This article describes the python list List of averaging and median instance, has a good reference value, we want to help. Xiao Bian together to follow up to see it
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import numpy as np
a = [2,4,6,8,10]
average_a = np.mean(a)
median_a = np.median(a)
Knowledge supplement: python- find two lists median
Description Title:
given size m and n are two ordered arrays and nums1 nums2.
Please find both the median and orderly array, and requires time complexity of the algorithm is O (log (m + n)).
You can not assume nums1 and nums2 both empty.
Example. 1:
nums1 = [. 1,. 3]
nums2 = [2]
The median is 2.0
Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3) / 2 = 2.5
# nums1 = [1,3,3]
# nums2 = [2,5,4]
def list_model(nums1,nums2):
nums = nums1 + nums2
nums.sort()
print("您输入的两列表为 :",nums)
n = len(nums)
# print((nums[int(n/2-1)] + nums[int(n/2)])/2)
# print(n)
if n%2==0:
model = (nums[int(n/2-1)] + nums[int(n/2)])/2
else:
model = nums[int((n+1)/2 - 1)]
return(model)
x = input("请输入第一个列表 :")
nums1 = x.split(',')
nums1 = [float(nums1[i]) for i in range(len(nums1))]
y = input("请输入第二个列表 :")
nums2 = y.split(',')
nums2 = [float(nums2[i]) for i in range(len(nums2))]
print("您所求的两列表的中位数为 ：",list_model(nums1,nums2))
请输入第一个列表 :1,2.5,4
请输入第二个列表 :2,3,4
您输入的两列表为 : [1.0, 2.0, 2.5, 3.0, 4.0, 4.0]
您所求的两列表的中位数为 ： 2.75
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