blocks|key|439289|text|from+math+import+ceil

num+=+0.1111111111000
num+=+ceil(num+*+100)+/+100.0|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|439290|请参见：|unstyled|439291|math.ceil+documentation|offset|length|style|CODE|439292|round+documentation+-无论如何，您都可能想要检查一下这篇文章，以便将来参考|439293|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/math.html#math.ceil|1|http://docs.python.org/library/functions.html#round^0|0|0|0|9|0|N|0|0|0|5|0|J|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|10|8|@$I|11|J|12|K|L]]|9|@$I|13|J|14|1|15]]|A|$]]|$1|M|3|N|5|F|7|16|8|@$I|17|J|18|K|L]]|9|@$I|19|J|1A|1|1B]]|A|$]]|$1|O|3|-4|5|F|7|1C|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]|W|$5|R|S|T|A|$U|X]]]]

<pre><code>from math import ceil

num = 0.1111111111000
num = ceil(num * 100) / 100.0
</code></pre>

See: 
<a href="http://docs.python.org/library/math.html#math.ceil" rel="noreferrer"><code>math.ceil</code> documentation</a> 
<a href="http://docs.python.org/library/functions.html#round" rel="noreferrer"><code>round</code> documentation</a> - You'll probably want to check this out anyway for future reference

blocks|key|439335|text|x+=+math.ceil(x+*+100.0)+/+100.0|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|439336|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|439395|text|从埃德温的答案推断：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|439396|from+math+import+ceil,+floor
def+float_round(num,+places+=+0,+direction+=+floor):
++++return+direction(num+*+(10**places))+/+float(10**places)|code-block|syntax|javascript|439397|要使用以下命令：|439398|>>>+float_round(0.21111,+3,+ceil)++#round+up
>>>+0.212
>>>+float_round(0.21111,+3)++++++++#round+down
>>>+0.211
>>>+float_round(0.21111,+3,+round)+#round+naturally
>>>+0.211|439399|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|$]]

Extrapolating from Edwin's answer:

<pre><code>from math import ceil, floor
def float_round(num, places = 0, direction = floor):
 return direction(num * (10**places)) / float(10**places)
</code></pre>

To use:

<pre><code>&gt;&gt;&gt; float_round(0.21111, 3, ceil) #round up
&gt;&gt;&gt; 0.212
&gt;&gt;&gt; float_round(0.21111, 3) #round down
&gt;&gt;&gt; 0.211
&gt;&gt;&gt; float_round(0.21111, 3, round) #round naturally
&gt;&gt;&gt; 0.211
</code></pre>

blocks|key|653906|text|请注意，ceil(num+*+100)+/+100技巧将在一些退化的输入上崩溃，如1e308。这可能不会经常出现，但我可以告诉你，这只是我花了几天的时间。为了避免这种情况，“如果”ceil()和floor()采用小数位参数就好了，就像round()那样……同时，有没有人知道一个干净的替代方案，不会像这样的输入崩溃？我曾对decimal包抱有一些希望，但它似乎也要破灭了：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|653907|>>>+from+math+import+ceil
>>>+from+decimal+import+Decimal,+ROUND_DOWN,+ROUND_UP
>>>+num+=+0.1111111111000
>>>+ceil(num+*+100)+/+100
0.12
>>>+float(Decimal(num).quantize(Decimal('.01'),+rounding=ROUND_UP))
0.12
>>>+num+=+1e308
>>>+ceil(num+*+100)+/+100
Traceback+(most+recent+call+last):
++File+"<string>",+line+301,+in+runcode
++File+"<interactive+input>",+line+1,+in+<module>
OverflowError:+cannot+convert+float+infinity+to+integer
>>>+float(Decimal(num).quantize(Decimal('.01'),+rounding=ROUND_UP))
Traceback+(most+recent+call+last):
++File+"<string>",+line+301,+in+runcode
++File+"<interactive+input>",+line+1,+in+<module>
decimal.InvalidOperation:+[<class+'decimal.InvalidOperation'>]|code-block|syntax|javascript|653908|当然，有人可能会说crash是这类输入上唯一合理的行为，但我认为导致问题的不是四舍五入，而是乘法(这就是为什么1e306不会崩溃)，一个更干净的四舍五入第n位fn的实现将避免乘法黑客。|653909|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.4/library/functions.html#round|1|https://docs.python.org/3.4/library/decimal.html^0|4|L|2I|6|2P|7|39|7|4H|7|39|7|0|4H|7|1|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]|$9|1A|A|1B|1|1C]]|E|$]]|$1|F|3|G|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1E|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

Note that the <code>ceil(num * 100) / 100</code> trick will crash on some degenerate inputs, like 1e308. This may not come up often but I can tell you it just cost me a couple of days. To avoid this, "it would be nice if" <code>ceil()</code> and <code>floor()</code> took a decimal places argument, like <a href="https://docs.python.org/3.4/library/functions.html#round" rel="noreferrer"><code>round()</code></a> does... Meanwhile, anyone know a clean alternative that won't crash on inputs like this? I had some hopes for the <a href="https://docs.python.org/3.4/library/decimal.html" rel="noreferrer"><code>decimal</code></a> package but it seems to die too:

