找到Python列表中最常见元素的有效方法是什么?

我的列表项可能不是可哈希的,所以不能使用字典。 同样,在抽取的情况下,应返回索引最低的项。例子:

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

当前回答

你可能不再需要这个了,但这是我对一个类似问题所做的。(因为评论,它看起来比实际要长。)

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

其他回答

从这里借鉴,这可以在Python 2.7中使用:

from collections import Counter

def Most_Common(lst):
    data = Counter(lst)
    return data.most_common(1)[0][0]

比Alex的解决方案快4-6倍,比newacct提出的一行程序快50倍。

在CPython 3.6+(任何Python 3.7+)上,上面将选择第一个看到的元素。如果你在旧的Python上运行,为了检索列表中第一个出现的元素,你需要进行两次传递来保持顺序:

# Only needed pre-3.6!
def most_common(lst):
    data = Counter(lst)
    return max(lst, key=data.get)

你可能不再需要这个了,但这是我对一个类似问题所做的。(因为评论,它看起来比实际要长。)

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

我这样做使用scipy统计模块和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))

结果:

 most_freq_val = 5

简单的一行解决方案

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

它会返回频率最高的元素。

简单的一行代码:

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