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
如果它们是不可哈希的,您可以对它们进行排序,并对结果进行一次循环,以计数项(相同的项将彼此相邻)。但是使它们可哈希并使用字典可能更快。
def most_common(lst):
cur_length = 0
max_length = 0
cur_i = 0
max_i = 0
cur_item = None
max_item = None
for i, item in sorted(enumerate(lst), key=lambda x: x[1]):
if cur_item is None or cur_item != item:
if cur_length > max_length or (cur_length == max_length and cur_i < max_i):
max_length = cur_length
max_i = cur_i
max_item = cur_item
cur_length = 1
cur_i = i
cur_item = item
else:
cur_length += 1
if cur_length > max_length or (cur_length == max_length and cur_i < max_i):
return cur_item
return max_item
其他回答
def mostCommonElement(list):
count = {} // dict holder
max = 0 // keep track of the count by key
result = None // holder when count is greater than max
for i in list:
if i not in count:
count[i] = 1
else:
count[i] += 1
if count[i] > max:
max = count[i]
result = i
return result
主要组合(“a”、“b”、“a”、“c”)->“a”
如果它们是不可哈希的,您可以对它们进行排序,并对结果进行一次循环,以计数项(相同的项将彼此相邻)。但是使它们可哈希并使用字典可能更快。
def most_common(lst):
cur_length = 0
max_length = 0
cur_i = 0
max_i = 0
cur_item = None
max_item = None
for i, item in sorted(enumerate(lst), key=lambda x: x[1]):
if cur_item is None or cur_item != item:
if cur_length > max_length or (cur_length == max_length and cur_i < max_i):
max_length = cur_length
max_i = cur_i
max_item = cur_item
cur_length = 1
cur_i = i
cur_item = item
else:
cur_length += 1
if cur_length > max_length or (cur_length == max_length and cur_i < max_i):
return cur_item
return max_item
简单的一行代码:
def most_common(lst):
return max(set(lst), key=lst.count)
如果没有最低索引的要求,您可以使用集合。计数器:
from collections import Counter
a = [1936, 2401, 2916, 4761, 9216, 9216, 9604, 9801]
c = Counter(a)
print(c.most_common(1)) # the one most common element... 2 would mean the 2 most common
[(9216, 2)] # a set containing the element, and it's count in 'a'
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