# Find the most common element in a list

Each Answer to this Q is separated by one/two green lines.

What is an efficient way to find the most common element in a Python list?

My list items may not be hashable so can’t use a dictionary.
Also in case of draws the item with the lowest index should be returned. Example:

``````>>> most_common(['duck', 'duck', 'goose'])
'duck'
>>> most_common(['goose', 'duck', 'duck', 'goose'])
'goose'
``````

A simpler one-liner:

``````def most_common(lst):
return max(set(lst), key=lst.count)
``````

Borrowing from here, this can be used with Python 2.7:

``````from collections import Counter

def Most_Common(lst):
data = Counter(lst)
return data.most_common(1)
``````

Works around 4-6 times faster than Alex’s solutions, and is 50 times faster than the one-liner proposed by newacct.

To retrieve the element that occurs first in the list in case of ties:

``````def most_common(lst):
data = Counter(lst)
return max(lst, key=data.get)
``````

With so many solutions proposed, I’m amazed nobody’s proposed what I’d consider an obvious one (for non-hashable but comparable elements) — [`itertools.groupby`]. `itertools` offers fast, reusable functionality, and lets you delegate some tricky logic to well-tested standard library components. Consider for example:

``````import itertools
import operator

def most_common(L):
# get an iterable of (item, iterable) pairs
SL = sorted((x, i) for i, x in enumerate(L))
# print 'SL:', SL
groups = itertools.groupby(SL, key=operator.itemgetter(0))
# auxiliary function to get "quality" for an item
def _auxfun(g):
item, iterable = g
count = 0
min_index = len(L)
for _, where in iterable:
count += 1
min_index = min(min_index, where)
# print 'item %r, count %r, minind %r' % (item, count, min_index)
return count, -min_index
# pick the highest-count/earliest item
return max(groups, key=_auxfun)
``````

This could be written more concisely, of course, but I’m aiming for maximal clarity. The two `print` statements can be uncommented to better see the machinery in action; for example, with prints uncommented:

``````print most_common(['goose', 'duck', 'duck', 'goose'])
``````

emits:

``````SL: [('duck', 1), ('duck', 2), ('goose', 0), ('goose', 3)]
item 'duck', count 2, minind 1
item 'goose', count 2, minind 0
goose
``````

As you see, `SL` is a list of pairs, each pair an item followed by the item’s index in the original list (to implement the key condition that, if the “most common” items with the same highest count are > 1, the result must be the earliest-occurring one).

`groupby` groups by the item only (via `operator.itemgetter`). The auxiliary function, called once per grouping during the `max` computation, receives and internally unpacks a group – a tuple with two items `(item, iterable)` where the iterable’s items are also two-item tuples, `(item, original index)` [[the items of `SL`]].

Then the auxiliary function uses a loop to determine both the count of entries in the group’s iterable, and the minimum original index; it returns those as combined “quality key”, with the min index sign-changed so the `max` operation will consider “better” those items that occurred earlier in the original list.

This code could be much simpler if it worried a little less about big-O issues in time and space, e.g….:

``````def most_common(L):
groups = itertools.groupby(sorted(L))
def _auxfun((item, iterable)):
return len(list(iterable)), -L.index(item)
return max(groups, key=_auxfun)
``````

same basic idea, just expressed more simply and compactly… but, alas, an extra O(N) auxiliary space (to embody the groups’ iterables to lists) and O(N squared) time (to get the `L.index` of every item). While premature optimization is the root of all evil in programming, deliberately picking an O(N squared) approach when an O(N log N) one is available just goes too much against the grain of scalability!-)

Finally, for those who prefer “oneliners” to clarity and performance, a bonus 1-liner version with suitably mangled names:-).

``````from itertools import groupby as g
def most_common_oneliner(L):
return max(g(sorted(L)), key=lambda(x, v):(len(list(v)),-L.index(x)))
``````

What you want is known in statistics as mode, and Python of course has a built-in function to do exactly that for you:

``````>>> from statistics import mode
>>> mode([1, 2, 2, 3, 3, 3, 3, 3, 4, 5, 6, 6, 6])
3
``````

Note that if there is no “most common element” such as cases where the top two are tied, this will raise `StatisticsError`, because statistically speaking, there is no mode in this case.

