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'
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
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)
其他回答
如果排序和哈希都不可行,这是一个明显的缓慢的解决方案(O(n²)),但相等比较(==)可用:
def most_common(items):
if not items:
raise ValueError
fitems = []
best_idx = 0
for item in items:
item_missing = True
i = 0
for fitem in fitems:
if fitem[0] == item:
fitem[1] += 1
d = fitem[1] - fitems[best_idx][1]
if d > 0 or (d == 0 and fitems[best_idx][2] > fitem[2]):
best_idx = i
item_missing = False
break
i += 1
if item_missing:
fitems.append([item, 1, i])
return items[best_idx]
但是,如果你的列表(n)的长度很大,那么让你的项目可哈希或可排序(正如其他答案所建议的那样)几乎总是能更快地找到最常见的元素。哈希时平均为O(n),排序时最差为O(n*log(n))。
对列表的一个副本排序并找到运行时间最长的。您可以在用每个元素的索引对列表排序之前对其进行修饰,然后在并列的情况下选择从最低索引开始的运行。
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
# use Decorate, Sort, Undecorate to solve the problem
def most_common(iterable):
# Make a list with tuples: (item, index)
# The index will be used later to break ties for most common item.
lst = [(x, i) for i, x in enumerate(iterable)]
lst.sort()
# lst_final will also be a list of tuples: (count, index, item)
# Sorting on this list will find us the most common item, and the index
# will break ties so the one listed first wins. Count is negative so
# largest count will have lowest value and sort first.
lst_final = []
# Get an iterator for our new list...
itr = iter(lst)
# ...and pop the first tuple off. Setup current state vars for loop.
count = 1
tup = next(itr)
x_cur, i_cur = tup
# Loop over sorted list of tuples, counting occurrences of item.
for tup in itr:
# Same item again?
if x_cur == tup[0]:
# Yes, same item; increment count
count += 1
else:
# No, new item, so write previous current item to lst_final...
t = (-count, i_cur, x_cur)
lst_final.append(t)
# ...and reset current state vars for loop.
x_cur, i_cur = tup
count = 1
# Write final item after loop ends
t = (-count, i_cur, x_cur)
lst_final.append(t)
lst_final.sort()
answer = lst_final[0][2]
return answer
print most_common(['x', 'e', 'a', 'e', 'a', 'e', 'e']) # prints 'e'
print most_common(['goose', 'duck', 'duck', 'goose']) # prints 'goose'
def most_common(lst):
if max([lst.count(i)for i in lst]) == 1:
return False
else:
return max(set(lst), key=lst.count)