I have a nested list g=[[2, 1],[1, 3],[8, 1]]
. I want to have a list in which:
every inner element of the list added up to at most another two inner elements. Here's the process and therefore desired output:
# 1= a single element like g[0][1]
# 2= a single element like g[0][0] or can be added by two inner elements like g[0][1]+g[1][0]
# 3= a single g[1][1], adding two elements g[0][0]+g[1][0] or at most three g[0][1]+g[1][0]+g[2][1]
.
.
.
# 13= g[0][0]+g[1][1]+g[2][0]
So the final result will be a list. Notice there can't be 7 in the results since there's no combination (without replacement) that can be add up to 7. Also from each element at most one inner value can be selected.
expected_result = [1,2,3,4,5,6,8,9,10,11,12,13]
This is what I've done but it doesn't contain 1 and 2 and it also contains 7:
g=[[2, 1],[1, 3],[8, 1]]
from itertools import product
maxx = []
# Getting all possible combination of adding at most 3 elements (=length of g)
for i in list((product([i for j in g for i in j], repeat=len(g)))):
maxx.append(sum(i))
# Narrowing the result so it doesn't exceed the maximum combination which is sum([2,3,8])
print([i for i in set(maxx) if i<=sum(max_list)])
>>> [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
Any help is appreciated.
It seems like the shape/format of the data is a red herring. Just flatten the list and looks for the unique sums of each combination of 1, 2, or 3 elements of the flattened list. Use itertools.chain
to flatten the list, itertools.combinations
to create the combinations, itertools.chain
again to combine the combinations, and a set
to return unique results:
>>> import itertools
>>> g = [[2, 1],[1, 3],[8, 1]]
>>> flattened = itertools.chain(*g)
>>> flattened
[2, 1, 1, 3, 8, 1]
>>> list(set(map(sum,
itertools.chain(itertools.combinations(flattened, 1),
itertools.combinations(flattened, 2),
itertools.combinations(flattened, 3)))))
[1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13]
Edit : The question was changed to include the following constraint:
Also from each element at most one inner value can be selected.
That means you can't flatten the list. Here's an updated solution that satisfies the new constraint:
>>> list(set(map(sum,
itertools.chain.from_iterable(itertools.combinations(p, n)
for n in range(1,4)
for p in itertools.product(*g)))))
[1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13]
>>> g = [[2, 1],[1, 3],[8, 1],[14, 15] # should not produce 29 as an answer
>>> list(set(map(sum,
itertools.chain.from_iterable(itertools.combinations(p, n)
for n in range(1,4)
for p in itertools.product(*g)))))
[1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26]
The simpliest implementation that I can think of now:
def fun(numbers):
all_numbers = [x for y in numbers for x in y]
output = list(set(all_numbers))
for x in all_numbers:
for y in all_numbers:
if x is not y:
output.append(x + y)
for z in all_numbers:
if y is not x and y is not z and x is not z:
output.append(x + y + z)
return list(set(output))
print(fun([[2, 1], [1, 3], [8, 1]]))
Above code prints
[1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13]
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