给定一个项目列表，回想一下，列表的众数是出现最频繁的项目。
我想知道如何创建一个函数，可以找到列表的众数，但如果列表没有众数（例如，列表中的所有项目仅出现一次），则显示消息。我想自己从头开始制作这个函数，而不引入任何函数。
27个回答
0
def mode(data):
lst =[]
hgh=0
for i in range(len(data)):
lst.append(data.count(data[i]))
m= max(lst)
ml = [x for x in data if data.count(x)==m ] #to find most frequent values
mode = []
for x in ml: #to remove duplicates of mode
if x not in mode:
mode.append(x)
return mode
print mode([1,2,2,2,2,7,7,5,5,5,5])
- Venkata Prasanth T
0
#function to find mode
def mode(data):
modecnt=0
#for count of number appearing
for i in range(len(data)):
icount=data.count(data[i])
#for storing count of each number in list will be stored
if icount>modecnt:
#the loop activates if current count if greater than the previous count
mode=data[i]
#here the mode of number is stored
modecnt=icount
#count of the appearance of number is stored
return mode
print mode(data1)
- user9283073
1
你应该在代码中添加注释或更多细节来解释你的答案。 - Michael
0
这里有一个简单的函数，用于获取列表中出现的第一个众数。它创建了一个字典，以列表元素作为键和出现次数，并读取字典值以获取众数。
def findMode(readList):
numCount={}
highestNum=0
for i in readList:
if i in numCount.keys(): numCount[i] += 1
else: numCount[i] = 1
for i in numCount.keys():
if numCount[i] > highestNum:
highestNum=numCount[i]
mode=i
if highestNum != 1: print(mode)
elif highestNum == 1: print("All elements of list appear once.")
- SMS von der Tann
0
如果您想要一种清晰的方法，适用于课堂教学，并且仅使用列表和字典理解，您可以这样做：
def mode(my_list):
# Form a new list with the unique elements
unique_list = sorted(list(set(my_list)))
# Create a comprehensive dictionary with the uniques and their count
appearance = {a:my_list.count(a) for a in unique_list}
# Calculate max number of appearances
max_app = max(appearance.values())
# Return the elements of the dictionary that appear that # of times
return {k: v for k, v in appearance.items() if v == max_app}
- María Frances Gaska
0
或许可以尝试以下方法。它的时间复杂度为O(n)，返回一个浮点数（或整数）列表。它经过了全面自动化测试。它使用了collections.defaultdict，但我认为您不会反对使用它。您也可以在https://stromberg.dnsalias.org/~strombrg/stddev.html找到它。
def compute_mode(list_: typing.List[float]) -> typing.List[float]:
"""
Compute the mode of list_.
Note that the return value is a list, because sometimes there is a tie for "most common value".
See https://dev59.com/jWgv5IYBdhLWcg3wKtvj
"""
if not list_:
raise ValueError('Empty list')
if len(list_) == 1:
raise ValueError('Single-element list')
value_to_count_dict: typing.DefaultDict[float, int] = collections.defaultdict(int)
for element in list_:
value_to_count_dict[element] += 1
count_to_values_dict = collections.defaultdict(list)
for value, count in value_to_count_dict.items():
count_to_values_dict[count].append(value)
counts = list(count_to_values_dict)
if len(counts) == 1:
raise ValueError('All elements in list are the same')
maximum_occurrence_count = max(counts)
if maximum_occurrence_count == 1:
raise ValueError('No element occurs more than once')
minimum_occurrence_count = min(counts)
if maximum_occurrence_count <= minimum_occurrence_count:
raise ValueError('Maximum count not greater than minimum count')
return count_to_values_dict[maximum_occurrence_count]
- dstromberg
0
def mode(inp_list):
sort_list = sorted(inp_list)
dict1 = {}
for i in sort_list:
count = sort_list.count(i)
if i not in dict1.keys():
dict1[i] = count
maximum = 0 #no. of occurences
max_key = -1 #element having the most occurences
for key in dict1:
if(dict1[key]>maximum):
maximum = dict1[key]
max_key = key
elif(dict1[key]==maximum):
if(key<max_key):
maximum = dict1[key]
max_key = key
return max_key
- akshaynagpal
0
import numpy as np
def get_mode(xs):
values, counts = np.unique(xs, return_counts=True)
max_count_index = np.argmax(counts) #return the index with max value counts
return values[max_count_index]
print(get_mode([1,7,2,5,3,3,8,3,2]))
- user1464878
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assert(mode[1, 1, 1]) == None和assert(mode[1, 2, 3, 4]) == None。一个数字要成为“众数”，它必须在列表中出现的次数比至少另一个数字多，并且它不能是列表中唯一的数字。 - lifebalance