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
def most_common(lst):
if max([lst.count(i)for i in lst]) == 1:
return False
else:
return max(set(lst), key=lst.count)
其他回答
这是一个很简单的解，时间复杂度是线性的
L =['鹅'，'鸭'，'鸭']
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
打印(most_common (L)
“鸭子”
number是列表中重复次数最多的元素吗
以Luiz的回答为基础，但满足“在抽取索引最低的项目时应返回”的条件:
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
例子:
>>> most_common(['a', 'b', 'b'])
'b'
>>> most_common([1, 2])
1
>>> most_common([])
StatisticsError: no mode for empty data
>>> 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'
这是O(n)解。
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用于确保它返回最低的索引项)
一行程序:
def most_common (lst):
return max(((item, lst.count(item)) for item in set(lst)), key=lambda a: a[1])[0]