blocks|key|81454|text|这应该能起到作用。我认为有一些聪明的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|81455|print+'70-78:'.ljust(10)+%2B+'*****'|code-block|syntax|javascript|81456|您还可以使用expandtabs()|offset|length|style|CODE|81457|print+('70-78'%2B'\t'%2B+'*****').expandtabs(10)|81458|entityMap^0|0|0|6|C|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|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|$]]

This should do the trick. I assume there are clever ways.

<pre><code>print '70-78:'.ljust(10) + '*****'
</code></pre>

You could also use <code>expandtabs()</code>

<pre><code>print ('70-78'+'\t'+ '*****').expandtabs(10)
</code></pre>

blocks|key|298730|text|简单的方法(在你的情况下)是放一个制表符而不是空格：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|298731|for+x+in+range(numClasses):
++++print('{0}-{1}:\t{2}'.format(lower[x],upper[x],"*"*num[x]))|code-block|syntax|javascript|298732|另一种方法是使用str.ljust|offset|length|style|CODE|298733|for+x+in+range(numClasses):
++++label+=+'{0}-{1}:'.format(lower[x],+upper[x])
++++print(label.ljust(10,+'+')+%2B+"*"+*+num[x])|298734|entityMap^0|0|0|8|9|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|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|$]]

Easy way (in your case) would be to put a tab instead of space:

<pre><code>for x in range(numClasses):
 print('{0}-{1}:\t{2}'.format(lower[x],upper[x],"*"*num[x]))
</code></pre>

Another way would be to use <code>str.ljust</code>:

<pre><code>for x in range(numClasses):
 label = '{0}-{1}:'.format(lower[x], upper[x])
 print(label.ljust(10, ' ') + "*" * num[x])
</code></pre>

blocks|key|298741|text|lower+=+[70,+79,+88,+97,+106]
upper+=+[78,+87,+105,+114,+123]
num+=+[5,+3,+4,+2,+6,+4]

for+l,+u,+n+in+zip(lower,+upper,+num):
++++print('{0:<9}+{1}'.format('{0}-{1}:'.format(l,+u),+'*'+*+n))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|298742|http://docs.python.org/3/library/string.html#format-specification-mini-language|unstyled|offset|length|298743|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|27|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|Q|8|@]|9|@$G|R|H|S|1|T]]|A|$]]|$1|I|3|-4|5|F|7|U|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|E]]]]

<pre><code>lower = [70, 79, 88, 97, 106]
upper = [78, 87, 105, 114, 123]
num = [5, 3, 4, 2, 6, 4]

for l, u, n in zip(lower, upper, num):
 print('{0:&lt;9} {1}'.format('{0}-{1}:'.format(l, u), '*' * n))
</code></pre>

<a href="http://docs.python.org/3/library/string.html#format-specification-mini-language" rel="nofollow">http://docs.python.org/3/library/string.html#format-specification-mini-language</a>

blocks|key|81517|text|好吧，虽然我使用的解决方案确实是即席的，但到目前为止，它的效果和可扩展性都比答案更好。基本上，这只是VR17建议的方法，但使用了更多的方法，以便选项卡大小与数据集一起缩放，而不仅仅是硬编码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|81518|首先，我创建了一个返回某个数字中的数字字符的方法。|81519|def+charNum(number):
++++return+math.floor(math.log(number,10)%2B1)|code-block|syntax|javascript|81520|然后，我在lower和upper数据集的最后一个点上使用了charNum()函数。每个列表只使用最后一个点，因为最后一个点是最大的数字。然后我计算不是数字的字符(破折号、分号和空格)，并进行相应的调整。|offset|length|style|CODE|81521|因此，最终的tabLength变量如下所示：|81522|tabLength+=+charNum(lower[-1])%2BcharNum(upper[-1])%2B3|81523|然后，我将tabLength变量插入到expandTab()函数中，以获得适当的间距。下面是一些示例输出：|81524|1-11:++*******
12-22:+*
23-33:+***
34-44:+**
45-55:+***
56-66:+*

99-249:+++*****
250-400:++****
401-551:++**
552-702:++**
703-853:++*
854-1004:+***

99-200079:++++++******
200080-400060:++**
400061-600041:++****
600042-800022:++**
800023-1000003:+*|81525|我能真正看到的唯一问题是，如果我想将它扩展到一个表或其他东西，那么这些选项卡都会很时髦。如果我这样做了，我可能会去看看我现在还不太熟悉的ljust和rjust。我会把这个问题搁置一段时间，以防有人想出更好的答案。|81526|entityMap^0|0|0|0|5|5|B|5|T|9|0|0|0|5|9|J|B|0|0|1W|5|22|5|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|13|8|@$K|14|L|15|M|N]|$K|16|L|17|M|N]|$K|18|L|19|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|1B|8|@]|9|@]|A|$G|H]]|$1|S|3|T|5|6|7|1C|8|@$K|1D|L|1E|M|N]|$K|1F|L|1G|M|N]]|9|@]|A|$]]|$1|U|3|V|5|F|7|1H|8|@]|9|@]|A|$G|H]]|$1|W|3|X|5|6|7|1I|8|@$K|1J|L|1K|M|N]|$K|1L|L|1M|M|N]]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]]|Z|$]]

Ok, while the solution I'm using is admittedly ad-hoc, it works, and scales better than the answers so far. Basically it's just the method VR17 suggested, but with a little more so that the tab size scales with the data set, and isn't just hard coded in. 

