Reputation: 11153
I have the following list of variables and a mastervariable
a = (1,5,7)
b = (1,3,5)
c = (2,2,2)
d = (5,2,8)
e = (5,5,8)
mastervariable = (3,2,5)
I'm trying to check if 2 elements in each variable exist in the master variable, such that the above would show B (3,5) and D (5,2) as being elements with at least 2 elements matching in the mastervariable. Also note that using sets would result in C showing up as matchign but I don't want to count C cause only 'one' of the elements in C are in mastervariable (i.e. 2 only shows up once in mastervariable not twice)
I currently have the very inefficient:
if current_variable[0]==mastervariable[0]:
if current_variable[1] = mastervariable[1]:
True
elif current_variable[2] = mastervariable[1]:
True
#### I don't use OR here because I need to know which variables match.
elif current_variable[1] == mastervariable[0]: ##<-- I'm now checking 2nd element
etc. etc.
I then continue to iterate like the above by checking each one at a time which is extremely inefficient. I did the above because using a FOR statement resulted in me checking the first element twice which was incorrect:
For i in a:
for j in a:
### this checked if 1 was in the master variable and not 1,5 or 1,7
Is there a way to use 2 FOR statement that allows me to check 2 elements in a list at once while skipping any element that has been used already? Alternatively, can you suggest an efficient way to do what I'm trying?
Edit: Mastervariable can have duplicates in it.
Upvotes: 0
Views: 99
Reputation: 304137
Seems like a good job for sets. Edit: sets aren't suitable since mastervariable can contain duplicates. Here is a version using Counters.
>>> a = (1,5,7)
>>>
>>> b = (1,3,5)
>>>
>>> c = (2,2,2)
>>>
>>> d = (5,2,8)
>>>
>>> e = (5,5,8)
>>> D=dict(a=a, b=b, c=c, d=d, e=e)
>>>
>>> from collections import Counter
>>> mastervariable = (5,5,3)
>>> mvc = Counter(mastervariable)
>>> for k,v in D.items():
... vc = Counter(v)
... if sum(min(count, vc[item]) for item, count in mvc.items())==2:
... print k
...
b
e
Upvotes: 1
Reputation: 35059
For the case where matching elements can be duplicated so that set
breaks, use Counter
as a multiset - the duplicates between a
and master
are found by:
count_a = Counter(a)
count_master = Counter(master)
count_both = count_a + count_master
dups = Counter({e : min((count_a[e], count_master[e])) for e in count_a if count_both[e] > count_a[e]})
The logic is reasonably intuitive: if there's more of an item in the combined count of a
and master
, then it is duplicated, and the multiplicity is however many of that item are in whichever of a
and master
has less of them.
It gives a Counter of all the duplicates, where the count is their multiplicity. If you want it back as a tuple, you can do tuple(dups.elements())
:
>>> a
(2, 2, 2)
>>> master
(1, 2, 2)
>>> dups = Counter({e : min((count_a[e], count_master[e])) for e in count_a if count_both[e] > count_a[e]})
>>> tuple(dups.elements())
(2, 2)
Upvotes: 1