How to count 1 to 9 on a single line in ruby

Question

I'm struggling to figure out how to loop numbers in a single line on ruby.

x = 0
while x <= 9
  puts x
  x +=1
end

This would give you

Each on different lines.

But what I want is to get this on a single line so like

01234567891011121314151617181920

Also not limited to just 0-9 more like 0 to infinity on a single line.

The purpose is to make an triangle of any size that follows this pattern.

Each of these would be on a different line. The formatting here won't let me put in on different lines.

Would really like to solve this. It is hurting my head.

http://stackoverflow.com/questions/5018633/what-is-the-difference-between-print-and-puts — kennytm, Oct 09 '16 at 06:28

score 4 · Accepted Answer · answered Oct 09 '16 at 06:26

4

try this:

(1..9).each { |n| print n }
puts

answered Oct 09 '16 at 06:26

seph

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You are a godsend! Thank you so much! – jackneedshelp Oct 09 '16 at 06:44

Cary Swoveland · Answer 2 · 2016-10-10T05:31:14.227

2

You said you want "to make a triangle of any size that follows this pattern", so you should not make assumptions about how that should be done. Here are two ways to do that.

#1

def print_triangle(n)
  (1..n).each.with_object('') { |i,s| puts s << i.to_s }
end

print_triangle(9)
1
12
123
1234
12345
123456
1234567
12345678
123456789

#2

def print_triangle(n)
  s = (1..n).to_a.join
  (1..n).each { |i| puts s[0,i] }
end

print_triangle(9)
1
12
123
1234
12345
123456
1234567
12345678
123456789

edited Oct 10 '16 at 05:31

answered Oct 09 '16 at 18:13

Cary Swoveland

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6
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What is the temp string here for? :) – Sergio Tulentsev Oct 09 '16 at 18:19
@sergio, I print the first character of `s`, then the first two characters of `s` and so on, rather than creating a string for each line (by creating an array then joining). Have I answered your question? Notice that I modified my answer. – Cary Swoveland Oct 09 '16 at 18:27
I don't have anything better to do, so I decided to profile different approaches here. You may find this interesting: http://pastie.org/10940329 – Sergio Tulentsev Oct 09 '16 at 18:36
@Sergio, if you mean `print 1`, `print 2` and so on, that doesn't seem as efficient as using strings. Let's see an answer! – Cary Swoveland Oct 09 '16 at 18:36
In my benchmarks, using prints is _slightly_ more efficient than that temp string approach. `each_with_object` beats it, though. – Sergio Tulentsev Oct 09 '16 at 18:38

score 1 · Answer 3 · answered Oct 09 '16 at 15:24

1

how about this solution:

last_num = 9
str      = (1..last_num).to_a.join # create string 123456789
0.upto(last_num-1){ |i| puts str[0..i] } # print line by line

answered Oct 09 '16 at 15:24

Hieu Pham

6,577
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50

undur_gongor · Answer 4 · 2016-10-09T18:32:09.317

0

puts (1..9).map(&:to_s).join

edited Oct 09 '16 at 18:32

answered Oct 09 '16 at 18:24

undur_gongor

15,657
5
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1

mapping to string is not needed. `.join` will call `.to_s` itself. – Sergio Tulentsev Oct 09 '16 at 18:26
@Sergio: Are you sure? Range does not seem to have a `join` method. The `map` makes an array out of it which does. – undur_gongor Oct 09 '16 at 18:35
Yes, you use `map` because it returns an array. Not because you need strings. `.to_a.join` will work just as well. – Sergio Tulentsev Oct 09 '16 at 18:37

score -1 · Answer 5 · edited Jun 20 '20 at 09:12

Regarding your final aim there are lots of (probably easier) ways, but here's one:

def print_nums k
  k.times { |n| puts (1..(n+1)).map { |i| i*10**(n+1-i) }.inject(:+) }
end

print_nums 9
#1
#12
#123
#1234
#12345
#123456
#1234567
#12345678
#123456789

This approach generates the actual numbers using units, tens, hundreds etc in relation to the line number i.

Thought Process

Looking at a basic example of four lines:

is the same as:

1*10**0                                   #=> 1
1*10**1 + 2*10**0                         #=> 12
1*10**2 + 2*10**1 + 3*10**0               #=> 123
1*10**3 + 2*10**2 + 3*10**1 + 4*10**0     #=> 1234

which in Ruby can be generated with:

(1..1).map { |i| i*10**(1-i) }.inject(:+) #=> 1
(1..2).map { |i| i*10**(2-i) }.inject(:+) #=> 12
(1..3).map { |i| i*10**(3-i) }.inject(:+) #=> 123
(1..4).map { |i| i*10**(4-i) }.inject(:+) #=> 1234

looking for a pattern we can generalise and put in a method:

def print_nums k
  k.times { |n| puts (1..(n+1)).map { |i| i*10**(n+1-i) }.inject(:+) }
end

You could (and should) of course ignore all of the above and just extend the excellent answer by @seph

3.times { |i| (1..(i+1)).each { |n| print n }; puts }
#1
#12
#123

I am preeeetty sure that you don't need each line to be an actual _number_. Therefore, no need for that "powers of 10" business. Just `each` and `print`. Alternatively, `.to_a` on a range + `.join` — Sergio Tulentsev, Oct 09 '16 at 17:22
@SergioTulentsev ha yes it's kind of over-the-top (and I do say as such in my answer). I had already started this train of thought and thought it interesting to follow it through. I enjoyed the process:) — Sagar Pandya, Oct 09 '16 at 19:02

Horacio · Answer 6 · 2016-10-09T09:13:22.863

-1

The simplest way if you want to start from 1

9.times {|n| puts n+1}

try if you want to start from 0

10.times {|n| puts n}

if you want pyramid format this is one way to do

9.times{|c| puts (1..c+1).to_a.join}

this is the ouput

2.3.0 :025 > 9.times{|c| puts (1..c+1).to_a.join}
1
12
123
1234
12345
123456
1234567
12345678
123456789

edited Oct 09 '16 at 09:13

answered Oct 09 '16 at 09:04

Horacio

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you first two lines don't do what OP wants. – Sergio Tulentsev Oct 09 '16 at 17:23
@SergioTulentsev I just explain the title of the question "How to count 1 to 9 on a single line in ruby" and then I explain one way to solve the problem of the piramid, sorry if I I hurt you <3 – Horacio Oct 14 '16 at 19:15

How to count 1 to 9 on a single line in ruby

6 Answers6

Thought Process