blocks|key|2451763|text|像这样怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2451764|def+round_accordingly(num):

++++if+num>=500:
++++++++return+str(round(num))
++++elif+num>=100:
++++++++return+str(round(num,+1))
++++elif+num>=5:
++++++++return+str(round(num,+2))


print(round_accordingly(550.73424))
print(round_accordingly(200.234245))
print(round_accordingly(7.564743))|code-block|syntax|javascript|2451765|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 something like this? 

<pre><code>def round_accordingly(num):

 if num&gt;=500:
 return str(round(num))
 elif num&gt;=100:
 return str(round(num, 1))
 elif num&gt;=5:
 return str(round(num, 2))


print(round_accordingly(550.73424))
print(round_accordingly(200.234245))
print(round_accordingly(7.564743))
</code></pre>

blocks|key|2451793|text|您可以为高度可定制的rounding使用Decimal+module|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2451794|import+decimal+

def+rnd(f,n,r=decimal.ROUND_HALF_DOWN):
++++pat='1.'%2B'0'*n
++++return+str(decimal.Decimal(str(f)).quantize(decimal.Decimal(pat),r))|code-block|syntax|javascript|2451795|现在来测试一下：|2451796|for+n+in+(500.99,100.39,5.019):
++++for+i+in+(0,1,2):
++++++++print('n={:10},+i={:2},+Down:+{:>10},+Closest:+{:>10}'.format(n,i,+rnd(n,i,decimal.ROUND_FLOOR),rnd(n,i)))+++|2451797|打印：|2451798|n=++++500.99,+i=+0,+Down:++++++++500,+Closest:++++++++501
n=++++500.99,+i=+1,+Down:++++++500.9,+Closest:++++++501.0
n=++++500.99,+i=+2,+Down:+++++500.99,+Closest:+++++500.99
n=++++100.39,+i=+0,+Down:++++++++100,+Closest:++++++++100
n=++++100.39,+i=+1,+Down:++++++100.3,+Closest:++++++100.4
n=++++100.39,+i=+2,+Down:+++++100.39,+Closest:+++++100.39
n=+++++5.019,+i=+0,+Down:++++++++++5,+Closest:++++++++++5
n=+++++5.019,+i=+1,+Down:++++++++5.0,+Closest:++++++++5.0
n=+++++5.019,+i=+2,+Down:+++++++5.01,+Closest:+++++++5.02|2451799|(注:这是1个十进制数的情况下的不同输出和100.39的输入。请说明在这种情况下100.2的期望输出是什么？输出不应该是100.3吗？)|style|CODE|2451800|2451801|有了这些说明，我认为您只需要使用fmod|2451802|import+math+

def+rnd(n,f,prec,direction='down'):
++++if+direction=='down':
++++++++if+prec==0:
++++++++++++return+str(int(n))
++++++++return+str(round(n-math.fmod(n,f),prec))
++++else:
++++++++if+prec==0:
++++++++++++return+str(int(round(n)))
++++++++n%2B=f
++++++++return+str(round(n-math.fmod(n,f),prec))|2451803|适用于所有给定的示例：|2451804|>>>+rnd(500.99,0.1,0)
'500'
>>>+rnd(500.99,0.1,0,'')
'501'
>>>+rnd(100.39,0.2,1,'')
'100.4'
>>>+rnd(100.39,0.2,1)
'100.2'
>>>+rnd(100.59,0.2,1,'')
'100.6'
>>>+rnd(100.59,0.2,1)
'100.4'
>>>+rnd(5.019,0.01,2)
'5.01'
>>>+rnd(5.019,0.01,2,'')
'5.02'|2451805|(使用'down'以外的任何字符串值调用带有direction=关键字参数的函数即可启动。根据您的喜好更改...)|2451806|2451807|编辑2|2451808|这将返回最接近输入数字的方法：|2451809|def+rnd(n,f,prec):
++++n1=int(n)+if+prec==0+else+round(n-math.fmod(n,f),prec)
++++n2=float(int(round(n)))+if+prec==0+else+round((n%2Bf)-math.fmod((n%2Bf),f),prec)
++++return+str(min((n1,n2),+key=lambda+e:+math.fabs(e-n)))|2451810|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.6/library/decimal.html#rounding-modes|1|https://docs.python.org/2/library/decimal.html^0|A|8|0|K|E|1|0|0|0|0|0|0|L|6|14|5|1O|5|0|0|G|4|0|0|0|0|3|6|M|A|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1M|8|@]|9|@$A|1N|B|1O|1|1P]|$A|1Q|B|1R|1|1S]]|C|$]]|$1|D|3|E|5|F|7|1T|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1U|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|1V|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|1W|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|1X|8|@]|9|@]|C|$G|H]]|$1|Q|3|R|5|6|7|1Y|8|@$A|1Z|B|20|S|T]|$A|21|B|22|S|T]|$A|23|B|24|S|T]]|9|@]|C|$]]|$1|U|3|-4|5|6|7|25|8|@]|9|@]|C|$]]|$1|V|3|W|5|6|7|26|8|@$A|27|B|28|S|T]]|9|@]|C|$]]|$1|X|3|Y|5|F|7|29|8|@]|9|@]|C|$G|H]]|$1|Z|3|10|5|6|7|2A|8|@]|9|@]|C|$]]|$1|11|3|12|5|F|7|2B|8|@]|9|@]|C|$G|H]]|$1|13|3|14|5|6|7|2C|8|@$A|2D|B|2E|S|T]|$A|2F|B|2G|S|T]]|9|@]|C|$]]|$1|15|3|-4|5|6|7|2H|8|@]|9|@]|C|$]]|$1|16|3|17|5|6|7|2I|8|@]|9|@]|C|$]]|$1|18|3|19|5|6|7|2J|8|@]|9|@]|C|$]]|$1|1A|3|1B|5|F|7|2K|8|@]|9|@]|C|$G|H]]|$1|1C|3|-4|5|6|7|2L|8|@]|9|@]|C|$]]]|1D|$1E|$5|1F|1G|1H|C|$1I|1J]]|1K|$5|1F|1G|1H|C|$1I|1L]]]]

