which outputs the same thing, but could be easily extended, e.g.
import itertools
a = ['A', 'B', 'C', 'D']
b = ['1', '2', '3', '4']
c = ['W', 'X', 'Y', 'Z']
gens = [itertools.permutations(lst) for lst in (a,b,c)] # add c
for perms in itertools.product(*gens): # no change
print(list(zip(*perms))) # ''
You can use recursion with yield for a no-import solution:
a = ['A', 'B', 'C', 'D']
b = ['1', '2', '3', '4']
def combinations(d, current = []):
if len(current) == 4:
yield current
elif filter(None, d):
for i in d[0]:
_d0, _d1 = [c for c in d[0] if c != i], [c for c in d[1] if c != d[1][0]]
yield from combinations([_d0, _d1] , current+[[i, d[1][0]]])
for i in combinations([a, b]):
print(i)
Output:
[['A', '1'], ['B', '2'], ['C', '3'], ['D', '4']]
[['A', '1'], ['B', '2'], ['D', '3'], ['C', '4']]
[['A', '1'], ['C', '2'], ['B', '3'], ['D', '4']]
[['A', '1'], ['C', '2'], ['D', '3'], ['B', '4']]
[['A', '1'], ['D', '2'], ['B', '3'], ['C', '4']]
[['A', '1'], ['D', '2'], ['C', '3'], ['B', '4']]
[['B', '1'], ['A', '2'], ['C', '3'], ['D', '4']]
[['B', '1'], ['A', '2'], ['D', '3'], ['C', '4']]
[['B', '1'], ['C', '2'], ['A', '3'], ['D', '4']]
[['B', '1'], ['C', '2'], ['D', '3'], ['A', '4']]
[['B', '1'], ['D', '2'], ['A', '3'], ['C', '4']]
[['B', '1'], ['D', '2'], ['C', '3'], ['A', '4']]
[['C', '1'], ['A', '2'], ['B', '3'], ['D', '4']]
[['C', '1'], ['A', '2'], ['D', '3'], ['B', '4']]
[['C', '1'], ['B', '2'], ['A', '3'], ['D', '4']]
[['C', '1'], ['B', '2'], ['D', '3'], ['A', '4']]
[['C', '1'], ['D', '2'], ['A', '3'], ['B', '4']]
[['C', '1'], ['D', '2'], ['B', '3'], ['A', '4']]
[['D', '1'], ['A', '2'], ['B', '3'], ['C', '4']]
[['D', '1'], ['A', '2'], ['C', '3'], ['B', '4']]
[['D', '1'], ['B', '2'], ['A', '3'], ['C', '4']]
[['D', '1'], ['B', '2'], ['C', '3'], ['A', '4']]
[['D', '1'], ['C', '2'], ['A', '3'], ['B', '4']]
[['D', '1'], ['C', '2'], ['B', '3'], ['A', '4']]
Given the following lists:
letters = ('a', 'b', 'c', 'd', 'e', 'f', 'g')
numbers = ('1', '2', '3', '4')
How can I produce an iterated list that produces the following:
output = [('a', '1'), ('b', '2'), ('c', '3'), ('d', '4'),
('e', '1'), ('f', '2'), ('g', '3'), ('a', '4'),
('b', '1'), ('c', '2'), ('d', '3'), ('e', '4'),
('f', '1'), ('g', '2')...]
I feel like I should be able to produce the desired output by using
output = (list(zip(letters, itertools.cycle(numbers))
But this produces the following:
output = [('a', '1'), ('b', '2'), ('c', '3'), ('d', '4'),
('e', '1'), ('f', '2'), ('g', '3')]
Any help would be greatly appreciated.
If you are looking for an infinite generator, you can use cycle with zip for both lists, in the form of zip(itertools.cycle(x), itertools.cycle(y)). That would supply you with the required generator:
>>> for x in zip(itertools.cycle(letters), itertools.cycle(numbers)):
... print(x)
...
('a', '1')
('b', '2')
('c', '3')
('d', '4')
('e', '1')
('f', '2')
('g', '3')
('a', '4')
('b', '1')
('c', '2')
('d', '3')
...
