blocks|key|1792281|text|您可以使用range()进行倒数|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1792282|for+i+in+range(6,+0,+-1):
++++for+j+in+range(1,+rows+%2B+1):
++++++++if(6-j+>=+i):+#+if+i+=+6,+doesn't+activate.+i=5,+activates+once.+i=4,+activates+twice,+etc.
++++++++++++print('+',+end+=+'+')
++++++++else:
++++++++++++print(i,+end+=+'+')
++++print()|code-block|syntax|javascript|1792283|entityMap^0|5|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can count backwards with <code>range()</code>:

<pre><code>for i in range(6, 0, -1):
 for j in range(1, rows + 1):
 if(6-j &gt;= i): # if i = 6, doesn't activate. i=5, activates once. i=4, activates twice, etc.
 print(' ', end = ' ')
 else:
 print(i, end = ' ')
 print()
</code></pre>

blocks|key|1792290|text|因此，从我所看到的情况来看，您正在尝试制作表单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1792291|+++++1
++++21
+++321
++4321
+54321
654321|code-block|syntax|javascript|1792292|因此，循环需要颠倒，您需要添加一个空间填充部分。|1792293|rows+=+6
for+i+in+range(1,+rows%2B1):
++++out+=+''
++++for+j+in+range(1,+rows):
++++++++out+%2B=+'+'

++++for+j+in+range(i,+0,+-1):
++++++++out+%2B=+str(j)
++++print(out)

++++rows+-=1|1792294|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

So from what I can see you are trying to make the form:

<pre><code> 1
 21
 321
 4321
 54321
654321
</code></pre>

So the loops need to be reversed and you need to add a space filler section.

<pre><code>rows = 6
for i in range(1, rows+1):
 out = ''
 for j in range(1, rows):
 out += ' '

 for j in range(i, 0, -1):
 out += str(j)
 print(out)

 rows -=1
</code></pre>

blocks|key|1577279|text|使用列表理解的1行语句如下|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1577280|pattern+=+'\n'.join(('+'+*+2+*+i)+%2B+'+'.join(str(n)+for+n+in+range(1,+num+%2B+1))+for+i,+num+in+enumerate(range(6,+-1,+-1)))|code-block|syntax|javascript|1577281|为了澄清，您可以查看在Python交互式终端上执行的以下命令。|1577282|>>>+pattern+=+'\n'.join(('+'+*+2+*+i)+%2B+'+'.join(str(n)+for+n+in+range(1,+num+%2B+1))+for+i,+num+in+enumerate(range(6,+-1,+-1)))
>>>+
>>>+print(pattern)++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
1+2+3+4+5+6
++1+2+3+4+5
++++1+2+3+4
++++++1+2+3
++++++++1+2
++++++++++1

>>>+|1577283|对于这种重复的工作，建议使用函数方法(如果您想尝试多个样本)。|1577284|def+print_num_triangle(n=6):
++++"""
++++1+2+3+4+5+6
++++++1+2+3+4+5
++++++++1+2+3+4
++++++++++1+2+3
++++++++++++1+2
++++++++++++++1
++++"""

++++pattern+=+'\n'.join(('+'+*+2+*+i)+%2B+'+'.join(str(n)+for+n+in+range(1,+num+%2B+1))+for+i,+num+in+enumerate(range(n,+-1,+-1)))
++++print(pattern)


if+__name__+==+"__main__":
++++print_num_triangle(10)

++++#+1+2+3+4+5+6+7+8+9+10
++++#+++1+2+3+4+5+6+7+8+9
++++#+++++1+2+3+4+5+6+7+8
++++#+++++++1+2+3+4+5+6+7
++++#+++++++++1+2+3+4+5+6
++++#+++++++++++1+2+3+4+5
++++#+++++++++++++1+2+3+4
++++#+++++++++++++++1+2+3
++++#+++++++++++++++++1+2
++++#+++++++++++++++++++1
++++#+++++++++++++++++++++

++++print_num_triangle(7)
++++#+1+2+3+4+5+6+7
++++#+++1+2+3+4+5+6
++++#+++++1+2+3+4+5
++++#+++++++1+2+3+4
++++#+++++++++1+2+3
++++#+++++++++++1+2
++++#+++++++++++++1

++++print_num_triangle()+#+default+->+6
++++#+1+2+3+4+5+6
++++#+++1+2+3+4+5
++++#+++++1+2+3+4
++++#+++++++1+2+3
++++#+++++++++1+2
++++#+++++++++++1|1577285|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

A 1 line statement using list comprehension would be 

<pre><code>pattern = '\n'.join((' ' * 2 * i) + ' '.join(str(n) for n in range(1, num + 1)) for i, num in enumerate(range(6, -1, -1)))
</code></pre>

For clarification, you can have a look at the below commands executed on Python interactive terminal.

