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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)[0][0]
Works around 4-6 times faster than Alex's solutions, and is 50 times faster than the one-liner proposed by newacct.
On CPython 3.6+ (any Python 3.7+) the above will select the first seen element in case of ties. If you're running on older Python, to retrieve the element that occurs first in the list in case of ties you need to do two passes to preserve order:
# Only needed pre-3.6!
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][1]. 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)[0]
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)[0]
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)))[0]
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 on Python
<=3.7, and on 3.8 onwards it will return the first one encountered.
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[1]):
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[1])[0]
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)[0][0]
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[0]:
# 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()
answer = lst_final[0][2]
return answer
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[0]:
return l[0]
# 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[0])
frequency = list_ordered_by_frequency.count(list_ordered_by_frequency[0])
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([1]))
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]
[1]
[1, 2]
[2]
[1, 2]
[2]
[2, 3]
[3, 4, 50, 60]
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[0] == item:
fitem[1] += 1
d = fitem[1] - fitems[best_idx][1]
if d > 0 or (d == 0 and fitems[best_idx][2] > fitem[2]):
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[1], x[0]), 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.
def most_frequent(List):
counter = 0
num = List[0]
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))
Hi this is a very simple solution, with linear time complexity
L = ['goose', 'duck', 'duck']
def most_common(L):
current_winner = 0
max_repeated = None
for i in L:
amount_times = L.count(i)
if amount_times > current_winner:
current_winner = amount_times
max_repeated = i
return max_repeated
print(most_common(L))
"duck"
Where number, is the element in the list that repeats most of the time
numbers = [1, 3, 7, 4, 3, 0, 3, 6, 3]
max_repeat_num = max(numbers, key=numbers.count) *# which number most* frequently
max_repeat = numbers.count(max_repeat_num) *#how many times*
print(f" the number {max_repeat_num} is repeated{max_repeat} times")
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"
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 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)
I have a tuple with tuples inside like this:
tup = ((1,2,3,'Joe'),(3,4,5,'Kevin'),(6,7,8,'Joe'),(10,11,12,'Donald'))
This goes on and on and the numbers don't matter here. The only data that matters are the names. What I need is to count how many times a given name occurs in the tuple and return a list where each item is a list and the number of times it occurs, like this:
list_that_i_want = [['Joe',2],['Kevin',1],['Donald',1]]
I don't want to use any modules or collections like Counter. I want to hard code this.
I actually wanted to hardcode the full solution and not even use the '.count()' method.
So far what I got is this:
def create_list(tuples):
new_list= list()
cont = 0
for tup in tuples:
for name in tup:
name = tup[3]
cont = tup.count(name)
if name not in new_list:
new_list.append(name)
new_list.append(cont)
return new_list
list_that_i_want = create_list(tup)
print(list_that_i_want)
And the output that I am been given is:
['Joe',1,'Kevin',1,'Donald',1]
Any help? Python newbie here.
You could. create a dictionary first and find the counts. Then convert the dictionary to a list of list.
tup = ((1,2,3,'Joe'),(3,4,5,'Kevin'),(6,7,8,'Joe'),(10,11,12,'Donald'))
dx = {}
for _,_,_,nm in tup:
if nm in dx: dx[nm] +=1
else: dx[nm] = 1
list_i_want = [[k,v] for k,v in dx.items()]
print (list_i_want)
You can replace the for_loop and the if statement section to this one line:
for _,_,_,nm in tup: dx[nm] = dx.get(nm, 0) + 1
The output will be
[['Joe', 2], ['Kevin', 1], ['Donald', 1]]
The updated code will be:
tup = ((1,2,3,'Joe'),(3,4,5,'Kevin'),(6,7,8,'Joe'),(10,11,12,'Donald'))
dx = {}
for _,_,_,nm in tup: dx[nm] = dx.get(nm, 0) + 1
list_i_want = [[k,v] for k,v in dx.items()]
print (list_i_want)
Output:
[['Joe', 2], ['Kevin', 1], ['Donald', 1]]
Using an intermediary dict:
def create_list(tuple_of_tuples):
results = {}
for tup in tuple_of_tuples:
name = tup[3]
if name not in results:
results[name] = 0
results[name] += 1
return list(results.items())
Of course, using defaultdict, or even Counter, would be the more Pythonic solution.
