Permutations between two lists of unequal length
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)]
itertools
permutations with unique values
class unique_element: def __init__(self,value,occurrences): self.value = value self.occurrences = occurrences def perm_unique(elements): eset=set(elements) listunique = [unique_element(i,elements.count(i)) for i in eset] u=len(elements) return perm_unique_helper(listunique,[0]*u,u-1) def perm_unique_helper(listunique,result_list,d): if d < 0: yield tuple(result_list) else: for i in listunique: if i.occurrences > 0: result_list[d]=i.value i.occurrences-=1 for g in perm_unique_helper(listunique,result_list,d-1): yield g i.occurrences+=1 a = list(perm_unique([1,1,2])) print(a) result: [(2, … Read more
How do I use itertools.groupby()?
IMPORTANT NOTE: You have to sort your data first. The part I didn’t get is that in the example construction groups = [] uniquekeys = [] for k, g in groupby(data, keyfunc): groups.append(list(g)) # Store group iterator as a list uniquekeys.append(k) k is the current grouping key, and g is an iterator that you can … Read more