If you want a finite list of elements, this should work
import itertools
letters = ('a', 'b', 'c', 'd', 'e', 'f', 'g')
numbers = ('1', '2', '3', '4')
max_elems = 10
list(itertools.islice((zip(itertools.cycle(letters), itertools.cycle(numbers))), max_elems))
results in
[('a', '1'), ('b', '2'), ('c', '3'), ('d', '4'), ('e', '1'), ('f', '2'), ('g', '3'), ('a', '4'), ('b', '1'), ('c', '2')]
Given a multidimensional list (a list of lists) I would like to get all possible combinations of the sub lists items.
For example an input of:
my_list = [
['a', 'b'], ['1', '2'], ['#', '&']
]
Would result in:
result = [
['a'],
['b'],
['1'],
['2'],
['#'],
['&'],
['a', '1'],
['a', '2'],
['a', '#'],
['a', '&']
['b', '1'],
['b', '2'],
['b', '#'],
['b', '&'],
['a', '1', '#'],
['a', '1', '&'],
['a', '2', '#'],
['a', '2', '&'],
...]
I tried using itertools.product(*list) but that results in a combination of all items without the smaller sets of combinations. It seems that itertools.combinations, itertools.permutations, etc don't quite give what I am looking for.
Is there a quick way of doing this?
In that case you first iterate over all possible lengths. For each length you pick all possible combinations of lists, and for each of these combinations you use itertools.product:
def weird_product(*data):
for i in range(1,len(data)+1):
for subdata in itertools.combinations(data,i):
for elem in itertools.product(*subdata):
yield elem
This generates:
>>> list(weird_product(*data))
[('a',), ('b',), ('1',), ('2',), ('#',), ('&',), ('a', '1'), ('a', '2'), ('b', '1'), ('b', '2'), ('a', '#'), ('a', '&'), ('b', '#'), ('b', '&'), ('1', '#'), ('1', '&'), ('2', '#'), ('2', '&'), ('a', '1', '#'), ('a', '1', '&'), ('a', '2', '#'), ('a', '2', '&'), ('b', '1', '#'), ('b', '1', '&'), ('b', '2', '#'), ('b', '2', '&')]
or more elegantly formatted:
>>> list(weird_product(*data))
[('a',),
('b',),
('1',),
('2',),
('#',),
('&',),
('a', '1'),
('a', '2'),
('b', '1'),
('b', '2'),
('a', '#'),
('a', '&'),
('b', '#'),
('b', '&'),
('1', '#'),
('1', '&'),
('2', '#'),
('2', '&'),
('a', '1', '#'),
('a', '1', '&'),
('a', '2', '#'),
('a', '2', '&'),
('b', '1', '#'),
('b', '1', '&'),
('b', '2', '#'),
('b', '2', '&')]
I want to iterate over a list of lists, and pull all the combinations from the lists.
I can do it in the following way:
list = [['1','2','3'],['a','b','c'],['x','y','z']]
for itemi in list[0]:
print itemi
for itemj in list[1]:
print itemi+itemj
for itemk in list[2]:
print itemi+itemj+itemk
My problem is that I want to do it over a varying number of lists in the list o' lists. There will be at first just one list, and (for now) end with 64 lists.
The above example is for 3 lists, and in reality all the lists contain the same values (zero to 255). I used the above examples just so you can see what the output should look like.
I figured there must be a better way to this, without having to build the nested for loops for each size of list of lists.
Use itertools.product():
>>> l = [['1','2','3'],['a','b','c'],['x','y','z']]
>>> import itertools
>>> list(itertools.product(*l))
[('1', 'a', 'x'), ('1', 'a', 'y'), ('1', 'a', 'z'), ('1', 'b', 'x'),
('1', 'b', 'y'), ('1', 'b', 'z'), ('1', 'c', 'x'), ('1', 'c', 'y'),
('1', 'c', 'z'), ('2', 'a', 'x'), ('2', 'a', 'y'), ('2', 'a', 'z'),
('2', 'b', 'x'), ('2', 'b', 'y'), ('2', 'b', 'z'), ('2', 'c', 'x'),
('2', 'c', 'y'), ('2', 'c', 'z'), ('3', 'a', 'x'), ('3', 'a', 'y'),
('3', 'a', 'z'), ('3', 'b', 'x'), ('3', 'b', 'y'), ('3', 'b', 'z'),
('3', 'c', 'x'), ('3', 'c', 'y'), ('3', 'c', 'z')]
This is not quite what you want yet, but it's easy to get there:
>>> for i in range(len(l)):
... print(list(itertools.product(*l[:i+1])))
...
