Reputation: 17069
I know how to generate combinations of a set and that's a builtin in Python (what I use), anyway. But how to generate combinations with replacements?
Suppose I have a set with, say, two identical elements - for example, AABCDE.
Combinations of 3 items could be:
"AAB"
"ABC"
"CDE"
However, the program would count ABC twice - once when using the first A, and the second one using the second A.
What is a good way to generate such combinations without duplicates?
Thanks.
Upvotes: 0
Views: 2844
Reputation: 1009
def stepper_w_w(l,stop):#stepper_with_while
"""l is a list of any size usually you would input [1,1,1,1...],
stop is the highest number you want to stop at so if you put in stop=5
the sequence would stop at [5,5,5,5...]
This stepper shows the first number that equals the last.
This generates combinations with replacement. """
numb1=1
while numb1<stop:
#print(numb1)
l[0]=numb1
NeL=0
while l[len(l)-1]<=numb1:
if l[NeL]==l[len(l)-1]:
l[NeL]+=1
l[(NeL+1):]=[1]*((len(l))-(NeL+1))
print(l)
"""iter_2s=NeL+1
while iter_2s<=(len(l)-1): #this is different from above
l[iter_2s]=2
iter_2s+=1
print(l)"""
NeL=-1
NeL+=1
numb1+=1
Upvotes: 0
Reputation: 304355
>>> import itertools
>>> ["".join(x) for x in (itertools.combinations(set("AABCDE"),3))]
['ACB', 'ACE', 'ACD', 'ABE', 'ABD', 'AED', 'CBE', 'CBD', 'CED', 'BED']
>>>
From your other comments, I think I misunderstood what you are asking.
>>> import itertools
>>> set("".join(x) for x in (itertools.combinations("AABCDE",3)))
set(['AAE', 'AAD', 'ABC', 'ABD', 'ABE', 'AAC', 'AAB', 'BCD', 'BCE', 'ACD', 'CDE', 'ACE', 'ADE', 'BDE'])
Upvotes: 2
Reputation: 319831
convert it to set
, that's the easiest way to get rid of duplicates.
Upvotes: 2