CODEWITHSUNDEEP
pythonstatistics
mgcy
mgcy

Reputation: 161

No unique mode; found 2 equally common values

I'm using statistics.mode([1, 1, 2, 2, 3]) to find the mode, but I get:

no unique mode; found 2 equally common values

When more than one mode is found, how can I output either 1 or 2?

Upvotes: 12

Views: 32310

Answers (8)

alquem
alquem

Reputation: 1

def MultiModeCalc(data):
    """
    set(data) -> discards duplicated values

    list(map(lambda x: data.count(x), set(data))))) -> counting how many of 
    each value there are

    dict(zip(set(data), list(map(lambda x: data.count(x), set(data))))) -> 
    making a dictionary by zipping all unique values(as keys) with their 
    frequency in the data(as values)

    return [i for i in multimode if multimode[i] == max(multimode.values())] 
    -> returning a list of the values that are most frequent in the data 
    """
    multimode = dict(zip(set(data), list(map(lambda x: data.count(x), set(data)))))
    return [i for i in multimode if multimode[i] == max(multimode.values())]
dataset = [1,2,5,6,3,7,2,3,9,1,10,3,1,2,9,9]
print(MultiModeCalc(dataset))
# [1, 2, 3, 9]

A DRY answer.

Upvotes: 0

Yury Tello
Yury Tello

Reputation: 51

Try this to find the max values as mode when no unique mode:

max([p[0] for p in statistics._counts([1, 1, 2, 2, 3])])

Upvotes: 5

Xavier Guihot
Xavier Guihot

Reputation: 61656

Note that in Python 3.8 the behaviour of statistics.mode has changed:

Changed in version 3.8: Now handles multimodal datasets by returning the first mode encountered. Formerly, it raised StatisticsError when more than one mode was found.

In your example:

from statistics import mode

mode([1, 1, 2, 2, 3])
# 1

Also starting in Python 3.8, you can alternatively use statistics.multimode to return the list of the most frequently occurring values in the order they were first encountered:

from statistics import multimode

multimode([1, 1, 2, 2, 3])
# [1, 2]

Upvotes: 11

Sai Akshay Erupaka
Sai Akshay Erupaka

Reputation: 93

from scipy import stats as s
a=[1,1,2,2,3]
print(int(s.mode(a)[0]))

Upvotes: 6

omargamal8
omargamal8

Reputation: 601

from collections import Counter
c = Counter([1,1,2,2,3])
c.most_common(1)
# output
>>> [(1,2)] #the key 1, 2 occurrences.

From the docs:

"most_common([n]): Returns a list of the n most common elements and their counts from the most common to the least. Elements with equal counts are ordered arbitrarily"

Upvotes: 7

Simon
Simon

Reputation: 1

I just faced the same issue. This how I solved it pretty simply:

def most_element(liste):
    numeral=[[liste.count(nb), nb] for nb in liste]
    numeral.sort(key=lambda x:x[0], reverse=True)
    return(numeral[0][1])

Not sure this the most elegant way but it does the job :). Hope it will help

Upvotes: -1

mgcy
mgcy

Reputation: 161

Try this function, which finds the max values as mode when no unique mode:

import statistics
def find_max_mode(list1):
    list_table = statistics._counts(list1)
    len_table = len(list_table)

    if len_table == 1:
        max_mode = statistics.mode(list1)
    else:
        new_list = []
        for i in range(len_table):
            new_list.append(list_table[i][0])
        max_mode = max(new_list) # use the max value here
    return max_mode

if __name__ == '__main__':
    a = [1,1,2,2,3]
    print(find_max_mode(a)) # print 2

Upvotes: 3

Jayhello
Jayhello

Reputation: 6592

For example:

lst = [1, 1, 2, 2, 3]

# test for count lst elements
lst_count = [[x, lst.count(x)] for x in set(lst)]
print lst_count
# [[1, 2], [2, 2], [3, 1]]

# remove count <= 1
lst_count = [x for x in set(lst) if lst.count(x) > 1]
print lst_count
# [1, 2]

# get 1 or 2 by index
print lst_count[0], lst_count[1]
# 1 2

Another way:

from collections import Counter

# change lst elements to str, for more readable
lst = ['a', 'a', 'b', 'b', 'c']

# get a dict, key is the elements in lst, value is count of the element
d_mem_count = Counter(lst)
print d_mem_count
# Counter({'a': 2, 'b': 2, 'c': 1})

for k in d_mem_count.keys():
    if d_mem_count[k] > 1:
        print k

# output as below
# a
# b

Upvotes: 0

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