You can use the <a href="https://docs.python.org/2/library/decimal.html" rel="nofollow noreferrer">Decimal module</a> for highly customizable <a href="https://docs.python.org/3.6/library/decimal.html#rounding-modes" rel="nofollow noreferrer">rounding</a>:

<pre><code>import decimal 

def rnd(f,n,r=decimal.ROUND_HALF_DOWN):
 pat='1.'+'0'*n
 return str(decimal.Decimal(str(f)).quantize(decimal.Decimal(pat),r))
</code></pre>

Now to test:

<pre><code>for n in (500.99,100.39,5.019):
 for i in (0,1,2):
 print('n={:10}, i={:2}, Down: {:&gt;10}, Closest: {:&gt;10}'.format(n,i, rnd(n,i,decimal.ROUND_FLOOR),rnd(n,i))) 
</code></pre>

Prints:

<pre><code>n= 500.99, i= 0, Down: 500, Closest: 501
n= 500.99, i= 1, Down: 500.9, Closest: 501.0
n= 500.99, i= 2, Down: 500.99, Closest: 500.99
n= 100.39, i= 0, Down: 100, Closest: 100
n= 100.39, i= 1, Down: 100.3, Closest: 100.4
n= 100.39, i= 2, Down: 100.39, Closest: 100.39
n= 5.019, i= 0, Down: 5, Closest: 5
n= 5.019, i= 1, Down: 5.0, Closest: 5.0
n= 5.019, i= 2, Down: 5.01, Closest: 5.02
</code></pre>

(Note: This is different output in the case of 1 decimal and the input of <code>100.39</code>. Please clarify the desired output of <code>100.2</code> in that case? Shouldn't the output be <code>100.3</code>?)

<hr>

With the clarification, I think you just need to use <code>fmod</code>:

<pre><code>import math 

def rnd(n,f,prec,direction='down'):
 if direction=='down':
 if prec==0:
 return str(int(n))
 return str(round(n-math.fmod(n,f),prec))
 else:
 if prec==0:
 return str(int(round(n)))
 n+=f
 return str(round(n-math.fmod(n,f),prec))
</code></pre>

Works with all examples given:

<pre><code>&gt;&gt;&gt; rnd(500.99,0.1,0)
'500'
&gt;&gt;&gt; rnd(500.99,0.1,0,'')
'501'
&gt;&gt;&gt; rnd(100.39,0.2,1,'')
'100.4'
&gt;&gt;&gt; rnd(100.39,0.2,1)
'100.2'
&gt;&gt;&gt; rnd(100.59,0.2,1,'')
'100.6'
&gt;&gt;&gt; rnd(100.59,0.2,1)
'100.4'
&gt;&gt;&gt; rnd(5.019,0.01,2)
'5.01'
&gt;&gt;&gt; rnd(5.019,0.01,2,'')
'5.02'
</code></pre>

(Call the function with the <code>direction=</code> keyword argument with any string value other than <code>'down'</code> to get up. Change to your liking...)

<hr>

<h1>Edit 2</h1>

This returns the method closest to the input number:

<pre><code>def rnd(n,f,prec):
 n1=int(n) if prec==0 else round(n-math.fmod(n,f),prec)
 n2=float(int(round(n))) if prec==0 else round((n+f)-math.fmod((n+f),f),prec)
 return str(min((n1,n2), key=lambda e: math.fabs(e-n)))
</code></pre>

I've found many posts regarding rounding in python, but from my understanding none so far solves my problem.

Two solutions I'm looking for:

<ul>
<li>Round closest: float/string numbers which need to be rounded differently depending on their size.</li>
<li>Round down: float/string numbers which need to be rounded down differently depending on their size.</li>
<li>Round up: float/string numbers which need to be rounded up differently depending on their size.</li>
</ul>

Example:

<ul>
<li>+500: 0 decimals (whole numbers) (e.g., Closest: 500.5 -> 501; Down: 500.99 -> 500; Up: 500.01 -> 501)</li>
<li>+100: 1 decimals (fraction: 0.2) (e.g., Closest: 100.29 -> 100.2; Down: 100.39 -> 100.2; Up: 100.21 -> 100.4). So, the value must end with .0, .2, .4, .6 or .8</li>
<li>+5: 2 decimals (fraction: 0.01) (e.g., Closest: 5.019 -> 5.02; Down: 5.019 -> 5.01; Up: 5.011 -> 5.02). So, the value must end with .x0, .x1, .x2, .x3, .x4, .x5, .x6, .x7, .x8 or .x9</li>
</ul>

The resulting variable should be a string.

The neat thing I'm, looking for is a function in which i can pass number of decimals and fraction

python: round/format float/string to specific decimal/fraction

我找到了很多关于python中四舍五入的帖子，但据我所知，到目前为止还没有解决我的问题。我正在寻找两个解决方案：示例：+ 500.5 :0小数(整数)(例如，最近:500.5 -> 501；向下: 500.99 -> 500；向上: 500.01 -> 501)+100: 1小数(分数: 0.2) (例如，最近: 10...

问python:四舍五入/格式化浮点数/字符串到特定的小数/分数
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