Without the requirement about the lowest index, you can use `collections.Counter` for this:

``````from collections import Counter

a = [1936, 2401, 2916, 4761, 9216, 9216, 9604, 9801]

c = Counter(a)

print(c.most_common(1)) # the one most common element... 2 would mean the 2 most common
[(9216, 2)] # a set containing the element, and it's count in 'a'
``````

If they are not hashable, you can sort them and do a single loop over the result counting the items (identical items will be next to each other). But it might be faster to make them hashable and use a dict.

``````def most_common(lst):
cur_length = 0
max_length = 0
cur_i = 0
max_i = 0
cur_item = None
max_item = None
for i, item in sorted(enumerate(lst), key=lambda x: x):
if cur_item is None or cur_item != item:
if cur_length > max_length or (cur_length == max_length and cur_i < max_i):
max_length = cur_length
max_i = cur_i
max_item = cur_item
cur_length = 1
cur_i = i
cur_item = item
else:
cur_length += 1
if cur_length > max_length or (cur_length == max_length and cur_i < max_i):
return cur_item
return max_item
``````

This is an O(n) solution.

``````mydict   = {}
cnt, itm = 0, ''
for item in reversed(lst):
mydict[item] = mydict.get(item, 0) + 1
if mydict[item] >= cnt :
cnt, itm = mydict[item], item

print itm
``````

(reversed is used to make sure that it returns the lowest index item)

Sort a copy of the list and find the longest run. You can decorate the list before sorting it with the index of each element, and then choose the run that starts with the lowest index in the case of a tie.

A one-liner:

``````def most_common (lst):
return max(((item, lst.count(item)) for item in set(lst)), key=lambda a: a)``````

I am doing this using scipy stat module and lambda:

``````import scipy.stats
lst = [1,2,3,4,5,6,7,5]
most_freq_val = lambda x: scipy.stats.mode(x)
print(most_freq_val(lst))
``````

Result:

`````` most_freq_val = 5
``````

``````# use Decorate, Sort, Undecorate to solve the problem

def most_common(iterable):
# Make a list with tuples: (item, index)
# The index will be used later to break ties for most common item.
lst = [(x, i) for i, x in enumerate(iterable)]
lst.sort()

# lst_final will also be a list of tuples: (count, index, item)
# Sorting on this list will find us the most common item, and the index
# will break ties so the one listed first wins.  Count is negative so
# largest count will have lowest value and sort first.
lst_final = []

# Get an iterator for our new list...
itr = iter(lst)

# ...and pop the first tuple off.  Setup current state vars for loop.
count = 1
tup = next(itr)
x_cur, i_cur = tup

# Loop over sorted list of tuples, counting occurrences of item.
for tup in itr:
# Same item again?
if x_cur == tup:
# Yes, same item; increment count
count += 1
else:
# No, new item, so write previous current item to lst_final...
t = (-count, i_cur, x_cur)
lst_final.append(t)
# ...and reset current state vars for loop.
x_cur, i_cur = tup
count = 1

# Write final item after loop ends
t = (-count, i_cur, x_cur)
lst_final.append(t)

lst_final.sort()

print most_common(['x', 'e', 'a', 'e', 'a', 'e', 'e']) # prints 'e'
print most_common(['goose', 'duck', 'duck', 'goose']) # prints 'goose'
``````

Simple one line solution

``````moc= max([(lst.count(chr),chr) for chr in set(lst)])
``````

It will return most frequent element with its frequency.

You probably don’t need this anymore, but this is what I did for a similar problem. (It looks longer than it is because of the comments.)

``````itemList = ['hi', 'hi', 'hello', 'bye']

counter = {}
maxItemCount = 0
for item in itemList:
try:
# Referencing this will cause a KeyError exception
# if it doesn't already exist
counter[item]
# ... meaning if we get this far it didn't happen so
# we'll increment
counter[item] += 1
except KeyError:
# If we got a KeyError we need to create the
# dictionary key
counter[item] = 1

# Keep overwriting maxItemCount with the latest number,
# if it's higher than the existing itemCount
if counter[item] > maxItemCount:
maxItemCount = counter[item]
mostPopularItem = item

print mostPopularItem
``````

Building on Luiz’s answer, but satisfying the “in case of draws the item with the lowest index should be returned” condition:

``````from statistics import mode, StatisticsError

def most_common(l):
try:
return mode(l)
except StatisticsError as e:
# will only return the first element if no unique mode found
if 'no unique mode' in e.args:
return l
# this is for "StatisticsError: no mode for empty data"
# after calling mode([])
raise
``````

Example:

``````>>> most_common(['a', 'b', 'b'])
'b'
>>> most_common([1, 2])
1
>>> most_common([])
StatisticsError: no mode for empty data
``````

``````ans  = [1, 1, 0, 0, 1, 1]
all_ans = {ans.count(ans[i]): ans[i] for i in range(len(ans))}
print(all_ans)
``````
``````all_ans={4: 1, 2: 0}
max_key = max(all_ans.keys())
``````

4

``````print(all_ans[max_key])
``````

1

``````#This will return the list sorted by frequency:

def orderByFrequency(list):

listUniqueValues = np.unique(list)
listQty = []
listOrderedByFrequency = []

for i in range(len(listUniqueValues)):
listQty.append(list.count(listUniqueValues[i]))
for i in range(len(listQty)):
index_bigger = np.argmax(listQty)
for j in range(listQty[index_bigger]):
listOrderedByFrequency.append(listUniqueValues[index_bigger])
listQty[index_bigger] = -1
return listOrderedByFrequency

