CODEWITHSUNDEEP
python
Denis Shvetsov
Denis Shvetsov

Reputation: 256

How do you find odd positive integers with python [no loops style]?

The problem I'm trying to solve is rather simple: write a function q2(n) that would return n odd positive integers. For example, q2(3) should return [1,3,5].

I have solved it, as you can see in the code below, using a for loop. However, there was a note to the problem that said 'try to not use loops'. So the question is: how do I do that without using loops?

Here's my version:

def q2(n):
    num100 = list(range(100))
    odd = []

    for i in num100:
        if (i%2) != 0:
            odd.append(i)
    result = odd[0:n]

    return result

Upvotes: 2

Views: 804

Answers (5)

Alain T.
Alain T.

Reputation: 42143

You can use the range function:

def q2(N): return list(range(1,N*2,2))

or recursion:

def q2(N,r=1): return [r] if N == 1 else [r] + q2(N-1,r+2)

or a list comprehension (but that may be considered a loop):

def q2(N): return [2*n+1 for n in range(N)]

Upvotes: 1

mohammed wazeem
mohammed wazeem

Reputation: 1328

You can also use the python built-in filter function.

>>> list(filter(lambda x: x%2 != 0, range(10*2)))
[1, 3, 5, 7, 9, 11, 13, 15, 17, 19]

Upvotes: 1

zabop
zabop

Reputation: 7852

You can use numpy's arange:

import numpy as np
def no_loop_odd_integers(n):
    return np.arange(1,2*n,2).tolist() #.tolist() in the end if want list not np array

Or without numpy, use the built-in range as suggested in the comments:

def no_loop_odd_integers2(n):
    return list(range(1,2*n,2))

Or, if you are into tensorflow, can use its range:

import tensorflow as tf
def no_loop_odd_integers3(n):
    # create a tensor, convert it to numpy array, then to list:
    return tf.range(1,2*n,2).numpy().tolist()

no_loop_odd_integers(4), no_loop_odd_integers2(4), no_loop_odd_integers3(4) all return:

[1, 3, 5, 7]

All of the above methods work in the same spirit: you specify from where you want to start your array (ie 1), then you say up to (not including) which number you want to be the upper limit (you want n odd numbers, so the nth odd number is 2n-1, so you say 2n, so all your numbers will be lower than this), then you specify what steps you want to make between these two limits. You want to make steps of 2 (1,3,5,...), so that is the third argument in the range functions in all namespaces.

Upvotes: 0

Taohidul Islam
Taohidul Islam

Reputation: 5414

You can use recursion,

def q2(n):
    result = []
    if n:
        result = result + q2(n - 1)
        result.append(n * 2 - 1)

    return result


result = q2(3)
print(result) # output : [1, 3, 5]

This is a good resource to start learning recursion.

Upvotes: 1

Rodrigo Ferreira
Rodrigo Ferreira

Reputation: 170

You should use a list comprehension, but it is also a kind of loop 

def q2(n):
        return [number for number in range(n) if number%2!=0]

Upvotes: 1

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