my code is this

idnt=[] 
idntfrq=[]           

for i in range(len(occ)):
    idnt.append([])
    idntfrq.append([])
    for j in range(len(occ[i])):
        for j2 in range(j,len(occ[i])):
            for d in occ[i][j]:
                idnt[i].append(d)
                idntfrq[i].append([j])
                occ[i][j].remove(d)
                for d2 in occ[i][j2]:
                    if d==d2:
                
                        idntfrq[i][-1].append(j2)
                        occ[i][j2].remove(d)
        

the list of lists is occ (50 lists inside with various lengths each)

the thought was to iterate over everything,store each value in the idnt[i] list and the index of the list in which it appears to the idntfrq[i] list and then remove the ellement from the list of the current itteration,the occ list should be empty after that but it is not,i uploaded a prntscr of the occ[0][0] to see what i mean

NOTE:each list inside a list contains every element only once , but i want to count the occurences across all the lists inside every occ[i] (for i in 50)and also keep the indexenter image description here

CodePudding user response：

This code will do what you are asking:

mainList = [[1,2,3],[3,4,5],[1,5,9]]
d = {}
for l in mainList:
    for item in l:
        if item in d.keys():
            d[item]  =1
        else:
            d[item] = 1
print(d)

Output:

{1: 2, 2: 1, 3: 2, 4: 1, 5: 2, 9: 1}

It gives the answer in a dictionary where keys are the items and the values is the number of appearances.

This output and be further formatted if needed.

CodePudding user response：

I think this works:

from collections import Counter
list_of_lists = [[1, 2, 3], [4, 1, 2], [3, 4, 5]]
counter = Counter()
for _list in list_of_lists:
    counter.update(_list)