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Hello I would like to know how to use a for loop to go through a list and insert an element after each other element in a new list.

I have looked at this link Insert element in Python list after every nth element

but when I tried that method it was giving me the exact same problem when implementing it in my code

"""
Created on Sat Mar 28 20:40:37 2020

@author: DeAngelo
"""

import math 
import matplotlib.pyplot as plt
import numpy as np
from numpy import sqrt
from quadpy import quad
import scipy.integrate as integrate
import scipy.special as special

experiment = [1,2,3,4,5]
count = len(experiment)

after = []
new = []
for i in range(0,count-1):
    average2 = (experiment[i] + experiment[i+1])/2
    new.append(average2)

print(experiment)

for i in range(0,count-1):
      just = experiment.insert(i+1,new[i])

print(new,'\n')      
print(experiment)

1st print(experiment) -> [1, 2, 3, 4, 5]

print(new,'\n') -> [1.5, 2.5, 3.5, 4.5] and 2nd, print(experiment) -> [1, 1.5, 2.5, 3.5, 4.5, 2, 3, 4, 5]

But I want it to be [1,1.5,2,2.5,3,3.5,4,4.5,5] I know I can use merge and sort, but I don't want to because I am working with a MUCH MUCH bigger list and it can't be sorted. This is just my baby list for right now.

I am really close I can feel it, it's like a sixth sense... Any help and guidance is much appreciated. Thanks a lot cutie

Jens
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Deangelo Wooley
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  • Take a look at the [`zip()`](https://docs.python.org/3/library/functions.html#zip) function. – Jens Mar 29 '20 at 06:35
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    Please reduce your code to the minimum to reproduce your problem - the imports are useless, for example. Also, explain clearly how you are supposed to build your output from your input. – Thierry Lathuille Mar 29 '20 at 06:37

5 Answers5

2

Inserting into the middle of a list a bunch of times is going to be very slow for a large dataset. It seems like you can just build a new list like you're doing:

first = [1, 2, 3, 4, 5]
second = []
for i in range(0, len(first) - 1):
    avg = (first[i] + first[i+1]) / 2
    second.append(first[i])
    second.append(avg)
second.append(first[-1])

print(second)

Which prints:

[1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5]
kingkupps
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  • I would just like to express my thanks to you. You really helped me. I was just curious what does the `second.append(first[-1])` do. I don't understand that but it works and that is all that matters. :) – Deangelo Wooley Mar 29 '20 at 17:21
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    No prob glad to help! The line `second.append(first[-1])` adds the last element of `first` to the end of `second`. Python indices can be negative numbers and when they are, they go from right to left (or back to front) so -1 is the last element -2 is the second to last element and so on. – kingkupps Mar 29 '20 at 20:33
1
experiment = [1, 2, 3, 4, 5]
count = len(experiment)

for i in range(count - 1):
    experiment.insert(2*i+1, (experiment[2*i] + experiment[2*i+1])/2)

print(experiment)
david
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1

As an attempt to generate the required output list in a single pass using list comprehension I offer this:

e = [1, 2, 3, 4]

r = [e[i] if j == 0 else sum(e[i:i+2])/2. for i in range(len(e)) for j in range(min(len(e) - i, 2))]

assert r == [1, 1.5, 2, 2.5, 3, 3.5, 4]

I use a nested for loop just to switch between passing the original element to output and calculating average of subsequent elements.

dlask
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0

For me it looks like task for roundrobin from itertools recipes that is:

from itertools import cycle, islice
def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    pending = len(iterables)
    nexts = cycle(iter(it).__next__ for it in iterables)
    while pending:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            pending -= 1
            nexts = cycle(islice(nexts, pending))

a = [1, 2, 3, 4, 5]
b = [10, 20, 30, 40]
ab = list(roundrobin(a,b))
print(ab)

Output:

[1, 10, 2, 20, 3, 30, 4, 40, 5]

Note that roundrobin might be used for interleaving more than 2 iterables and always terminates on longest one. As you are working with a MUCH MUCH bigger list it has additional benefit of creating generator rather than list.

Daweo
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0

you can use this one-liner list comprehension

from itertools import zip_longest

a = [1,2,3,4,5]
b = [1.5, 2.5, 3.5, 4.5]
flat_list = [val for sublist in zip_longest(a,b) for val in sublist if val]

output = [1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5]

ishmam
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