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
简单的一行代码:
def most_common(lst):
return max(set(lst), key=lst.count)
其他回答
numbers = [1, 3, 7, 4, 3, 0, 3, 6, 3]
max_repeat_num = max(numbers, key=numbers.count) *# which number most* frequently
max_repeat = numbers.count(max_repeat_num) *#how many times*
print(f" the number {max_repeat_num} is repeated{max_repeat} times")
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
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。
最常见的元素应该是在数组中出现超过N/2次的元素,其中N是len(数组)。下面的技术将以O(n)个时间复杂度完成,只消耗O(1)个辅助空间。
from collections import Counter
def majorityElement(arr):
majority_elem = Counter(arr)
size = len(arr)
for key, val in majority_elem.items():
if val > size/2:
return key
return -1
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