blocks|key|3380599|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3380600|(1..9).each+{+%7Cn%7C+print+n+}
puts|code-block|syntax|javascript|3380601|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

try this:

<pre><code>(1..9).each { |n| print n }
puts
</code></pre>

blocks|key|1951964|text|你说你想要“形成一个任何大小的三角形，遵循这个模式”，所以你不应该假设应该如何去做。有两种方法可以做到这一点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1951965|#1|offset|length|style|BOLD|1951966|def+print_triangle(n)
++(1..n).each.with_object('')+{+%7Ci,s%7C+puts+s+<<+i.to_s+}
end

print_triangle(9)
1
12
123
1234
12345
123456
1234567
12345678
123456789|code-block|syntax|javascript|1951967|#2|1951968|def+print_triangle(n)
++s+=+(1..n).to_a.join
++(1..n).each+{+%7Ci%7C+puts+s[0,i]+}
end

print_triangle(9)
1
12
123
1234
12345
123456
1234567
12345678
123456789|1951969|entityMap^0|0|0|2|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

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

<pre><code>def print_triangle(n)
 (1..n).each.with_object('') { |i,s| puts s &lt;&lt; i.to_s }
end

print_triangle(9)
1
12
123
1234
12345
123456
1234567
12345678
123456789
</code></pre>

#2

<pre><code>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
</code></pre>

blocks|key|3380680|text|这个解决方案怎么样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3380681|last_num+=+9
str++++++=+(1..last_num).to_a.join+#+create+string+123456789
0.upto(last_num-1){+%7Ci%7C+puts+str[0..i]+}+#+print+line+by+line|code-block|syntax|javascript|3380682|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

how about this solution:

<pre><code>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
</code></pre>

blocks|key|1951971|text|puts+(1..9).map(&:to_s).join|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1951972|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

blocks|key|1951903|text|关于你的最终目标，有很多(可能更容易)的方法，但这里有一个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1951904|def+print_nums+k
++k.times+{+%7Cn%7C+puts+(1..(n%2B1)).map+{+%7Ci%7C+i*10**(n%2B1-i)+}.inject(:%2B)+}
end

print_nums+9
#1
#12
#123
#1234
#12345
#123456
#1234567
#12345678
#123456789|code-block|syntax|javascript|1951905|这种方法使用单位、数十个、数百个与行号i相关的单位生成实际数字。|offset|length|style|CODE|1951906|思维过程|1951907|看一个四行的基本示例：|1951908|1
12
123
1234|1951909|与以下相同：|1951910|1*10**0+++++++++++++++++++++++++++++++++++#=>+1
1*10**1+%2B+2*10**0+++++++++++++++++++++++++#=>+12
1*10**2+%2B+2*10**1+%2B+3*10**0+++++++++++++++#=>+123
1*10**3+%2B+2*10**2+%2B+3*10**1+%2B+4*10**0+++++#=>+1234|1951911|在Ruby中可以使用以下方法生成：|1951912|(1..1).map+{+%7Ci%7C+i*10**(1-i)+}.inject(:%2B)+#=>+1
(1..2).map+{+%7Ci%7C+i*10**(2-i)+}.inject(:%2B)+#=>+12
(1..3).map+{+%7Ci%7C+i*10**(3-i)+}.inject(:%2B)+#=>+123
(1..4).map+{+%7Ci%7C+i*10**(4-i)+}.inject(:%2B)+#=>+1234|1951913|寻找一种模式，我们可以概括并放入一种方法：|1951914|def+print_nums+k
++k.times+{+%7Cn%7C+puts+(1..(n%2B1)).map+{+%7Ci%7C+i*10**(n%2B1-i)+}.inject(:%2B)+}
end|1951915|当然，您可以(而且应该)忽略上述所有内容，只需扩展@seph的优秀答案即可。|1951916|3.times+{+%7Ci%7C+(1..(i%2B1)).each+{+%7Cn%7C+print+n+};+puts+}
#1
#12
#123|1951917|entityMap^0|0|0|J|1|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1C|8|@$I|1D|J|1E|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1F|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1G|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1I|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1J|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1K|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1L|8|@]|9|@]|A|$E|F]]|$1|10|3|11|5|6|7|1M|8|@]|9|@]|A|$]]|$1|12|3|13|5|D|7|1N|8|@]|9|@]|A|$E|F]]|$1|14|3|15|5|6|7|1O|8|@]|9|@]|A|$]]|$1|16|3|17|5|D|7|1P|8|@]|9|@]|A|$E|F]]|$1|18|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]]|19|$]]

