Python函数来显示出现的最大数字,如果两者都有平局返回。num=[1,2,3,2,1,2,2,4,5,4,4,4]
我编写了这段Python代码来将键和值存储在字典中Python函数来显示出现的最大数字,如果两者都有平局返回。num=[1,2,3,2,1,2,2,4,5,4,4,4],python,arrays,list,dictionary,word-frequency,Python,Arrays,List,Dictionary,Word Frequency,我编写了这段Python代码来将键和值存储在字典中 num = [1,2,3,2,1,2,2,4,5,4,4,4] dict1 = {} for val in num: if val in dict1: dict1[val] = dict1[val] + 1 else: dict1[val] = 1; print(max(dict1,key=dict1.get)) Max函数返回它遇到的第一个值。请告知打领带时该怎么办 谢谢 我建议使用计数器 f
num = [1,2,3,2,1,2,2,4,5,4,4,4]
dict1 = {}
for val in num:
if val in dict1:
dict1[val] = dict1[val] + 1
else:
dict1[val] = 1;
print(max(dict1,key=dict1.get))
Max函数返回它遇到的第一个值。请告知打领带时该怎么办
谢谢 我建议使用
计数器
from collections import Counter
def frequency(myList: list):
counter = Counter(myList)
most_common = counter.most_common()
return dict(most_common)
我建议使用计数器
from collections import Counter
def frequency(myList: list):
counter = Counter(myList)
most_common = counter.most_common()
return dict(most_common)
您可以添加此行以获取列表中所有所需的值
num = [1,2,3,2,1,2,2,4,5,4,4,4]
dict1 = {}
for val in num:
if val in dict1:
dict1[val] = dict1[val] + 1
else:
dict1[val] = 1;
maxkey = max(dict1,key=dict1.get)
print([key for key in dict1 if dict1[key]==dict1[maxkey]])
输出:
[2, 4]
您可以添加此行以获取列表中所有所需的值
num = [1,2,3,2,1,2,2,4,5,4,4,4]
dict1 = {}
for val in num:
if val in dict1:
dict1[val] = dict1[val] + 1
else:
dict1[val] = 1;
maxkey = max(dict1,key=dict1.get)
print([key for key in dict1 if dict1[key]==dict1[maxkey]])
输出:
[2, 4]
我基本上是存储最大频率的值,并在字典上重复以打印值与最大频率匹配的所有键
我基本上是存储最大频率的值,并在字典上重复以打印所有值与最大频率匹配的键。在这里使用字典可能会过度拉伸。您可以在线性时间内实现此功能,而无需额外的模块
nums = [1,2,3,2,1,2,2,4,5,5,5,4,4]
def max_in_list(nums):
max_list = []
max_value = 0
for i in nums:
if i > max_value:
max_list = [i]
max_value = i
elif i == max_value:
max_list.append(i)
return max_list
assert max_in_list(nums) == [5,5,5]
print(max_in_list(nums))
在这里使用字典可能会有点过分。您可以在线性时间内实现此功能,而无需额外的模块
nums = [1,2,3,2,1,2,2,4,5,5,5,4,4]
def max_in_list(nums):
max_list = []
max_value = 0
for i in nums:
if i > max_value:
max_list = [i]
max_value = i
elif i == max_value:
max_list.append(i)
return max_list
assert max_in_list(nums) == [5,5,5]
print(max_in_list(nums))
我想我一定没有抓住重点,因为我看到了一些非常复杂的解决方案,在我看来,这是一个简单的问题
def maxes(nums):
return [m for m in nums if m == max(nums)]
>>> nums = [1,2,3,2,1,2,2,4,5,5,5,4,4]
>>> maxes(nums)
[5, 5, 5]
>>> nums = [1,2,3,2,1,2,2,4,5,5,5,4,4,6]
>>> maxes(nums)
[6]
我想我一定没有抓住重点,因为我看到了一些非常复杂的解决方案,在我看来,这是一个简单的问题
def maxes(nums):
return [m for m in nums if m == max(nums)]
>>> nums = [1,2,3,2,1,2,2,4,5,5,5,4,4]
>>> maxes(nums)
[5, 5, 5]
>>> nums = [1,2,3,2,1,2,2,4,5,5,5,4,4,6]
>>> maxes(nums)
[6]
max
仅返回一个值max
仅返回一个值