You can try with this approach:
tuples = ((1,2,3,'Joe'),(3,4,5,'Kevin'),(6,7,8,'Joe'),(10,11,12,'Donald'))
results = {}
for tup in tuples:
if tup[-1] not in results:
results[tup[-1]] = 1
else:
results[tup[-1]] += 1
new_list = [[key,val] for key,val in results.items()]
Here, a no-counter solution:
results = {}
for t in tup:
results[t[-1]] = results[t[-1]]+1 if (t[-1] in results) else 1
results.items()
#dict_items([('Joe', 2), ('Kevin', 1), ('Donald', 1)])
I have a sample of n elements. I want to sub-sample m unique pairs from n.
Is there a simple off-the-self method to do this in python ?
For example, if n = [1,2,3,4,5,6,7] and m = 3, one such sample will be [(1,2),(3,4),(5,6)]
The random module has a sample function which will pick n unique items from a collection. You can then pair them up to create your desired output:
import random
import itertools
data = [1,2,3,4,5,6,7,8,9,10]
m = 3
def pairwise(iterable):
# from the itertools cookbook: https://docs.python.org/3/library/itertools.html#itertools-recipes
a, b = itertools.tee(iterable)
next(b, None)
return zip(a, b)
chosen = random.sample(data, m * 2)
result = pairwise(chosen)
Depending on what you mean precisely with 'random', the answer will differ!
For uniform sampling of unique pairs, assuming all elements of your list are distinct:
import itertools, random
n, m = [1,2,3,4,5,6,7], 3
x = random.sample( list(itertools.permutations(n,2)), m)
print(x) #e.g. [(1,2),(3,4),(5,6)]
I'm searching for a list with the keys of the minimum values within a dict.
The goal is to get the keys of the minimum value for each i.
#Dict:
p[i,j] = {(0,0):10, (0,1):11, (0,2):12, (1,0):20, (1,1):21, (1,2):22, (2,0):30, (2,1):31, (2,2):32}
#The list should look like this:
list = [(0,0), (1,0), (2,0)]
Can i do it with a loop?
for i in range(3):
...
Or how can i generate this list?
Sorry, I don't know exactly what you mean. Maybe you want to get the key of P which is the minimum value and this key belongs to the list. You can try the code below.
p = {(0,0):10, (0,1):11, (0,2):12,
(1,0):20, (1,1):21, (1,2):22,
(2,0):30, (2,1):31, (2,2):32}
list = [(0,0), (1,1), (2,2)]
listRes = {}
# get listRes 10:(0,0) 21:(1,1) 32:(2,2)
for j in range(len(list)):
listRes[p[list[j]]] = list[j]
# (10, (0, 0))
print(min(listRes.items(), key=lambda x: x[0]))
OK, maybe I see. See the code below.
