calculate the mean of every nth list in a list of lists using a loop

Question

I am trying to calculate the mean of every nth list in a list of lists using a loop. I have been able to do so without a loop, but this will prove laborious to do when the list of lists gets longer.

Im struggling to explain this so heres the code

import numpy as np
import matplotlib.pyplot as plt 


list = []
t_r = np.arange(0,8)
i = 0
a =[[0.98,1.93,2.99,4.01,4.92,6.00,7.08,7.67, 8.00],[0,0,0,1,2,3,3,2,3],[0.93,1.99,2.99,3.91,4.82,6.03,7.01,8.00],[0,1,2,3,4,5,5,6],[0.88,2.09,3.01,4.11,5.65,7.12,8.00],[4,5,6,7,8,7,6]]
#########################
for t in t_r:
    b1 = np.array(a[0]) <= t
    b2 = np.array(a[2]) <= t 
    b3 = np.array(a[4]) <= t 

    ind1 = [(np.count_nonzero(b1))]
    ind2 = [(np.count_nonzero(b2))]
    ind3 = [(np.count_nonzero(b3))]

    x1 = np.array(a[1])
    x_mean1 = x1[ind1]
    x2 = np.array(a[3])
    x_mean2 = x2[ind2]
    x3 = np.array(a[5])
    x_mean3 = x3[ind2]

    x_mean_list = [x_mean1, x_mean2, x_mean3]
    x_average = np.mean(x_mean_list)    
    list.append(x_average)
#########################



no_of_sim = 3
counter = 0
while counter <= ((no_of_sim*2)-1):
    plt.plot(a[counter],a[counter+1], lw = 0.5)
    plt.plot(list, color = "black")
    plt.plot(x_average)
    plt.xlabel('Time (s)')
    plt.ylabel('copy no.')
    counter += 2
plt.show()

The bit inbetween the hashtags is the bit I'm trying to write a loop for so I don't have to manually change it when the list if lists gets much longer

Does this answer your question? [Pythonic way to return list of every nth item in a larger list](https://stackoverflow.com/questions/1403674/pythonic-way-to-return-list-of-every-nth-item-in-a-larger-list) — Georgy, May 28 '20 at 13:58
Can you try to simplify your question? You can use a simpler example of what you want to do. — sam, May 28 '20 at 14:16
Yah sure (the bits in bewteen hashtags is what im sturggling with): as the timepoints are not the same between lists (i.e. the [0], [2], [4] list in the lists of lists) , the bp and ind bit basically makes the first time point the same by marking the first element in that time list as true. It then takes the first element of the value list (i.e. the [1], [3], [5] list in the list of lists) and averages them, and appends the average appended to "list". It loops through a for the 2nd, 3rd etc element of each list — Danyn Patel, May 28 '20 at 15:34
I just want it to loop through [0] to [2] etc wihtout having to type it out each time — Danyn Patel, May 28 '20 at 15:35

score 1 · Answer 1 · answered May 28 '20 at 14:42

Your code between the hashes are equivalent to the following:

data = [np.array(x) for x in a[::2]]
idx = [np.array(x) for x in a[1::2]]

lst = [np.mean([x[(d<=t).sum()] for x,d in zip(idx,data)]) for t in t_r ]

Output:

[1.3333333333333333,
 2.0,
 2.3333333333333335,
 3.3333333333333335,
 4.0,
 5.333333333333333,
 5.0,
 5.0]

score 0 · Answer 2 · answered May 28 '20 at 14:00

check out this code:

a =[[0.98,1.93,2.99,4.01,4.92,6.00,7.08,7.67, 8.00],[0,0,0,1,2,3,3,2,3],[0.93,1.99,2.99,3.91,4.82,6.03,7.01,8.00],[0,1,2,3,4,5,5,6],[0.88,2.09,3.01,4.11,5.65,7.12,8.00],[4,5,6,7,8,7,6]]

i = 0
for li in a:
    print("Mean of list at index", i, "is:", sum(li)/len(li))
    i += 1

calculate the mean of every nth list in a list of lists using a loop

2 Answers2