Create a simple number pattern in python?

Question

I'm trying to get this number pattern

But I can't figure out how to reset the digits when I reach over 8 in my function. There is my code:

def afficherPatron(n):
triangle = ''

for i in range(0, n):
    triangle = triangle + (str(i))
    print(triangle)
    i+=1

Thanks in advance to all of you!

You don't need i+=1, the for loop increments the variable automatically. — Barmar
– Barmar, Commented Jan 21, 2016 at 17:07
Is 8 related to n or just some sort of cyclic limitation you wanted to be? — Iron Fist
– Iron Fist, Commented Jan 21, 2016 at 17:08
Oops, I forgot to delete it from the code! It's because I was using a while loop before;) — Francis-Olivier Couture
– Francis-Olivier Couture, Commented Jan 21, 2016 at 17:10

Assem · Accepted Answer · 2016-01-21 17:06:21Z

9

Use i mod 8 (i%8) because it is cyclic 0 to 7 :

for i in range(0, n):
    triangle = triangle + str(i%8)
    print(triangle)

answered Jan 21, 2016 at 17:06

Assem

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1 Comment

Francis-Olivier Couture Over a year ago

I'll try this, I'm pretty sure this is what I missed! Thanks a lot:D

kpie · Accepted Answer · 2016-01-21 17:36:28Z

0

I Like methods...

def yourPattern(b,s):
    r = ""
    q = [str(k) for k in range(1,b+1)*(s/b)+range(1,b+1)[0:s%b]]
    for k in q:#    ^Iterator[^Array ^Scale^add^Array   ^Portion]
        r += k # Put string together
    return(r)

def pattern(b,l):
    r = ""
    for k in range(1,l+1):
        r+=yourPattern(b,k)+"\n"
    return(r)
print pattern(8,18)

o(4n)

edited Jan 21, 2016 at 17:36

answered Jan 21, 2016 at 17:20

kpie

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