I’m having trouble wrapping my head around a algorithm I’m try to implement. I have two lists and want to take particular combinations from the two lists.
Here’s an example.
names = ['a', 'b'] numbers = [1, 2]
the output in this case would be:
[('a', 1), ('b', 2)]
[('b', 1), ('a', 2)]
I might have more names than numbers, i.e. len(names) >= len(numbers). Here’s an example with 3 names and 2 numbers:
names = ['a', 'b', 'c'] numbers = [1, 2]
output:
[('a', 1), ('b', 2)]
[('b', 1), ('a', 2)]
[('a', 1), ('c', 2)]
[('c', 1), ('a', 2)]
[('b', 1), ('c', 2)]
[('c', 1), ('b', 2)]
Answers:
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Method 1
The simplest way is to use itertools.product:
a = ["foo", "melon"]
b = [True, False]
c = list(itertools.product(a, b))
>> [("foo", True), ("foo", False), ("melon", True), ("melon", False)]
Method 2
May be simpler than the simplest one above:
>>> a = ["foo", "bar"]
>>> b = [1, 2, 3]
>>> [(x,y) for x in a for y in b] # for a list
[('foo', 1), ('foo', 2), ('foo', 3), ('bar', 1), ('bar', 2), ('bar', 3)]
>>> ((x,y) for x in a for y in b) # for a generator if you worry about memory or time complexity.
<generator object <genexpr> at 0x1048de850>
without any import
Method 3
Note: This answer is for the specific question asked above. If you are here from Google and just looking for a way to get a Cartesian product in Python, itertools.product or a simple list comprehension may be what you are looking for – see the other answers.
Suppose len(list1) >= len(list2). Then what you appear to want is to take all permutations of length len(list2) from list1 and match them with items from list2. In python:
import itertools list1=['a','b','c'] list2=[1,2] [list(zip(x,list2)) for x in itertools.permutations(list1,len(list2))]
Returns
[[('a', 1), ('b', 2)], [('a', 1), ('c', 2)], [('b', 1), ('a', 2)], [('b', 1), ('c', 2)], [('c', 1), ('a', 2)], [('c', 1), ('b', 2)]]
Method 4
I was looking for a list multiplied by itself with only unique combinations, which is provided as this function.
import itertools itertools.combinations(list, n_times)
Here as an excerpt from the Python docs on itertools That might help you find what your looking for.
Combinatoric generators:
Iterator | Results
-----------------------------------------+----------------------------------------
product(p, q, ... [repeat=1]) | cartesian product, equivalent to a
| nested for-loop
-----------------------------------------+----------------------------------------
permutations(p[, r]) | r-length tuples, all possible
| orderings, no repeated elements
-----------------------------------------+----------------------------------------
combinations(p, r) | r-length tuples, in sorted order, no
| repeated elements
-----------------------------------------+----------------------------------------
combinations_with_replacement(p, r) | r-length tuples, in sorted order,
| with repeated elements
-----------------------------------------+----------------------------------------
product('ABCD', repeat=2) | AA AB AC AD BA BB BC BD CA CB CC CD DA DB DC DD
permutations('ABCD', 2) | AB AC AD BA BC BD CA CB CD DA DB DC
combinations('ABCD', 2) | AB AC AD BC BD CD
combinations_with_replacement('ABCD', 2) | AA AB AC AD BB BC BD CC CD DD
Method 5
the best way to find out all the combinations for large number of lists is:
import itertools
from pprint import pprint
inputdata = [
['a', 'b', 'c'],
['d'],
['e', 'f'],
]
result = list(itertools.product(*inputdata))
pprint(result)
the result will be:
[('a', 'd', 'e'),
('a', 'd', 'f'),
('b', 'd', 'e'),
('b', 'd', 'f'),
('c', 'd', 'e'),
('c', 'd', 'f')]
Method 6
Or the KISS answer for short lists:
[(i, j) for i in list1 for j in list2]
Not as performant as itertools but you’re using python so performance is already not your top concern…
I like all the other answers too!
Method 7
You might want to try a one line list comprehension:
>>> [name+number for name in 'ab' for number in '12'] ['a1', 'a2', 'b1', 'b2'] >>> [name+number for name in 'abc' for number in '12'] ['a1', 'a2', 'b1', 'b2', 'c1', 'c2']
Method 8
a tiny improvement for the answer from interjay, to make the result as a flatten list.
>>> list3 = [zip(x,list2) for x in itertools.permutations(list1,len(list2))]
>>> import itertools
>>> chain = itertools.chain(*list3)
>>> list4 = list(chain)
[('a', 1), ('b', 2), ('a', 1), ('c', 2), ('b', 1), ('a', 2), ('b', 1), ('c', 2), ('c', 1), ('a', 2), ('c', 1), ('b', 2)]
reference from this link
Method 9
Without itertools as a flattened list:
[(list1[i], list2[j]) for i in range(len(list1)) for j in range(len(list2))]
or in Python 2:
[(list1[i], list2[j]) for i in xrange(len(list1)) for j in xrange(len(list2))]
Method 10
The better answers to this only work for specific lengths of lists that are provided.
Here’s a version that works for any lengths of input. It also makes the algorithm clear in terms of the mathematical concepts of combination and permutation.
from itertools import combinations, permutations
list1 = ['1', '2']
list2 = ['A', 'B', 'C']
num_elements = min(len(list1), len(list2))
list1_combs = list(combinations(list1, num_elements))
list2_perms = list(permutations(list2, num_elements))
result = [
tuple(zip(perm, comb))
for comb in list1_combs
for perm in list2_perms
]
for idx, ((l11, l12), (l21, l22)) in enumerate(result):
print(f'{idx}: {l11}{l12} {l21}{l22}')
This outputs:
0: A1 B2 1: A1 C2 2: B1 A2 3: B1 C2 4: C1 A2 5: C1 B2
Method 11
Answering the question “given two lists, find all possible permutations of pairs of one item from each list” and using basic Python functionality (i.e., without itertools) and, hence, making it easy to replicate for other programming languages:
def rec(a, b, ll, size):
ret = []
for i,e in enumerate(a):
for j,f in enumerate(b):
l = [e+f]
new_l = rec(a[i+1:], b[:j]+b[j+1:], ll, size)
if not new_l:
ret.append(l)
for k in new_l:
l_k = l + k
ret.append(l_k)
if len(l_k) == size:
ll.append(l_k)
return ret
a = ['a','b','c']
b = ['1','2']
ll = []
rec(a,b,ll, min(len(a),len(b)))
print(ll)
Returns
[['a1', 'b2'], ['a1', 'c2'], ['a2', 'b1'], ['a2', 'c1'], ['b1', 'c2'], ['b2', 'c1']]
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0