输入一个列表,计算并输出只出现一次的元素
时间: 2024-05-14 22:19:46 浏览: 10
好的,对于您的问题,我可以给出以下 Python 代码实现:
```python
def unique_elements(lst):
"""
计算列表中只出现过一次的元素
"""
result = []
for element in lst:
if lst.count(element) == 1:
result.append(element)
return result
```
您可以将要计算的列表作为函数的参数传入,函数会返回仅出现过一次的元素的列表。希望这个答案能够帮到您!
相关问题
输入一个整数列表,计算并输出只出现一次的元素
。好的,这是一个计算问题。可以这样实现:
def find_unique(nums):
counter = dict()
for num in nums:
counter[num] = counter.get(num, 0) + 1
unique_nums = [num for num, count in counter.items() if count == 1]
return unique_nums
输入一个整数列表:
[1, 2, 3, 2, 1, 4, 5, 3]
只出现一次的元素是:
[4, 5]
请问还有什么其它问题吗?
请输入一个列表,快速计算并输出列表中只出现一次的元素
Sure, here's a solution in Python:
```
from collections import Counter
def find_unique(lst):
# Count the occurrences of each element in the list
count = Counter(lst)
# Find the elements that only appear once
unique = [elem for elem, cnt in count.items() if cnt == 1]
# Return the list of unique elements
return unique
my_list = [1, 2, 3, 4, 4, 5, 6, 7, 8, 7, 9]
unique_elements = find_unique(my_list)
print(unique_elements)
```
This code uses the `Counter` class from the `collections` module to count the occurrences of each element in the list. Then it filters out the elements that appear more than once, leaving only the unique elements. Finally, it returns the list of unique elements, which can then be printed or used for further processing.