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 program that should have the output that looks like this.

1
22
333
4444
55555
666666
7777777
88888888
999999999 

With the above code I am pretty close, but the way my count is working it just literally counts up and up. I just need help getting it to only count to 9 but display like above. Thanks.

enter image description here

• Jeremy A. West: I came here to make sure I was indenting properly.

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()

• iMaxPrime: Yea this was per an assignment that required nested loops. So I needed it the second way. Thanks a bunch!

Answer 2

This works in both python2 and python3:

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

Answer 3

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

Answer 4

The simple mistake in your code is the placement of count = count + 1. It should be placed after the second for loop block. I have made a simple change in your own code to obtain the output you want.

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

This will give the output you want with the code you wrote. :)

Answer 5

This is one line solution. A little bit long:

print ('\n'.join([str(i)*i for i in range(1,10)]))

Answer 6

Change print(count, end='') to print(i + 1, end='') and remove count. Just make sure you understand why it works.

• Matt Bryant: I suspect you mean i, not j.
• David Ehrmann: @MattBryant fixed it as you were commenting.
• iMaxPrime: That helped @David, but it starts my lines off at 2 instead of one now.
• David Ehrmann: @iMaxPrime also start your inner range at 1, not 0. for j in range(1, i): Again, the important thing here is debugging a bit on your own and thinking things through, not just copy and pasting the code of others.

Answer 7

Is this what you want:

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

Answer 8

"""2. 111 222 333 printing"""
for l in range (1,10):
 for k in range(l):
 print(l,end='')
print()

• Jens: Please describe in short what is happening in your code for the benefit of other readers.

Answer 9

count = 1
for i in range(9):
 for j in range (-1, i):
 print (count, end = '')
 count = count + 1
 print (" ")

• thewaywewere: Please read this how-to-answer for providing a quality answer. Just post the code is not the best answer.

Answer 10

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 11

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 12

This can be your one line code to problem

print(''.join([str(x)*x+ '\n' for x in range(1,10)]))

Answer 13

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.