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.

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

I came here to make sure I was indenting properly. – Jeremy A. West Jul 18 '18 at 01:37 — Jeremy A. West, Jul 18 '18 at 01:37

Dan Oberlam · Accepted Answer · 2015-05-14T21:13:59.287

10

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()

edited May 14 '15 at 21:13

answered Jul 06 '14 at 01:36

Dan Oberlam

2,435
9
36
54

Yea this was per an assignment that required nested loops. So I needed it the second way. Thanks a bunch! – iMaxPrime Jul 06 '14 at 01:40

score 2 · Answer 2 · answered Jul 06 '14 at 01:34

2

This works in both python2 and python3:

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

answered Jul 06 '14 at 01:34

darkcharl

58
6

score 2 · Answer 3 · answered May 28 '15 at 14:24

2

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

answered May 28 '15 at 14:24

Raghav Chadha

117
1
5

score 1 · Answer 4 · answered Mar 21 '17 at 16:33

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. :)

score 1 · Answer 5 · answered Nov 02 '17 at 07:51

1

This is one line solution. A little bit long:

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

answered Nov 02 '17 at 07:51

Coskun Soysal

21
4

score 0 · Answer 6 · answered Jul 06 '14 at 01:29

0

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

answered Jul 06 '14 at 01:29

David Ehrmann

7,366
2
31
40

I suspect you mean `i`, not `j`. – Matt Bryant Jul 06 '14 at 01:29
@MattBryant fixed it as you were commenting. – David Ehrmann Jul 06 '14 at 01:30
That helped @David, but it starts my lines off at 2 instead of one now. – iMaxPrime Jul 06 '14 at 01:34
@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. – David Ehrmann Jul 06 '14 at 01:39

score 0 · Answer 7 · answered Jul 06 '14 at 01:36

0

Is this what you want:

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

answered Jul 06 '14 at 01:36

Zacrath

521
3
8

score 0 · Answer 8 · answered Dec 11 '14 at 10:57

0

"""2. 111 222 333 printing"""

for l in range (1,10):
    for k in range(l):
        print(l,end='')
print()

answered Dec 11 '14 at 10:57

Harit Singh

91
1
3

Please describe in short what is happening in your code for the benefit of other readers. – Jens Dec 11 '14 at 12:30

score 0 · Answer 9 · edited Nov 02 '17 at 08:16

0

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))

edited Nov 02 '17 at 08:16

Marian Nasry

821
9
22

answered Jul 28 '15 at 02:07

Fredrick Kofi Tam

1
1

score 0 · Answer 10 · edited Dec 20 '18 at 13:40

0

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 :)

edited Dec 20 '18 at 13:40

Zian

559
4
16

answered Jan 12 '17 at 20:02

Tejumade

1
5

score 0 · Answer 11 · answered May 27 '17 at 17:08

0

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

answered May 27 '17 at 17:08

Arvind T

145
2
2
7

Please read this [how-to-answer](http://stackoverflow.com/help/how-to-answer) for providing a quality answer. Just post the code is not the best answer. – thewaywewere May 27 '17 at 17:28

score 0 · Answer 12 · answered May 04 '19 at 22:21

0

This can be your one line code to problem

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

answered May 04 '19 at 22:21

HamzaMushtaq

1,660
13
13

score -1 · Answer 13 · answered Mar 29 '16 at 14:13

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.

Nested Loop Python

13 Answers13

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