Regarding your final aim there are lots of (probably easier) ways, but here's one:
<pre><code>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
</code></pre>
This approach generates the actual numbers using units, tens, hundreds etc in relation to the line number <code>i</code>.
<hr />
<h3>Thought Process</h3>
Looking at a basic example of four lines:
<pre><code>1
12
123
1234
</code></pre>
is the same as:
<pre><code>1*10**0 #=&gt; 1
1*10**1 + 2*10**0 #=&gt; 12
1*10**2 + 2*10**1 + 3*10**0 #=&gt; 123
1*10**3 + 2*10**2 + 3*10**1 + 4*10**0 #=&gt; 1234
</code></pre>
which in Ruby can be generated with:
<pre><code>(1..1).map { |i| i*10**(1-i) }.inject(:+) #=&gt; 1
(1..2).map { |i| i*10**(2-i) }.inject(:+) #=&gt; 12
(1..3).map { |i| i*10**(3-i) }.inject(:+) #=&gt; 123
(1..4).map { |i| i*10**(4-i) }.inject(:+) #=&gt; 1234
</code></pre>
looking for a pattern we can generalise and put in a method:
<pre><code>def print_nums k
 k.times { |n| puts (1..(n+1)).map { |i| i*10**(n+1-i) }.inject(:+) }
end
</code></pre>
<hr />
You could (and should) of course ignore all of the above and just extend the excellent answer by @seph
<pre><code>3.times { |i| (1..(i+1)).each { |n| print n }; puts }
#1
#12
#123
</code></pre>

blocks|key|3380661|text|如果你想从1开始，最简单的方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3380662|9.times+{%7Cn%7C+puts+n%2B1}|code-block|syntax|javascript|3380663|如果你想从0开始试一试|3380664|10.times+{%7Cn%7C+puts+n}|3380665|如果您想要金字塔格式，这是一种方法。|3380666|9.times{%7Cc%7C+puts+(1..c%2B1).to_a.join}|3380667|这是输出|3380668|2.3.0+:025+>+9.times{%7Cc%7C+puts+(1..c%2B1).to_a.join}
1
12
123
1234
12345
123456
1234567
12345678
123456789|3380669|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

The simplest way if you want to start from 1

<pre><code>9.times {|n| puts n+1}
</code></pre>

try if you want to start from 0

<pre><code>10.times {|n| puts n}
</code></pre>

if you want pyramid format this is one way to do

<pre><code>9.times{|c| puts (1..c+1).to_a.join}
</code></pre>

this is the ouput

<pre><code>2.3.0 :025 &gt; 9.times{|c| puts (1..c+1).to_a.join}
1
12
123
1234
12345
123456
1234567
12345678
123456789
</code></pre>

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

<pre><code>x = 0
while x &lt;= 9
 puts x
 x +=1
end
</code></pre>

This would give you

<pre><code>0
1
2
3
4
5
6
7
8
9
</code></pre>

Each on different lines.

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

<pre><code>01234567891011121314151617181920
</code></pre>

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.

<pre><code>1
12
123
1234
12345
123456
</code></pre>

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.

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

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我很难弄清楚如何在红宝石上用一行来循环数字。x = 0while x <= 9 puts x x +=1end这会给你0123456789每个都在不同的线上。但我想要的是把这个放在一条线上01234567891011121314151617181920也不局限于单行上的0到9，更像0到无穷大。其目的是建立一个任何大小的三角形，遵循这个模式。112123123412345123456每一条都在不

问如何计算红宝石中单行的1到9
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回答 6

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问如何计算红宝石中单行的1到9EN

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问如何计算红宝石中单行的1到9
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