Array generation from a list in python

Question

I have a list of numbers L.

L = [233.0, 0.084, 4.308, 0.0, 2.6208547, 5.1109018, 54.694964, 0.0010984, 54.5804842, 43.473086, 2763.5231162, 28212.5694463, 0.1387176, 0.4591909, 233.0, 0.084, 4.308, 0.0, 2.6208547, 5.1109018, 54.694964, 0.0010984, 54.5804842, 43.473086, 2763.5231162, 28212.5694463, 0.1387176, 0.4591909, 233.0, 0.084, 4.308, 0.0, 2.6208547, 5.1109018, 54.694964, 0.0010984, 54.5804842, 43.473086, 2763.5231162, 28212.5694463, 0.1387176, 0.4591909, 233.0, 0.084, 4.308, 0.0, 2.6208547, 5.1109018, 54.694964, 0.0010984, 54.5804842, 43.473086, 2763.5231162, 28212.5694463, 0.1387176, 0.4591909]

This size of the list is based upon a certain variable, for example if n=4, the list will have 56 elements, if n = 5 it will have 70 elements and so on.

I want to generate an array 'a' of length = n. I want a[0] = L[1], a[1] = L[14], a[2]=L[28] and so on. I tried using a nested for loop but did not work out well. Any help would be highly appreciated.

Don't you mean a[0] = L[0]? Otherwise, you won't get the last element. — AKS
– AKS, Commented May 19, 2016 at 4:21
@AKS Yes, thats precisely what I was looking for. Stand corrected. — Aspro
– Aspro, Commented May 19, 2016 at 6:41

Joe T. Boka · Accepted Answer · 2016-05-19 04:46:59Z

4

You can slice L and create a new list new_list:

new_list = L[::14]
print (new_list)

output:

[233.0, 233.0, 233.0, 233.0]

answered May 19, 2016 at 4:46

Joe T. Boka

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Comments

Ricardo Silveira · Accepted Answer · 2016-05-19 04:51:34Z

1

What about:

indices_list = [i*n for i in xrange(n)]
a = [L[index] for index in indices_list]

Is that what you need?

answered May 19, 2016 at 4:51

Ricardo Silveira

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Radhamohan Parashar · Accepted Answer · 2016-05-19 04:55:58Z

1

#!/usr/bin/python
from array import *
count = 1
list = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28]
size_of_list = len(list)
my_array = array('i',[])
i = 0
    while (count < size_of_list):
        my_array.append(list[count])
        count = (i+1) *14
        i = i+1
    print my_array
    print "Good bye!"

answered May 19, 2016 at 4:55

Radhamohan Parashar

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Comments

Iron Fist · Accepted Answer · 2016-05-19 05:07:49Z

0

IMHO, easiest one is using slice notation, as mentioned by Joe R's answer. Other options are:

You could've also used the step parameter in range function:

>>> def f(lst, step):
        new_list = []
        for i in range(0,len(lst), step):
            new_list.append(lst[i]))
        return new_list

>>> f(L,14)
[233.0, 233.0, 233.0, 233.0]

And if you like List comprehensions, then :

>>> [L[i] for i in range(0,len(L), 14)]
[233.0, 233.0, 233.0, 233.0]

Alternative to that is using slice object:

>>> s = slice(0,len(L), 14)
>>> L[s]
[233.0, 233.0, 233.0, 233.0]

edited May 19, 2016 at 5:07

answered May 19, 2016 at 5:02

Iron Fist

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2 Comments

Aspro Over a year ago

This last solution works best since I do need to create other rarrays in the same systematic recursive form from the list. Thank you.

Iron Fist Over a year ago

@Arya .. No pblm...Glad to be of a help :)

konstov · Accepted Answer · 2016-05-19 05:13:04Z

0

If @AKS is right and you mean a[0] = L[0] answer is:

a = [elem for i, elem in enumerate(L) if i % 14 == 0]

edited May 19, 2016 at 5:13

answered May 19, 2016 at 4:53

konstov

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2 Comments

Iron Fist Over a year ago

NOP...That's not what OP wanted...read question carefully and update your answer with correct fix

konstov Over a year ago

@IronFist fix. Really boner.

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