In Python, I have a list:
L = [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67]
I want to identify the item that occurred the highest number of times. I am able to solve it but I need the fastest way to do so. I know there is a nice Pythonic answer to this.
Answers:
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Method 1
I am surprised no-one has mentioned the simplest solution,max() with the key list.count:
max(lst,key=lst.count)
Example:
>>> lst = [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67] >>> max(lst,key=lst.count) 4
This works in Python 3 or 2, but note that it only returns the most frequent item and not also the frequency. Also, in the case of a draw (i.e. joint most frequent item) only a single item is returned.
Although the time complexity of using max() is worse than using Counter.most_common(1) as PM 2Ring comments, the approach benefits from a rapid C implementation and I find this approach is fastest for short lists but slower for larger ones (Python 3.6 timings shown in IPython 5.3):
In [1]: from collections import Counter ...: ...: def f1(lst): ...: return max(lst, key = lst.count) ...: ...: def f2(lst): ...: return Counter(lst).most_common(1) ...: ...: lst0 = [1,2,3,4,3] ...: lst1 = lst0[:] * 100 ...: In [2]: %timeit -n 10 f1(lst0) 10 loops, best of 3: 3.32 us per loop In [3]: %timeit -n 10 f2(lst0) 10 loops, best of 3: 26 us per loop In [4]: %timeit -n 10 f1(lst1) 10 loops, best of 3: 4.04 ms per loop In [5]: %timeit -n 10 f2(lst1) 10 loops, best of 3: 75.6 us per loop
Method 2
from collections import Counter most_common,num_most_common = Counter(L).most_common(1)[0] # 4, 6 times
For older Python versions (< 2.7), you can use this recipe to create the Counter class.
Method 3
In your question, you asked for the fastest way to do it. As has been demonstrated repeatedly, particularly with Python, intuition is not a reliable guide: you need to measure.
Here’s a simple test of several different implementations:
import sys
from collections import Counter, defaultdict
from itertools import groupby
from operator import itemgetter
from timeit import timeit
L = [1,2,45,55,5,4,4,4,4,4,4,5456,56,6,7,67]
def max_occurrences_1a(seq=L):
"dict iteritems"
c = dict()
for item in seq:
c[item] = c.get(item, 0) + 1
return max(c.iteritems(), key=itemgetter(1))
def max_occurrences_1b(seq=L):
"dict items"
c = dict()
for item in seq:
c[item] = c.get(item, 0) + 1
return max(c.items(), key=itemgetter(1))
def max_occurrences_2(seq=L):
"defaultdict iteritems"
c = defaultdict(int)
for item in seq:
c[item] += 1
return max(c.iteritems(), key=itemgetter(1))
def max_occurrences_3a(seq=L):
"sort groupby generator expression"
return max(((k, sum(1 for i in g)) for k, g in groupby(sorted(seq))), key=itemgetter(1))
def max_occurrences_3b(seq=L):
"sort groupby list comprehension"
return max([(k, sum(1 for i in g)) for k, g in groupby(sorted(seq))], key=itemgetter(1))
def max_occurrences_4(seq=L):
"counter"
return Counter(L).most_common(1)[0]
versions = [max_occurrences_1a, max_occurrences_1b, max_occurrences_2, max_occurrences_3a, max_occurrences_3b, max_occurrences_4]
print sys.version, "n"
for vers in versions:
print vers.__doc__, vers(), timeit(vers, number=20000)
The results on my machine:
2.7.2 (v2.7.2:8527427914a2, Jun 11 2011, 15:22:34) [GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] dict iteritems (4, 6) 0.202214956284 dict items (4, 6) 0.208412885666 defaultdict iteritems (4, 6) 0.221301078796 sort groupby generator expression (4, 6) 0.383440971375 sort groupby list comprehension (4, 6) 0.402786016464 counter (4, 6) 0.564319133759
So it appears that the Counter solution is not the fastest. And, in this case at least, groupby is faster. defaultdict is good but you pay a little bit for its convenience; it’s slightly faster to use a regular dict with a get.
What happens if the list is much bigger? Adding L *= 10000 to the test above and reducing the repeat count to 200:
dict iteritems (4, 60000) 10.3451900482 dict items (4, 60000) 10.2988479137 defaultdict iteritems (4, 60000) 5.52838587761 sort groupby generator expression (4, 60000) 11.9538850784 sort groupby list comprehension (4, 60000) 12.1327362061 counter (4, 60000) 14.7495789528
Now defaultdict is the clear winner. So perhaps the cost of the ‘get’ method and the loss of the inplace add adds up (an examination of the generated code is left as an exercise).
But with the modified test data, the number of unique item values did not change so presumably dict and defaultdict have an advantage there over the other implementations. So what happens if we use the bigger list but substantially increase the number of unique items? Replacing the initialization of L with:
LL = [1,2,45,55,5,4,4,4,4,4,4,5456,56,6,7,67]
L = []
for i in xrange(1,10001):
L.extend(l * i for l in LL)
dict iteritems (2520, 13) 17.9935798645
dict items (2520, 13) 21.8974409103
defaultdict iteritems (2520, 13) 16.8289561272
sort groupby generator expression (2520, 13) 33.853593111
sort groupby list comprehension (2520, 13) 36.1303369999
counter (2520, 13) 22.626899004
So now Counter is clearly faster than the groupby solutions but still slower than the iteritems versions of dict and defaultdict.
The point of these examples isn’t to produce an optimal solution. The point is that there often isn’t one optimal general solution. Plus there are other performance criteria. The memory requirements will differ substantially among the solutions and, as the size of the input goes up, memory requirements may become the overriding factor in algorithm selection.
