[英]Find the most common element in a list
在 Python 列表中查找最常见元素的有效方法是什么?
我的列表项可能无法散列,因此无法使用字典。 此外,在绘制的情况下,应返回索引最低的项目。 例子:
>>> most_common(['duck', 'duck', 'goose'])
'duck'
>>> most_common(['goose', 'duck', 'duck', 'goose'])
'goose'
一个更简单的单行:
def most_common(lst):
return max(set(lst), key=lst.count)
从这里借用,这可以与 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]
工作速度比 Alex 的解决方案快 4-6 倍,比 newacct 提出的 one-liner 快 50 倍。
要在出现关系时检索列表中首先出现的元素:
def most_common(lst):
data = Counter(lst)
return max(lst, key=data.get)
提出了这么多解决方案,我很惊讶没有人提出我认为显而易见的解决方案(对于不可散列但可比较的元素)——[ itertools.groupby
][1]。 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良好测试的标准库组件。 考虑例如:
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,但我的目标是最大限度地清晰。 这两个print
语句可以取消注释,以便更好地查看运行中的机器; 例如,打印未注释:
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
是一个对的列表,每对一个项目后跟项目在原始列表中的索引(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)。
groupby
仅按项目分组(通过operator.itemgetter
)。 辅助函数,在max
计算期间每个分组调用一次,接收并在内部解包一个组 - 一个包含两个项目(item, iterable)
的元组,其中可迭代的项目也是两个项目的元组, (item, original index)
[[the SL
的项目]]。
然后辅助函数使用循环来确定组的可迭代项中的条目数和最小原始索引; 它将这些作为组合的“质量键”返回,最小索引符号已更改,因此max
操作将考虑“更好”原始列表中较早出现的那些项目。
如果这段代码不太担心时间和空间上的大 O 问题,它可能会简单得多,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]
相同的基本思想,只是表达得更简单和紧凑......但是,唉,额外的 O(N) 辅助空间(以体现组的可迭代列表)和 O(N 平方)时间(以获得L.index
每个项目)。 虽然过早的优化是编程中万恶之源,但当 O(N log N) 可用时故意选择 O(N 平方) 方法,这对可扩展性来说太过分了!-)
最后,对于那些更喜欢“oneliners”而不是清晰度和性能的人,还有一个额外的 1-liner 版本,带有适当的名称:-)。
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]
你想要的在统计学中被称为模式,Python当然有一个内置函数可以为你做这件事:
>>> 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”,例如前两个并列的情况,这将引发StatisticsError
,因为从统计学上讲,这种情况下没有模式。
如果没有最低索引的要求,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'
如果它们不可散列,您可以对它们进行排序并在结果上执行一个循环,计算项目(相同的项目将彼此相邻)。 但是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
这是一个 O(n) 的解决方案。
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
(reversed 用于确保它返回最低索引项)
对列表的副本进行排序并找到最长的运行。 您可以在使用每个元素的索引对其进行排序之前装饰列表,然后在平局的情况下选择从最低索引开始的运行。
单线:
def most_common (lst):
return max(((item, lst.count(item)) for item in set(lst)), key=lambda a: a[1])[0]
我正在使用 scipy stat 模块和 lambda 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
# 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'
简单的一条线解决方案
moc= max([(lst.count(chr),chr) for chr in set(lst)])
它将以其频率返回最频繁的元素。
您可能不再需要这个,但这是我为类似问题所做的。 (由于评论,它看起来比实际更长。)
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
以Luiz 的回答为基础,但满足“在绘制时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
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
#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]
如果排序和散列都不可行,但相等比较 ( ==
) 可用,则这是明显的慢速解决方案 (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,如果您的列表 (n) 的长度很大,则使您的项目可散列或可排序(如其他答案所建议的那样)几乎总是可以更快地找到最常见的元素。 散列平均为 O(n),排序最坏为 O(n*log(n))。
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
我有一种模糊的感觉,标准库中的某处有一种方法可以为您提供每个元素的计数,但我找不到它。
>>> 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'
我需要在最近的一个程序中这样做。 我承认,我无法理解亚历克斯的回答,所以这就是我最终得到的。
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 \
or (curCount==mpCount and i<mpIndex):
mpEl=curEl
mpIndex=i
mpCount=curCount
return mpEl, mpCount, mpIndex
我根据 Alex 的解决方案对其进行了计时,对于短列表,它的速度大约快了 10-15%,但是一旦你超过 100 个或更多元素(测试高达 200000),它就会慢大约 20%。
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','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(最常见的(L))
“鸭”
其中 number 是列表中大部分时间重复的元素
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(array)
。 下面的技术将在O(n)
时间复杂度内完成,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
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))
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"
def most_common(lst):
if max([lst.count(i)for i in lst]) == 1:
return False
else:
return max(set(lst), key=lst.count)
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)
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