#And this will return a list with the most frequent values in a list:

def getMostFrequentValues(list):

if (len(list) <= 1):
return list

list_most_frequent = []
list_ordered_by_frequency = orderByFrequency(list)

list_most_frequent.append(list_ordered_by_frequency)
frequency = list_ordered_by_frequency.count(list_ordered_by_frequency)

index = 0
while(index < len(list_ordered_by_frequency)):
index = index + frequency

if(index < len(list_ordered_by_frequency)):
testValue = list_ordered_by_frequency[index]
testValueFrequency = list_ordered_by_frequency.count(testValue)

if (testValueFrequency == frequency):
list_most_frequent.append(testValue)
else:
break

return list_most_frequent

#tests:
print(getMostFrequentValues([]))
print(getMostFrequentValues())
print(getMostFrequentValues([1,1]))
print(getMostFrequentValues([2,1]))
print(getMostFrequentValues([2,2,1]))
print(getMostFrequentValues([1,2,1,2]))
print(getMostFrequentValues([1,2,1,2,2]))
print(getMostFrequentValues([3,2,3,5,6,3,2,2]))
print(getMostFrequentValues([1,2,2,60,50,3,3,50,3,4,50,4,4,60,60]))

Results:
[]


[1, 2]

[1, 2]

[2, 3]
[3, 4, 50, 60]
``````

``````def most_frequent(List):

counter = 0

num = List

for i in List:

curr_frequency = List.count(i)

if(curr_frequency> counter):

counter = curr_frequency

num = i

return num

List = [2, 1, 2, 2, 1, 3]

print(most_frequent(List))
``````

Here:

``````def most_common(l):
max = 0
maxitem = None
for x in set(l):
count =  l.count(x)
if count > max:
max = count
maxitem = x
return maxitem
``````

I have a vague feeling there is a method somewhere in the standard library that will give you the count of each element, but I can’t find it.

This is the obvious slow solution (O(n^2)) if neither sorting nor hashing is feasible, but equality comparison (`==`) is available:

``````def most_common(items):
if not items:
raise ValueError
fitems = []
best_idx = 0
for item in items:
item_missing = True
i = 0
for fitem in fitems:
if fitem == item:
fitem += 1
d = fitem - fitems[best_idx]
if d > 0 or (d == 0 and fitems[best_idx] > fitem):
best_idx = i
item_missing = False
break
i += 1
if item_missing:
fitems.append([item, 1, i])
return items[best_idx]
``````

But making your items hashable or sortable (as recommended by other answers) would almost always make finding the most common element faster if the length of your list (n) is large. O(n) on average with hashing, and O(n*log(n)) at worst for sorting.

``````>>> li  = ['goose', 'duck', 'duck']

>>> def foo(li):
st = set(li)
mx = -1
for each in st:
temp = li.count(each):
if mx < temp:
mx = temp
h = each
return h

>>> foo(li)
'duck'
``````

I needed to do this in a recent program. I’ll admit it, I couldn’t understand Alex’s answer, so this is what I ended up with.

``````def mostPopular(l):
mpEl=None
mpIndex=0
mpCount=0
curEl=None
curCount=0
for i, el in sorted(enumerate(l), key=lambda x: (x, x), reverse=True):
curCount=curCount+1 if el==curEl else 1
curEl=el
if curCount>mpCount \
or (curCount==mpCount and i<mpIndex):
mpEl=curEl
mpIndex=i
mpCount=curCount
return mpEl, mpCount, mpIndex
``````

I timed it against Alex’s solution and it’s about 10-15% faster for short lists, but once you go over 100 elements or more (tested up to 200000) it’s about 20% slower.

Hi this is a very simple solution with big O(n)

``````L = [1, 4, 7, 5, 5, 4, 5]

def mode_f(L):
counter = 0
number = L
for i in L:
amount_times = L.count(i)
if amount_times > counter:
counter = amount_times
number = i

return number
``````

Where number the element in the list that repeats most of the time

The most common element should be the one which is appearing more than `N/2` times in the array where `N` being the `len(array)`. The below technique will do it in `O(n)` time complexity, with just consuming `O(1)` auxiliary space.

``````from collections import Counter

def majorityElement(arr):
majority_elem = Counter(arr)
size = len(arr)
for key, val in majority_elem.items():
if val > size/2:
return key
return -1
``````

``````def mostCommonElement(list):
count = {} // dict holder
max = 0 // keep track of the count by key
result = None // holder when count is greater than max
for i in list:
if i not in count:
count[i] = 1
else:
count[i] += 1
if count[i] > max:
max = count[i]
result = i
return result
``````

mostCommonElement([“a”,”b”,”a”,”c”]) -> “a”

`````` def most_common(lst):
if max([lst.count(i)for i in lst]) == 1:
return False
else:
return max(set(lst), key=lst.count)
``````

``````def popular(L):
C={}
for a in L:
C[a]=L.count(a)
for b in C.keys():
if C[b]==max(C.values()):
return b
L=[2,3,5,3,6,3,6,3,6,3,7,467,4,7,4]
print popular(L)
`````` The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .