python 中两个列表中重叠值的平均值

Question

我有 4 个列表，其中包含 UNIX 中的时间数据和相应的速度。一组比另一组大。我想在较小列表的每个匹配时间戳处找到较大列表的找到值前后几个值的平均速度。

t1 = [2, 5, 7]

t2 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

v1 = [0.5, 1, 0.7]

v2 = [0.1, 0.5, 1, 1.3, 1.4, 1.8, 0.9, 2, 1.5, 1.2]

所以如果 t1 和 t2 匹配，我想得到那个时间之前和之后的 x 值的平均速度。

假设 x = 1 在这种情况下，它应该给我 (0.1, 0.5, 1) (1.3, 1.4, 1.8) 和 (1.8, 0.9, 2)

的平均值

Answer 1

尝试

import numpy as np

for num in t1: #for each value in t1
    try:
        match_index=t2.index(num) #check if match exists
    except:
        match_index=-1 #otherwise set it to -1

    if(match_index!=-1): #if match was found
        my_list=[]       #make an empty list 
        if(match_index==0): #if index was 0, then nearby 3 elements can't be found, only 2 can
            my_list=v2[:match_index+2]
        elif(match_index==len(v2)-1): #similarly if index was for last element
            my_list=v2[match_index-1:]
        else:
            my_list=v2[match_index-1:match_index+2] #normal case slice the required values
        print("average of :",my_list," is ",np.average(my_list))

#average of : [0.1, 0.5, 1]  is  0.5333333333333333
#average of : [1.3, 1.4, 1.8]  is  1.5
#average of : [1.8, 0.9, 2]  is  1.5666666666666667

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