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
#This will return the list sorted by frequency:
def orderByFrequency(list):
listUniqueValues = np.unique(list)
listQty = []
listOrderedByFrequency = []
for i in range(len(listUniqueValues)):
listQty.append(list.count(listUniqueValues[i]))
for i in range(len(listQty)):
index_bigger = np.argmax(listQty)
for j in range(listQty[index_bigger]):
listOrderedByFrequency.append(listUniqueValues[index_bigger])
listQty[index_bigger] = -1
return listOrderedByFrequency
#And this will return a list with the most frequent values in a list:
def getMostFrequentValues(list):
if (len(list) <= 1):
return list
list_most_frequent = []
list_ordered_by_frequency = orderByFrequency(list)
list_most_frequent.append(list_ordered_by_frequency[0])
frequency = list_ordered_by_frequency.count(list_ordered_by_frequency[0])
index = 0
while(index < len(list_ordered_by_frequency)):
index = index + frequency
if(index < len(list_ordered_by_frequency)):
testValue = list_ordered_by_frequency[index]
testValueFrequency = list_ordered_by_frequency.count(testValue)
if (testValueFrequency == frequency):
list_most_frequent.append(testValue)
else:
break
return list_most_frequent
#tests:
print(getMostFrequentValues([]))
print(getMostFrequentValues([1]))
print(getMostFrequentValues([1,1]))
print(getMostFrequentValues([2,1]))
print(getMostFrequentValues([2,2,1]))
print(getMostFrequentValues([1,2,1,2]))
print(getMostFrequentValues([1,2,1,2,2]))
print(getMostFrequentValues([3,2,3,5,6,3,2,2]))
print(getMostFrequentValues([1,2,2,60,50,3,3,50,3,4,50,4,4,60,60]))
Results:
[]
[1]
[1]
[1, 2]
[2]
[1, 2]
[2]
[2, 3]
[3, 4, 50, 60]
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)
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
简单的一行代码:
def most_common(lst):
return max(set(lst), key=lst.count)
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")
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