import sys
p = {(0,0):10, (0,1):11, (0,2):12,
(1,0):20, (1,1):21, (1,2):22,
(2,0):30, (2,1):31, (2,2):32}
res = []
tuple1 = ()
for i in range(3):
minValue = sys.maxsize
for j in range(3):
if minValue>p[(i,j)]:
tuple1 = (i,j)
minValue = p[(i,j)]
res.append(tuple1)
print(res)
I am trying to find conditional mutual information between three discrete random variable using pyitlib package for python with the help of the formula:
I(X;Y|Z)=H(X|Z)+H(Y|Z)-H(X,Y|Z)
The expected Conditional Mutual information value is= 0.011
My 1st code:
import numpy as np
from pyitlib import discrete_random_variable as drv
X=[0,1,1,0,1,0,1,0,0,1,0,0]
Y=[0,1,1,0,0,0,1,0,0,1,1,0]
Z=[1,0,0,1,1,0,0,1,1,0,0,1]
a=drv.entropy_conditional(X,Z)
##print(a)
b=drv.entropy_conditional(Y,Z)
##print(b)
c=drv.entropy_conditional(X,Y,Z)
##print(c)
p=a+b-c
print(p)
The answer i am getting here is=0.4632245116328402
My 2nd code:
import numpy as np
from pyitlib import discrete_random_variable as drv
X=[0,1,1,0,1,0,1,0,0,1,0,0]
Y=[0,1,1,0,0,0,1,0,0,1,1,0]
Z=[1,0,0,1,1,0,0,1,1,0,0,1]
a=drv.information_mutual_conditional(X,Y,Z)
print(a)
The answer i am getting here is=0.1583445441575102
While the expected result is=0.011
Can anybody help? I am in big trouble right now. Any kind of help will be appreciable.
Thanks in advance.
I think that the library function entropy_conditional(x,y,z) has some errors. I also test my samples, the same problem happens.
however, the function entropy_conditional with two variables is ok.
So I code my entropy_conditional(x,y,z) as entropy(x,y,z), the results is correct.
the code may be not beautiful.
def gen_dict(x):
dict_z = {}
for key in x:
dict_z[key] = dict_z.get(key, 0) + 1
return dict_z
def entropy(x,y,z):
x = np.array([x,y,z]).T
x = x[x[:,-1].argsort()] # sorted by the last column
w = x[:,-3]
y = x[:,-2]
z = x[:,-1]
# dict_w = gen_dict(w)
# dict_y = gen_dict(y)
dict_z = gen_dict(z)
list_z = [dict_z[i] for i in set(z)]
p_z = np.array(list_z)/sum(list_z)
pos = 0
ent = 0
for i in range(len(list_z)):
w = x[pos:pos+list_z[i],-3]
y = x[pos:pos+list_z[i],-2]
z = x[pos:pos+list_z[i],-1]
pos += list_z[i]
list_wy = np.zeros((len(set(w)),len(set(y))), dtype = float , order ="C")
list_w = list(set(w))
list_y = list(set(y))
for j in range(len(w)):
pos_w = list_w.index(w[j])
pos_y = list_y.index(y[j])
list_wy[pos_w,pos_y] += 1
#print(pos_w)
#print(pos_y)
list_p = list_wy.flatten()
list_p = np.array([k for k in list_p if k>0]/sum(list_p))
ent_t = 0
for j in list_p:
ent_t += -j * math.log2(j)
#print(ent_t)
ent += p_z[i]* ent_t
return ent
X=[0,1,1,0,1,0,1,0,0,1,0,0]
Y=[0,1,1,0,0,0,1,0,0,1,1,0]
Z=[1,0,0,1,1,0,0,1,1,0,0,1]
a=drv.entropy_conditional(X,Z)
##print(a)
b=drv.entropy_conditional(Y,Z)
c = entropy(X, Y, Z)
p=a+b-c
print(p)
0.15834454415751043
Based on the definitions of conditional entropy, calculating in bits (i.e. base 2) I obtain H(X|Z)=0.784159, H(Y|Z)=0.325011, H(X,Y|Z) = 0.950826. Based on the definition of conditional mutual information you provide above, I obtain I(X;Y|Z)=H(X|Z)+H(Y|Z)-H(X,Y|Z)= 0.158344. Noting that pyitlib uses base 2 by default, drv.information_mutual_conditional(X,Y,Z) appears to be computing the correct result.
Note that your use of drv.entropy_conditional(X,Y,Z) in your first example to compute conditional entropy is incorrect, you can however use drv.entropy_conditional(XY,Z), where XY is a 1D array representing the joint observations about X and Y, for example XY = [2*xy[0] + xy[1] for xy in zip(X,Y)].