<pre><code>&gt;&gt;&gt; pattern = '\n'.join((' ' * 2 * i) + ' '.join(str(n) for n in range(1, num + 1)) for i, num in enumerate(range(6, -1, -1)))
&gt;&gt;&gt; 
&gt;&gt;&gt; print(pattern) 
1 2 3 4 5 6
 1 2 3 4 5
 1 2 3 4
 1 2 3
 1 2
 1

&gt;&gt;&gt; 
</code></pre>

It is suggested to use functional approach for this kind of repetitive work (If you want to try with multiple samples).

<pre><code>def print_num_triangle(n=6):
 """
 1 2 3 4 5 6
 1 2 3 4 5
 1 2 3 4
 1 2 3
 1 2
 1
 """

 pattern = '\n'.join((' ' * 2 * i) + ' '.join(str(n) for n in range(1, num + 1)) for i, num in enumerate(range(n, -1, -1)))
 print(pattern)


if __name__ == "__main__":
 print_num_triangle(10)

 # 1 2 3 4 5 6 7 8 9 10
 # 1 2 3 4 5 6 7 8 9
 # 1 2 3 4 5 6 7 8
 # 1 2 3 4 5 6 7
 # 1 2 3 4 5 6
 # 1 2 3 4 5
 # 1 2 3 4
 # 1 2 3
 # 1 2
 # 1
 # 

 print_num_triangle(7)
 # 1 2 3 4 5 6 7
 # 1 2 3 4 5 6
 # 1 2 3 4 5
 # 1 2 3 4
 # 1 2 3
 # 1 2
 # 1

 print_num_triangle() # default -&gt; 6
 # 1 2 3 4 5 6
 # 1 2 3 4 5
 # 1 2 3 4
 # 1 2 3
 # 1 2
 # 1
</code></pre>

blocks|key|1577296|text|此解决方案类似于您的解决方案，但有一些更改|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1577297|它首先打印出三角形所需的空间数量|1577298|然后计数到count，然后移到下一行|offset|length|style|CODE|1577299|num+=+6
for+i+in+range(num,0,-1):
++++print('++'*(num+-+i),+end='')
++++for+j+in+range(i):
++++++++print(j+%2B+1,+end='+')
++++print()|code-block|syntax|javascript|1577300|如果使用大于10的值，则会产生奇怪的效果|1577301|1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+
++1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+
++++1+2+3+4+5+6+7+8+9+10+11+12+13+14+
++++++1+2+3+4+5+6+7+8+9+10+11+12+13+
++++++++1+2+3+4+5+6+7+8+9+10+11+12+
++++++++++1+2+3+4+5+6+7+8+9+10+11+
++++++++++++1+2+3+4+5+6+7+8+9+10+
++++++++++++++1+2+3+4+5+6+7+8+9+
++++++++++++++++1+2+3+4+5+6+7+8+
++++++++++++++++++1+2+3+4+5+6+7+
++++++++++++++++++++1+2+3+4+5+6+
++++++++++++++++++++++1+2+3+4+5+
++++++++++++++++++++++++1+2+3+4+
++++++++++++++++++++++++++1+2+3+
++++++++++++++++++++++++++++1+2+
++++++++++++++++++++++++++++++1+|1577302|您可以使用tabulate将所有内容保留在其列中。在这里，我还使用了列表理解来减少代码大小。|1577303|代码：|1577304|from+tabulate+import+tabulate
count+=+16
numbers+=+[['']+*+(count+-+i)+%2B+[j%2B1+for+j+in+range(i)]+for+i+in+range(count,+0,+-1)]
print(tabulate(numbers))|1577305|输出：|1577306|-++-++-++-++-++-++-++-++-++--++--++--++--++--++--++--
1++2++3++4++5++6++7++8++9++10++11++12++13++14++15++16
+++1++2++3++4++5++6++7++8++9+++10++11++12++13++14++15
++++++1++2++3++4++5++6++7++8+++9+++10++11++12++13++14
+++++++++1++2++3++4++5++6++7+++8+++9+++10++11++12++13
++++++++++++1++2++3++4++5++6+++7+++8+++9+++10++11++12
+++++++++++++++1++2++3++4++5+++6+++7+++8+++9+++10++11
++++++++++++++++++1++2++3++4+++5+++6+++7+++8+++9+++10
+++++++++++++++++++++1++2++3+++4+++5+++6+++7+++8++++9
++++++++++++++++++++++++1++2+++3+++4+++5+++6+++7++++8
+++++++++++++++++++++++++++1+++2+++3+++4+++5+++6++++7
+++++++++++++++++++++++++++++++1+++2+++3+++4+++5++++6
+++++++++++++++++++++++++++++++++++1+++2+++3+++4++++5
+++++++++++++++++++++++++++++++++++++++1+++2+++3++++4
+++++++++++++++++++++++++++++++++++++++++++1+++2++++3
+++++++++++++++++++++++++++++++++++++++++++++++1++++2
++++++++++++++++++++++++++++++++++++++++++++++++++++1
-++-++-++-++-++-++-++-++-++--++--++--++--++--++--++--|1577307|entityMap^0|0|0|5|5|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|16|8|@$F|17|G|18|H|I]]|9|@]|A|$]]|$1|J|3|K|5|L|7|19|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Q|3|R|5|L|7|1B|8|@]|9|@]|A|$M|N]]|$1|S|3|T|5|6|7|1C|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1D|8|@]|9|@]|A|$]]|$1|W|3|X|5|L|7|1E|8|@]|9|@]|A|$M|N]]|$1|Y|3|Z|5|6|7|1F|8|@]|9|@]|A|$]]|$1|10|3|11|5|L|7|1G|8|@]|9|@]|A|$M|N]]|$1|12|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|13|$]]

