I have a nested list structure as below. Each of the 4 nested structures represents some free positions for me. I want to find which elements are present in all 4 nested lists.
ary=[ [[0, 4], [5, 11]], [[0, 2], [0, 4], [5,10]], [[0, 4], [0, 14], [5,11]], [[0, 4], [0, 14], [5,11]] ]
As in above, in the first nested list [[0, 4], [5, 11]], the [0,4] is present in all but [5,11] is not. Hence, my answer should be [[0,4]] or even just [0,4].
I did this in two ways. Solution1:
ary1=ary
newlist = [item for items in ary for item in items]
x=[i for i in ary[0] if newlist.count(i)== len(ary1)]
#OUTPUT is x= [[0,4]]
Solution2:
x=[]
for u in ary[0]:
n=[]
n=[1 for t in range(1,len(ary)) if u in ary[t]]
if len(ary)-1==len(n):
x.append(u)
#OUTPUT is x= [[0,4]]
These two seem to take similar computational time when checked using line profiler. And this is the only point of heavy computation in my hundreds of liens of code and I want to reduce this. So, can you suggest any other Python commands/code that can do the task better than these two solutions?
You can try to convert each nested array at the second level into the set of tuples, where each lowest level array (i.e. [0,4]) is an element of the set. The conversion into tuples is required because lists are not hashable. Once you have each nested list of lists as a set, simply find their intersection.
set.intersection(*[set(tuple(elem) for elem in sublist) for sublist in ary])