Bottom line: it all depends and you need to measure.
Method 4
Here is a defaultdict solution that will work with Python versions 2.5 and above:
from collections import defaultdict
L = [1,2,45,55,5,4,4,4,4,4,4,5456,56,6,7,67]
d = defaultdict(int)
for i in L:
d[i] += 1
result = max(d.iteritems(), key=lambda x: x[1])
print result
# (4, 6)
# The number 4 occurs 6 times
Note if L = [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 7, 7, 7, 7, 7, 56, 6, 7, 67]
then there are six 4s and six 7s. However, the result will be (4, 6) i.e. six 4s.
Method 5
If you’re using Python 3.8 or above, you can use either statistics.mode() to return the first mode encountered or statistics.multimode() to return all the modes.
>>> import statistics >>> data = [1, 2, 2, 3, 3, 4] >>> statistics.mode(data) 2 >>> statistics.multimode(data) [2, 3]
If the list is empty, statistics.mode() throws a statistics.StatisticsError and statistics.multimode() returns an empty list.
Note before Python 3.8, statistics.mode() (introduced in 3.4) would additionally throw a statistics.StatisticsError if there is not exactly one most common value.
Method 6
Perhaps the most_common() method
Method 7
A simple way without any libraries or sets
def mcount(l):
n = [] #To store count of each elements
for x in l:
count = 0
for i in range(len(l)):
if x == l[i]:
count+=1
n.append(count)
a = max(n) #largest in counts list
for i in range(len(n)):
if n[i] == a:
return(l[i],a) #element,frequency
return #if something goes wrong
Method 8
I obtained the best results with groupby from itertools module with this function using Python 3.5.2:
from itertools import groupby
a = [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67]
def occurrence():
occurrence, num_times = 0, 0
for key, values in groupby(a, lambda x : x):
val = len(list(values))
if val >= occurrence:
occurrence, num_times = key, val
return occurrence, num_times
occurrence, num_times = occurrence()
print("%d occurred %d times which is the highest number of times" % (occurrence, num_times))
Output:
4 occurred 6 times which is the highest number of times
Test with timeit from timeit module.
I used this script for my test with number= 20000:
from itertools import groupby
def occurrence():
a = [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67]
occurrence, num_times = 0, 0
for key, values in groupby(a, lambda x : x):
val = len(list(values))
if val >= occurrence:
occurrence, num_times = key, val
return occurrence, num_times
if __name__ == '__main__':
from timeit import timeit
print(timeit("occurrence()", setup = "from __main__ import occurrence", number = 20000))
Output (The best one):
0.1893607140000313
Method 9
Simple and best code:
def max_occ(lst,x):
count=0
for i in lst:
if (i==x):
count=count+1
return count
lst=[1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67]
x=max(lst,key=lst.count)
print(x,"occurs ",max_occ(lst,x),"times")
Output: 4 occurs 6 times
Method 10
if you are using numpy in your solution for faster computation use this:
import numpy as np x = np.array([2,5,77,77,77,77,77,77,77,9,0,3,3,3,3,3]) y = np.bincount(x,minlength = max(x)) y = np.argmax(y) print(y) #outputs 77
Method 11
I want to throw in another solution that looks nice and is fast for short lists.
def mc(seq=L):
"max/count"
max_element = max(seq, key=seq.count)
return (max_element, seq.count(max_element))
You can benchmark this with the code provided by Ned Deily which will give you these results for the smallest test case:
3.5.2 (default, Nov 7 2016, 11:31:36) [GCC 6.2.1 20160830] dict iteritems (4, 6) 0.2069783889998289 dict items (4, 6) 0.20462976200065896 defaultdict iteritems (4, 6) 0.2095775119996688 sort groupby generator expression (4, 6) 0.4473949929997616 sort groupby list comprehension (4, 6) 0.4367636879997008 counter (4, 6) 0.3618192010007988 max/count (4, 6) 0.20328268999946886
But beware, it is inefficient and thus gets really slow for large lists!
Method 12
Following is the solution which I came up with if there are multiple characters in the string all having the highest frequency.
mystr = input("enter string: ")
#define dictionary to store characters and their frequencies
mydict = {}
#get the unique characters
unique_chars = sorted(set(mystr),key = mystr.index)
#store the characters and their respective frequencies in the dictionary
for c in unique_chars:
ctr = 0
for d in mystr:
if d != " " and d == c:
ctr = ctr + 1
mydict[c] = ctr
print(mydict)
#store the maximum frequency
max_freq = max(mydict.values())
print("the highest frequency of occurence: ",max_freq)
#print all characters with highest frequency
print("the characters are:")
for k,v in mydict.items():
if v == max_freq:
print(k)
Input: “hello people”
Output:
{'o': 2, 'p': 2, 'h': 1, ' ': 0, 'e': 3, 'l': 3}
the highest frequency of occurence: 3
the characters are:
e l
Method 13
My (simply) code (three months studying Python):
def more_frequent_item(lst):
new_lst = []
times = 0
for item in lst:
count_num = lst.count(item)
new_lst.append(count_num)
times = max(new_lst)
key = max(lst, key=lst.count)
print("In the list: ")
print(lst)
print("The most frequent item is " + str(key) + ". Appears " + str(times) + " times in this list.")
more_frequent_item([1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67])
The output will be:
In the list: [1, 2, 45, 55, 5, 4, 4, 4, 4, 4, 4, 5456, 56, 6, 7, 67] The most frequent item is 4. Appears 6 times in this list.
Method 14
may something like this:
testList = [1, 2, 3, 4, 2, 2, 1, 4, 4]
print(max(set(testList), key = testList.count))
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0