CODEWITHSUNDEEP
pythonsequence
user1460818
user1460818

Reputation: 491

Generate a sequence of numbers in Python

How can I generate the sequence of numbers "1,2,5,6,9,10......" and so until 100 in Python? I even need the comma (',') included, but this is not the main problem.

The sequence: every number from 1..100, divisible by 4 with remainder 1 or 2.

Upvotes: 47

Views: 241964

Answers (9)

Gunjan Thareja
Gunjan Thareja

Reputation: 37

Write a function that takes a number as an argument and prints the Fibonacci series till that number

def Series(n):  
    a = 0  
    b = 1  
    print(a)  
    print(b)  
    S = 0  
    for i in range(0,n):  
        if S <= n-1:  
            S = a + b  
            print(S)  
            a = b  
            b = S

Upvotes: -1

sushmit
sushmit

Reputation: 4603

using numpy and list comprehension you can do the

import numpy as np
[num for num in np.arange(1,101) if (num%4 == 1 or num%4 == 2)]

Upvotes: 2

Oleh Prypin
Oleh Prypin

Reputation: 34116

>>> ','.join('{},{}'.format(i, i + 1) for i in range(1, 100, 4))
'1,2,5,6,9,10,13,14,17,18,21,22,25,26,29,30,33,34,37,38,41,42,45,46,49,50,53,54,57,58,61,62,65,66,69,70,73,74,77,78,81,82,85,86,89,90,93,94,97,98'

That was a quick and quite dirty solution.

Now, for a solution that is suitable for different kinds of progression problems:

def deltas():
    while True:
        yield 1
        yield 3
def numbers(start, deltas, max):
    i = start
    while i <= max:
        yield i
        i += next(deltas)
print(','.join(str(i) for i in numbers(1, deltas(), 100)))

And here are similar ideas implemented using itertools:

from itertools import cycle, takewhile, accumulate, chain

def numbers(start, deltas, max):
    deltas = cycle(deltas)
    numbers = accumulate(chain([start], deltas))
    return takewhile(lambda x: x <= max, numbers)

print(','.join(str(x) for x in numbers(1, [1, 3], 100)))

Upvotes: 20

FieryBanditry
FieryBanditry

Reputation: 21

This works by exploiting the % properties of the list rather than the increments.

for num in range(1,100):
    if num % 4 == 1 or num % 4 ==2:
        n.append(num)
        continue
    pass

Upvotes: 2

Aleksei astynax Pirogov
Aleksei astynax Pirogov

Reputation: 2491

Every number from 1,2,5,6,9,10... is divisible by 4 with remainder 1 or 2.

>>> ','.join(str(i) for i in xrange(100) if i % 4 in (1,2))
'1,2,5,6,9,10,13,14,...'

Upvotes: 41

indiag
indiag

Reputation: 233

In python 3.1 you can produce a list in a way

     lst=list(range(100))
     for i in range(100)
        print (lst[i],',',end='')

In python 2.7 you can do it as 

     lst=range(100)
     for i in range(100)
        print lst[i]+','

Upvotes: -2

Facundo Casco
Facundo Casco

Reputation: 10585

Assuming your sequence alternates increments between 1 and 3

numbers = [1]
while numbers[-1] < 100:
    numbers.append(numbers[-1] + 1)
    numbers.append(numbers[-1] + 3)

print ', '.join(map(str, numbers))

This could be easier to modify if your sequence is different but I think poke or BlaXpirit are nicer answers than mine.

Upvotes: 0

poke
poke

Reputation: 387527

Includes some guessing on the exact sequence you are expecting:

>>> l = list(range(1, 100, 4)) + list(range(2, 100, 4))
>>> l.sort()
>>> ','.join(map(str, l))
'1,2,5,6,9,10,13,14,17,18,21,22,25,26,29,30,33,34,37,38,41,42,45,46,49,50,53,54,57,58,61,62,65,66,69,70,73,74,77,78,81,82,85,86,89,90,93,94,97,98'

As one-liner:

>>> ','.join(map(str, sorted(list(range(1, 100, 4))) + list(range(2, 100, 4))))

(btw. this is Python 3 compatible)

Upvotes: 15

Xion
Xion

Reputation: 22770

Assuming I've guessed the pattern correctly (alternating increments of 1 and 3), this should produce the desired result:

def sequence():
    res = []
    diff = 1
    x = 1
    while x <= 100:
        res.append(x)
        x += diff
        diff = 3 if diff == 1 else 1
    return ', '.join(res)

Upvotes: 0

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