查找列表中最常见的元素
- 2024-12-26 08:43:00
- admin 原创
- 120
问题描述:
在 Python 列表中查找最常见元素的有效方法是什么?
我的列表项可能不可哈希,因此无法使用字典。此外,在绘制时应返回索引最低的项目。示例:
>>> most_common(['duck', 'duck', 'goose'])
'duck'
>>> most_common(['goose', 'duck', 'duck', 'goose'])
'goose'
解决方案 1:
更简单的一行:
def most_common(lst):
return max(set(lst), key=lst.count)
解决方案 2:
借用这里,它可以与 Python 2.7 一起使用:
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 提出的单行代码快 50 倍。
在 CPython 3.6+(任何 Python 3.7+)上,如果出现平局,上述操作将选择第一个看到的元素。如果您在较旧的 Python 上运行,则要在出现平局的情况下检索列表中首先出现的元素,您需要执行两次传递以保持顺序:
# Only needed pre-3.6!
def most_common(lst):
data = Counter(lst)
return max(lst, key=data.get)
解决方案 3:
有这么多的解决方案被提出,我很惊讶没有人提出我认为显而易见的解决方案(对于不可哈希但可比较的元素)—— itertools.groupby
。 itertools
提供快速、可重用的功能,并允许您将一些棘手的逻辑委托给经过充分测试的标准库组件。例如:
import itertools
import operator
def most_common(L):
# get an iterable of (item, iterable) pairs
SL = sorted((x, i) for i, x in enumerate(L))
# print 'SL:', SL
groups = itertools.groupby(SL, key=operator.itemgetter(0))
# auxiliary function to get "quality" for an item
def _auxfun(g):
item, iterable = g
count = 0
min_index = len(L)
for _, where in iterable:
count += 1
min_index = min(min_index, where)
# print 'item %r, count %r, minind %r' % (item, count, min_index)
return count, -min_index
# pick the highest-count/earliest item
return max(groups, key=_auxfun)[0]
当然,这可以写得更简洁,但我的目标是尽可能清晰。print
可以取消注释这两个语句,以便更好地查看机器的运行情况;例如,取消注释 prints:
print most_common(['goose', 'duck', 'duck', 'goose'])
发出:
SL: [('duck', 1), ('duck', 2), ('goose', 0), ('goose', 3)]
item 'duck', count 2, minind 1
item 'goose', count 2, minind 0
goose
如您所见,SL
是一个成对的列表,每对一个项目后跟原始列表中该项目的索引(以实现关键条件,即如果具有相同最高计数的“最常见”项目> 1,则结果必须是最早出现的项目)。
groupby
仅按项目分组(通过operator.itemgetter
)。辅助函数在max
计算过程中每次分组调用一次,接收并在内部解包一个组 - 一个包含两个项目的元组(item, iterable)
,其中可迭代对象的项目也是两个项目的元组,(item, original index)
[[的项目SL
]]。
然后,辅助函数使用循环来确定组的可迭代中的条目数和最小原始索引;它将它们作为组合的“质量键”返回,并将最小索引符号更改,因此操作max
将“更好地”考虑原始列表中较早出现的那些项目。
如果这段代码不那么担心时间和空间中的大 O 问题,它会简单得多,例如……:
def most_common(L):
groups = itertools.groupby(sorted(L))
def _auxfun((item, iterable)):
return len(list(iterable)), -L.index(item)
return max(groups, key=_auxfun)[0]
基本思想相同,只是表达得更简单、更紧凑……但是,唉,额外的 O(N) 辅助空间(用于将组的可迭代项体现到列表中)和 O(N 平方) 时间(用于获取L.index
每个项目的)。虽然过早优化是编程中所有罪恶的根源,但在 O(N log N) 方法可用时故意选择 O(N 平方) 方法,这太违背可扩展性了!-)
最后,对于那些更喜欢“单行代码”而不是清晰度和性能的人来说,还有一个带有适当混乱名称的额外单行版本:-)。
from itertools import groupby as g
def most_common_oneliner(L):
return max(g(sorted(L)), key=lambda(x, v):(len(list(v)),-L.index(x)))[0]
解决方案 4:
您想要的在统计学中称为模式,Python 当然有一个内置函数可以为您完成这一点:
>>> from statistics import mode
>>> mode([1, 2, 2, 3, 3, 3, 3, 3, 4, 5, 6, 6, 6])
3
请注意,如果没有“最常见的元素”,例如前两个元素并列的情况,则在 Python <=3.7 中会引发此问题StatisticsError
,而在 3.8 及更高版本中,它将返回遇到的第一个元素。
解决方案 5:
如果没有关于最低索引的要求,您可以使用collections.Counter
:
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'
解决方案 6:
如果它们不可散列,您可以对它们进行排序,然后对结果进行一次循环,计算项目数(相同的项目将彼此相邻)。但使它们可散列并使用字典可能会更快。
