How do you calculate the greatest number of repetitions in a list?

Question

If I have a list in Python like

[1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1]

How do I calculate the greatest number of repeats for any element? In this case 2 is repeated a maximum of 4 times and 1 is repeated a maximum of 3 times.

Is there a way to do this but also record the index at which the longest run began?

Answer 1

Usegroupby , it group elements by value:

from itertools import groupby
group = groupby([1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1])
print max(group, key=lambda k: len(list(k[1])))

And here is the code in action:

>>> group = groupby([1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1])
>>> print max(group, key=lambda k: len(list(k[1])))
(2, <itertools._grouper object at 0xb779f1cc>)
>>> group = groupby([1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 3, 3, 3, 3])
>>> print max(group, key=lambda k: len(list(k[1])))
(3, <itertools._grouper object at 0xb7df95ec>)

From python documentation:

The operation of groupby() is similar to the uniq filter in Unix. It generates a break or new group every time the value of the key function changes

# [k for k, g in groupby('AAAABBBCCDAABBB')] --> A B C D A B
# [list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D

If you also want the index of the longest run you can do the following:

group = groupby([1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 3, 3, 3, 3])
result = []
index = 0
for k, g in group:
   length = len(list(g))
   result.append((k, length, index))
   index += length

print max(result, key=lambda a:a[1])

Answer 2

Loop through the list, keep track of the current number, how many times it has been repeated, and compare that to the most times youve seen that number repeated.

Counts={}
Current=0
Current_Count=0
LIST = [1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1]
for i in LIST:
    if Current == i:
        Current_Count++
    else:
        Current_Count=1
        Current=i
    if Current_Count>Counts[i]:
        Counts[i]=Current_Count
print Counts

Answer 3

This is my solution:

def longest_repetition(l):
    if l == []:
        return None

    element = l[0]
    new = []
    lar = []

    for e in l:            
        if e == element:
            new.append(e)
        else:
            if len(new) > len(lar):
                lar = new
            new = []
            new.append(e)
            element = e
    if len(new) > len(lar):
        lar = new    
    return lar[0]

Answer 4

-You can make new copy of the list but with unique values and a corresponding hits list.

-Then get the Max of hits list and get from it's index your most repeated item.

oldlist = ["A", "B", "E", "C","A", "C","D","A", "E"]
newlist=[]
hits=[]
for i in range(len(oldlist)):
    if oldlist[i] in newlist:
        hits[newlist.index(oldlist[i])]+= 1
    else:
        newlist.append(oldlist[i])
        hits.append(1);
#find the most repeated item
temp_max_hits=max(hits)
temp_max_hits_index=hits.index(temp_max_hits)
print(newlist[temp_max_hits_index])
print(temp_max_hits)

But I don't know is this the fastest way to do that or there are faster solution. If you think there are faster or more efficient solution, kindly inform us.

Answer 5

If you want it for just any element (ie the element with the most repetitions), you could use:

def f((v, l, m), x):
    nl = l+1 if x==v else 1
    return (x, nl, max(m,nl))

maxrep = reduce(f, l, (0,0,0))[2];

This only counts continuous repetitions (Result for [1,2,2,2,1,2] would be 3 ) and only records the element with the the maximum number.

Edit : Made definition of fa bit shorter ...

Answer 6

i write this code and working easly:

lst = [4,7,2,7,7,7,3,12,57]
maximum=0
for i in lst:
    count = lst.count(i)  
    if count>maximum:
        maximum=count
        indexx = lst.index(i)
print(lst[indexx])

Answer 7

This code seems to work:

l = [1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1]
previous = None

# value/repetition pair
greatest = (-1, -1)
reps = 1

for e in l:
    if e == previous:
        reps += 1
    else:
        if reps > greatest[1]:
            greatest = (previous, reps)

        previous = e
        reps = 1

if reps > greatest[1]:
    greatest = (previous, reps)

print greatest

Answer 8

I'd use a hashmap of item to counter.

Every time you see a 'key' succession, increment its counter value. If you hit a new element, set the counter to 1 and keep going. At the end of this linear search, you should have the maximum succession count for each number.