Nested Loop Python

Question

count = 1
for i in range(10):
    for j in range(0, i):
        print(count, end=\'\')
        count = count +1
    print()
input()

I am writing a

Answer 1

You're incrementing count in the inner loop which is why you keep getting larger numbers before you want to

You could just do this.

>>> for i in range(1, 10):
        print str(i) * i


1
22
333
4444
55555
666666
7777777
88888888
999999999

or if you want the nested loop for some reason

from __future__ import print_function

for i in range(1, 10):
    for j in range(i):
        print(i, end='')
    print()

Answer 2

I realised that the problem is solved but here's how you wanted your code to look like.

count=0
for i in range(10):
    for j in range(0, i):
        print (count, end='')
count +=1
print()

i think @Dannnno answer is shorter and straight to the point :)

Answer 3

What you are trying to do involves a mathematical concept called repunit numbers

you could also do it as follows:

for i in range(1,n):
    print (int(i*((10**i)-1)/9))

Answer 4

Others have suggested some interesting solutions but this can also be done mathematically using a simple observation. Notice that:

1 - 1*1

22 - 2*11

333 - 3*111

4444 - 4*1111

and so on ....

We can have a general formula for producing 1,11,111,1111,... at every iteration. Observe that:

1 = 9/9 = (10 - 1)/9 = (10^1 - 1)/9

11 = 99/9 = (100 - 1)/9 = (10^2 - 1)/9

111 = 999/9 = (1000 - 1)/9 = (10^3 - 1)/9

......

that is we have (10^i - 1)/9 for the ith iteration.

Now it is simple enough to implement. We will multiply i with the above formula in each iteration. Hence the overall formula is:

i*(10^i - 1)/9 (for every ith iteration). Here's the python code:

for i in xrange(1,10):
    print i*(10**i-1)/9

Hope this helps.

Answer 5

This works in both python2 and python3:

for i in range(10):
  print(str(i) * i)

Answer 6

for i in range(1,10):
    for j in range(0,i):
        print i,
print "\n"