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
解决方案 7:
这是一个 O(n) 解决方案。
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
(使用反转来确保返回最低索引项)
解决方案 8:
单行:
def most_common (lst):
return max(((item, lst.count(item)) for item in set(lst)), key=lambda a: a[1])[0]
解决方案 9:
对列表副本进行排序并找出最长的运行。您可以在使用每个元素的索引对列表进行排序之前对其进行修饰,然后在出现平局的情况下选择以最低索引开始的运行。
解决方案 10:
我使用 scipy stat 模块和 lambda 来执行此操作:
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
解决方案 11:
# 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'
解决方案 12:
基于Luiz 的回答,但满足“在抽取的情况下,应返回索引最低的项目”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
解决方案 13:
简单的一行解决方案
moc= max([(lst.count(chr),chr) for chr in set(lst)])
它将返回出现频率最高的元素及其频率。
解决方案 14:
你可能不再需要这个了,但这是我为类似问题所做的。(由于有评论,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
解决方案 15:
ans = [1, 1, 0, 0, 1, 1]
all_ans = {ans.count(ans[i]): ans[i] for i in range(len(ans))}
print(all_ans)
all_ans={4: 1, 2: 0}
max_key = max(all_ans.keys())
4
print(all_ans[max_key])
1
解决方案 16:
这里:
def most_common(l):
max = 0
maxitem = None
for x in set(l):
count = l.count(x)
if count > max:
max = count
maxitem = x
return maxitem
我隐约觉得标准库中某处有一种方法可以给出每个元素的数量,但我找不到它。
解决方案 17:
如果排序和散列都不可行,但==
可以进行相等性比较(),则这显然是一个缓慢的解决方案(O(n^2)):
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,如果列表的长度 (n) 很大,则使项目可散列或可排序 (如其他答案所建议) 几乎总是可以更快地找到最常见的元素。散列平均为 O(n),排序最差为 O(n*log(n))。
解决方案 18:
>>> 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'
解决方案 19:
我需要在最近的程序中执行此操作。我承认,我无法理解 Alex 的答案,所以这就是我最终的结果。
def mostPopular(l):
mpEl=None
mpIndex=0
mpCount=0
curEl=None
curCount=0
for i, el in sorted(enumerate(l), key=lambda x: (x[1], x[0]), reverse=True):
curCount=curCount+1 if el==curEl else 1
curEl=el
if curCount>mpCount \n or (curCount==mpCount and i<mpIndex):
mpEl=curEl
mpIndex=i
mpCount=curCount
return mpEl, mpCount, mpIndex
我将其与 Alex 的解决方案进行了计时,对于短列表,它大约快 10-15%,但是一旦超过 100 个元素或更多(测试多达 200000 个),它就会慢 20% 左右。
解决方案 20:
#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]
解决方案 21:
def most_frequent(List):
counter = 0
num = List[0]
for i in List:
curr_frequency = List.count(i)
if(curr_frequency> counter):
counter = curr_frequency
num = i
return num
List = [2, 1, 2, 2, 1, 3]
print(most_frequent(List))
解决方案 22:
嗨,这是一个非常简单的解决方案,具有线性时间复杂度
L = ['鹅', '鸭子', '鸭子']
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”
其中 number 是列表中重复次数最多的元素
解决方案 23:
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")
解决方案 24:
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
mostCommonElement(["a","b","a","c"]) -> "a"
解决方案 25:
N/2
最常见的元素应该是在数组中出现次数超过的元素,其中N
是len(array)
。下面的技术将在O(n)
时间复杂度内完成此操作,仅消耗O(1)
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
解决方案 26:
def most_common(lst):
if max([lst.count(i)for i in lst]) == 1:
return False
else:
return max(set(lst), key=lst.count)
解决方案 27:
def popular(L):
C={}
for a in L:
C[a]=L.count(a)
for b in C.keys():
if C[b]==max(C.values()):
return b
L=[2,3,5,3,6,3,6,3,6,3,7,467,4,7,4]
print popular(L)