This solution resembles yours with a few changes 
It first prints out the number of spaces needed for the triangle shape 
Then it counts up to <code>count</code>
and then it moves to the next line

<pre><code>num = 6
for i in range(num,0,-1):
 print(' '*(num - i), end='')
 for j in range(i):
 print(j + 1, end=' ')
 print()
</code></pre>

this does have an odd effect if you use a value greater than 10

<pre class="lang-none prettyprint-override"><code>1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 
 1 2 3 4 5 6 7 8 9 10 11 12 13 
 1 2 3 4 5 6 7 8 9 10 11 12 
 1 2 3 4 5 6 7 8 9 10 11 
 1 2 3 4 5 6 7 8 9 10 
 1 2 3 4 5 6 7 8 9 
 1 2 3 4 5 6 7 8 
 1 2 3 4 5 6 7 
 1 2 3 4 5 6 
 1 2 3 4 5 
 1 2 3 4 
 1 2 3 
 1 2 
 1 
</code></pre>

You can use tabulate to keep everything in its column. Here I also used a list comprehension to reduce code size. 
code:

<pre><code>from tabulate import tabulate
count = 16
numbers = [[''] * (count - i) + [j+1 for j in range(i)] for i in range(count, 0, -1)]
print(tabulate(numbers))
</code></pre>

output:

<pre class="lang-none prettyprint-override"><code>- - - - - - - - - -- -- -- -- -- -- --
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
 1 2 3 4 5 6 7 8 9 10 11 12 13 14
 1 2 3 4 5 6 7 8 9 10 11 12 13
 1 2 3 4 5 6 7 8 9 10 11 12
 1 2 3 4 5 6 7 8 9 10 11
 1 2 3 4 5 6 7 8 9 10
 1 2 3 4 5 6 7 8 9
 1 2 3 4 5 6 7 8
 1 2 3 4 5 6 7
 1 2 3 4 5 6
 1 2 3 4 5
 1 2 3 4
 1 2 3
 1 2
 1
- - - - - - - - - -- -- -- -- -- -- --
</code></pre>

I need to produce the following pattern:

<pre><code>1 2 3 4 5 6
 1 2 3 4 5
 1 2 3 4
 1 2 3
 1 2
 1
</code></pre>

I have written a code that produces the same but in a right side up form. i don't understand how to flip it over.

<pre><code>for i in range(1, 6 + 1):
 for j in range(1, rows + 1):
 if(j &lt; i):
 print(' ', end = ' ')
 else:
 print(i, end = ' ')
 print()
</code></pre>

I need to write a program to print the following pattern:

我需要生成以下模式：1 2 3 4 5 6  1 2 3 4 5    1 2 3 4      1 2 3        1 2          1我已经写了一个代码，产生相同的，但在右边向上的形式。我不知道怎么把它翻过来。for i in range(1, 6 + 1):    for j in range(1,...

问我需要编写一个程序来打印以下模式：
EN

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问我需要编写一个程序来打印以下模式：EN

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问我需要编写一个程序来打印以下模式：
EN