First I made a method that returns the number characters in some number.

<pre><code>def charNum(number):
 return math.floor(math.log(number,10)+1)
</code></pre>

Then I used the <code>charNum()</code> function on the last point of my <code>lower</code> and <code>upper</code> data sets. Only the last point had to be used on each list because the last point is the biggest number. I then counted the character that weren't numbers(the dash, semicolon, and space), and adjusted accordingly. 
So the final tabLength variable looks like this:

<pre><code>tabLength = charNum(lower[-1])+charNum(upper[-1])+3
</code></pre>

I then plugged the <code>tabLength</code> variable into the <code>expandTab()</code> function to get proper spacing. Here's some example output:

<pre><code>1-11: *******
12-22: *
23-33: ***
34-44: **
45-55: ***
56-66: *

99-249: *****
250-400: ****
401-551: **
552-702: **
703-853: *
854-1004: ***

99-200079: ******
200080-400060: **
400061-600041: ****
600042-800022: **
800023-1000003: *
</code></pre>

The only problem I can really see with this is that if I wanted to expand this to a table or something the tabs would be all funky. If I did that though, I'd probably look into <code>ljust</code> and <code>rjust</code>which I'm not all that familiar with right now. I'll leave the question open for a little while in case someone comes up with a better answer.

blocks|key|81568|text|str.format已经可以指定对齐方式。您可以使用{0:>5}来完成此操作；这将使参数0向右对齐5个字符。然后，我们可以使用相等地显示所有数字所需的最大位数来动态构建格式字符串：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|81569|>>>+lower+=+[70,+79,+88,+97,+106,+115]
>>>+upper+=+[78,+87,+96,+105,+114,+123]
>>>+num+=+[5,+3,+4,+2,+6,+4]
>>>+digits+=+len(str(max(lower+%2B+upper)))
>>>+digits
3
>>>+f+=+'{0:>%25d}-{1:>%25d}:+{2}'+%25+(digits,+digits)
>>>+f
'{0:>3}-{1:>3}:+{2}'
>>>+for+i+in+range(len(num)):
++++++++print(f.format(lower[i],+upper[i],+'*'+*+num[i]))

+70-+78:+*****
+79-+87:+***
+88-+96:+****
+97-105:+**
106-114:+******
115-123:+****|code-block|syntax|javascript|81570|实际上，您甚至可以在此处使用带有嵌套字段的单一格式字符串：|81571|>>>+for+i+in+range(len(num)):
++++++++print('{0:>{numLength}}-{1:>{numLength}}:+{2}'.format(lower[i],+upper[i],+'*'+*+num[i],+numLength=digits))|81572|entityMap^0|0|A|Q|6|18|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

<code>str.format</code> already has the possibility to specify alignment. You can do that using <code>{0:&gt;5}</code>; this would align parameter <code>0</code> to the right for 5 characters. We can then dynamically build a format string using the maximum number of digits necessary to display all numbers equally:

<pre><code>&gt;&gt;&gt; lower = [70, 79, 88, 97, 106, 115]
&gt;&gt;&gt; upper = [78, 87, 96, 105, 114, 123]
&gt;&gt;&gt; num = [5, 3, 4, 2, 6, 4]
&gt;&gt;&gt; digits = len(str(max(lower + upper)))
&gt;&gt;&gt; digits
3
&gt;&gt;&gt; f = '{0:&gt;%d}-{1:&gt;%d}: {2}' % (digits, digits)
&gt;&gt;&gt; f
'{0:&gt;3}-{1:&gt;3}: {2}'
&gt;&gt;&gt; for i in range(len(num)):
 print(f.format(lower[i], upper[i], '*' * num[i]))

 70- 78: *****
 79- 87: ***
 88- 96: ****
 97-105: **
106-114: ******
115-123: ****
</code></pre>

Actually, you could even use a single format string here with nested fields:

<pre><code>&gt;&gt;&gt; for i in range(len(num)):
 print('{0:&gt;{numLength}}-{1:&gt;{numLength}}: {2}'.format(lower[i], upper[i], '*' * num[i], numLength=digits))
</code></pre>

So I've got a function which creates a little star table based on some data collected elsewhere in the program. While the table produces the correct output, since the number of characters in each number changes, it un-aligns the table. For example,

<pre><code>70-78: *****
79-87: ***
88-96: ****
97-105: **
106-114: ******
115-123: ****
</code></pre>

Is there any way to make the stars align (hehe) so that the output is something like this:

<pre><code>70-78: *****
79-87: ***
88-96: ****
97-105: **
106-114: ******
115-123: ****
</code></pre>

Here's how I currently print the table.

<pre><code>for x in range(numClasses):
 print('{0}-{1}: {2}'.format(lower[x],upper[x],"*"*num[x]))
</code></pre>

How do I align text output in python?

所以我有一个函数，它根据程序中其他地方收集的数据创建一个小小的星表。虽然该表生成了正确的输出，但由于每个数字中的字符数发生了变化，因此它取消了该表的对齐。例如,70-78: *****79-87: ***88-96: ****97-105: **106-114: ******115-123: ****有没有办法让星星对...

问如何在python中对齐文本输出？
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问如何在python中对齐文本输出？
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