Question

我想在python中使用statistics.mode([1,1,2,2,3])函数来计算模式，它返回no unique mode; found 2 equally common values。这样，我想输出1或2。有人有什么好主意吗？

Answer 1

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

从文档中：

“ most_common（[n]）：返回n个最常见元素的列表，以及他们的人数从最常见到最少。相等的元素计数是任意排序的”

Answer 2

请注意，在Python 3.8中，statistics.mode的行为已更改：

在3.8版中进行了更改：现在通过返回遇到的第一个模式来处理多峰数据集。以前，当发现多个模式时会引发StatisticsError。

在您的示例中：

from statistics import mode

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

也可以从Python 3.8开始，也可以使用statistics.multimode以最先出现的值的顺序返回它们的列表：

from statistics import multimode

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

Answer 3

尝试以下功能，当没有唯一模式时，该最大值将作为模式找到：

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

Answer 4

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

Answer 5

当没有唯一模式时，试试这个以找到最大值作为模式：

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

Answer 6

例如：

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

另一种方式：

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

Answer 7

我只是遇到了同样的问题。这是我很简单地解决它的方法：

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])

不确定这是最优雅的方法，但它确实可以完成:)。希望对您有帮助

Answer 8

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]

一个枯燥的答案。

没有独特的模式；找到2个相同的共同值

8 个答案: