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.
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9 Answers
41
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,...'
Aleksei astynax Pirogov
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2A Python 3 version is supplied by Oleh Prypin – dǝɥɔS ʇoıןןƎ Nov 07 '17 at 03:09
20
>>> ','.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)))
Oleh Prypin
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One liner, thanks! That's really nice! Initially I though something of this 'odd = []' 'even = []' 'for c in range(0,100):' 'if c%2==0:' 'odd.append(c)' 'for c in range(0,100):' 'if c%2==1:' 'even.append(c)' 'for c in range(0, 50):' 'if c % 2 == 0:' 'print(even[c],end=",")' 'else:' 'print(odd[c],end=",")' Yeah, I'm a newcomer. But I knew there must be better ways! Thanks guys! – user1460818 Jun 16 '12 at 16:19
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Further description of the final function would be greatly appreciated. – Wolfpack'08 Dec 10 '18 at 01:37
14
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)
poke
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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
FieryBanditry
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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)]
sushmit
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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)
Xion
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0
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.
Facundo Casco
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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
kdauria
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Gunjan Thareja
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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]+','
indiag
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Why are you iterating over a separate range object? You should just iterate over the list items. – poke Jun 16 '12 at 16:10