求列表list1=[9,7,8,3,2,1,55,6]中元素的个数、 最大值、最小值,以及元素之和、平均值。
时间: 2023-06-27 13:01:57 浏览: 46
列表list1=[9,7,8,3,2,1,55,6]中元素的个数为8,最大值为55,最小值为1,元素之和为91,平均值为11.375。可以使用Python的内置函数和方法来进行计算:
```python
list1 = [9, 7, 8, 3, 2, 1, 55, 6]
# 计算元素个数
count = len(list1)
print("元素个数为:", count)
# 计算最大值
max_value = max(list1)
print("最大值为:", max_value)
# 计算最小值
min_value = min(list1)
print("最小值为:", min_value)
# 计算元素之和
sum_value = sum(list1)
print("元素之和为:", sum_value)
# 计算平均值
avg_value = sum_value / count
print("平均值为:", avg_value)
```
输出结果为:
```python
元素个数为: 8
最大值为: 55
最小值为: 1
元素之和为: 91
平均值为: 11.375
```
相关问题
编写程序,求列表 s=[9,8,7,3,2,1,55,6]中元素的个数,最大值、最小值、元素之和、平均值。请至少写出三种实现方法。提示:①for循环;②while循环;③list的函数
方法一:for循环
s = [9,8,7,3,2,1,55,6]
count = 0
total = 0
max_num = s[0]
min_num = s[0]
for num in s:
count += 1
total += num
if num > max_num:
max_num = num
if num < min_num:
min_num = num
average = total / count
print("元素个数:", count)
print("最大值:", max_num)
print("最小值:", min_num)
print("元素之和:", total)
print("平均值:", average)
方法二:while循环
s = [9,8,7,3,2,1,55,6]
count = 0
total = 0
max_num = s[0]
min_num = s[0]
i = 0
while i < len(s):
count += 1
total += s[i]
if s[i] > max_num:
max_num = s[i]
if s[i] < min_num:
min_num = s[i]
i += 1
average = total / count
print("元素个数:", count)
print("最大值:", max_num)
print("最小值:", min_num)
print("元素之和:", total)
print("平均值:", average)
方法三:list函数
s = [9,8,7,3,2,1,55,6]
count = len(s)
total = sum(s)
max_num = max(s)
min_num = min(s)
average = total / count
print("元素个数:", count)
print("最大值:", max_num)
print("最小值:", min_num)
print("元素之和:", total)
print("平均值:", average)
list=[1,2,3,4,5,6,7,8,9]把list中每个元素相乘
当然可以,将列表中的每个元素相乘,可以使用for循环和乘法操作符实现:
```python
list = [1, 2, 3, 4, 5, 6, 7, 8, 9]
product = 1
for num in list:
product *= num
print(product)
```
输出结果为362880,即1x2x3x4x5x6x7x8x9的结果。