[('1',), ('2',), ('3',)]
[('1', 'a'), ('1', 'b'), ('1', 'c'), ('2', 'a'), ('2', 'b'), ('2', 'c'),
('3', 'a'), ('3', 'b'), ('3', 'c')]
[('1', 'a', 'x'), ('1', 'a', 'y'), ('1', 'a', 'z'), ('1', 'b', 'x'),
('1', 'b', 'y'), ('1', 'b', 'z'), ('1', 'c', 'x'), ('1', 'c', 'y'),
('1', 'c', 'z'), ('2', 'a', 'x'), ('2', 'a', 'y'), ('2', 'a', 'z'),
('2', 'b', 'x'), ('2', 'b', 'y'), ('2', 'b', 'z'), ('2', 'c', 'x'),
('2', 'c', 'y'), ('2', 'c', 'z'), ('3', 'a', 'x'), ('3', 'a', 'y'),
('3', 'a', 'z'), ('3', 'b', 'x'), ('3', 'b', 'y'), ('3', 'b', 'z'),
('3', 'c', 'x'), ('3', 'c', 'y'), ('3', 'c', 'z')]
Get everything in one list:
>>> result = []
>>> for i in range(len(l)):
... result.extend(list(itertools.product(*l[:i+1])))
...
>>> result
[('1',), ('2',), ('3',), ('1', 'a'), ('1', 'b'), ('1', 'c'), ('2', 'a'),
('2', 'b'), ('2', 'c'), ('3', 'a'), ('3', 'b'), ('3', 'c'), ('1', 'a', 'x'),
('1', 'a', 'y'), ('1', 'a', 'z'), ('1', 'b', 'x'), ('1', 'b', 'y'),
('1', 'b', 'z'), ('1', 'c', 'x'), ('1', 'c', 'y'), ('1', 'c', 'z'),
('2', 'a', 'x'), ('2', 'a', 'y'), ('2', 'a', 'z'), ('2', 'b', 'x'),
('2', 'b', 'y'), ('2', 'b', 'z'), ('2', 'c', 'x'), ('2', 'c', 'y'),
('2', 'c', 'z'), ('3', 'a', 'x'), ('3', 'a', 'y'), ('3', 'a', 'z'),
('3', 'b', 'x'), ('3', 'b', 'y'), ('3', 'b', 'z'), ('3', 'c', 'x'),
('3', 'c', 'y'), ('3', 'c', 'z')]
Get it in the exact shape you want:
>>> sorted(result)
[('1',), ('1', 'a'), ('1', 'a', 'x'), ('1', 'a', 'y'), ('1', 'a', 'z'),
('1', 'b'), ('1', 'b', 'x'), ('1', 'b', 'y'), ('1', 'b', 'z'), ('1', 'c'),
('1', 'c', 'x'), ('1', 'c', 'y'), ('1', 'c', 'z'), ('2',), ('2', 'a'),
('2', 'a', 'x'), ('2', 'a', 'y'), ('2', 'a', 'z'), ('2', 'b'), ('2', 'b', 'x'),
('2', 'b', 'y'), ('2', 'b', 'z'), ('2', 'c'), ('2', 'c', 'x'), ('2', 'c', 'y'),
('2', 'c', 'z'), ('3',), ('3', 'a'), ('3', 'a', 'x'), ('3', 'a', 'y'),
('3', 'a', 'z'), ('3', 'b'), ('3', 'b', 'x'), ('3', 'b', 'y'), ('3', 'b', 'z'),
('3', 'c'), ('3', 'c', 'x'), ('3', 'c', 'y'), ('3', 'c', 'z')]