<pre><code>&gt;&gt;&gt; from math import ceil
&gt;&gt;&gt; from decimal import Decimal, ROUND_DOWN, ROUND_UP
&gt;&gt;&gt; num = 0.1111111111000
&gt;&gt;&gt; ceil(num * 100) / 100
0.12
&gt;&gt;&gt; float(Decimal(num).quantize(Decimal('.01'), rounding=ROUND_UP))
0.12
&gt;&gt;&gt; num = 1e308
&gt;&gt;&gt; ceil(num * 100) / 100
Traceback (most recent call last):
 File "&lt;string&gt;", line 301, in runcode
 File "&lt;interactive input&gt;", line 1, in &lt;module&gt;
OverflowError: cannot convert float infinity to integer
&gt;&gt;&gt; float(Decimal(num).quantize(Decimal('.01'), rounding=ROUND_UP))
Traceback (most recent call last):
 File "&lt;string&gt;", line 301, in runcode
 File "&lt;interactive input&gt;", line 1, in &lt;module&gt;
decimal.InvalidOperation: [&lt;class 'decimal.InvalidOperation'&gt;]
</code></pre>

Of course one might say that crashing is the only sane behavior on such inputs, but I would argue that it's not the rounding but the multiplication that's causing the problem (that's why, eg, 1e306 doesn't crash), and a cleaner implementation of the round-up-nth-place fn would avoid the multiplication hack.

blocks|key|439724|text|下面是一个适用于任何数字的更通用的一行代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|439725|import+math
def+ceil(number,+digits)+->+float:+return+math.ceil((10.0+**+digits)+*+number)+/+(10.0+**+digits)|code-block|syntax|javascript|439726|示例用法：|439727|>>>+ceil(1.111111,+2)
1.12|439728|警告:如nimeshkiranverma所述：|439729|>>>+ceil(1.11,+2)+
1.12++#Because:+1.11+*+100.0+has+value+111.00000000000001|439730|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|$]]

Here is a more general one-liner that works for any digits:
<pre><code>import math
def ceil(number, digits) -&gt; float: return math.ceil((10.0 ** digits) * number) / (10.0 ** digits)
</code></pre>
Example usage:
<pre><code>&gt;&gt;&gt; ceil(1.111111, 2)
1.12
</code></pre>
Caveat: as stated by nimeshkiranverma:
<pre><code>&gt;&gt;&gt; ceil(1.11, 2) 
1.12 #Because: 1.11 * 100.0 has value 111.00000000000001
</code></pre>

blocks|key|439522|text|def+round_up(number,+ndigits=None):
++++#+start+by+just+rounding+the+number,+as+sometimes+this+rounds+it+up
++++result+=+round(number,+ndigits+if+ndigits+else+0)
++++if+result+<+number:
++++++++#+whoops,+the+number+was+rounded+down+instead,+so+correct+for+that
++++++++if+ndigits:
++++++++++++#+use+the+type+of+number+provided,+e.g.+float,+decimal,+fraction
++++++++++++Numerical+=+type(number)
++++++++++++#+add+the+digit+1+in+the+correct+decimal+place
++++++++++++result+%2B=+Numerical(10)+**+-ndigits
++++++++++++#+may+need+to+be+tweaked+slightly+if+the+addition+was+inexact
++++++++++++result+=+round(result,+ndigits)
++++++++else:
++++++++++++result+%2B=+1+#+same+as+10+**+-0+for+precision+of+zero+digits
++++return+result

assert+round_up(0.022499999999999999,+2)+==+0.03
assert+round_up(0.1111111111111000,+2)+==+0.12

assert+round_up(1.11,+2)+==+1.11
assert+round_up(1e308,+2)+==+1e308|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|439523|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|$]]

<pre><code>def round_up(number, ndigits=None):
 # start by just rounding the number, as sometimes this rounds it up
 result = round(number, ndigits if ndigits else 0)
 if result &lt; number:
 # whoops, the number was rounded down instead, so correct for that
 if ndigits:
 # use the type of number provided, e.g. float, decimal, fraction
 Numerical = type(number)
 # add the digit 1 in the correct decimal place
 result += Numerical(10) ** -ndigits
 # may need to be tweaked slightly if the addition was inexact
 result = round(result, ndigits)
 else:
 result += 1 # same as 10 ** -0 for precision of zero digits
 return result

assert round_up(0.022499999999999999, 2) == 0.03
assert round_up(0.1111111111111000, 2) == 0.12

assert round_up(1.11, 2) == 1.11
assert round_up(1e308, 2) == 1e308
</code></pre>

blocks|key|439589|text|python+round函数的取整方式可能会超出您的预期。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|439590|通过使用Decimal.quantize，您可以更具体地了解舍入方法|439591|例如：|439592|from+decimal+import+Decimal,+ROUND_HALF_UP
res+=+Decimal('0.25').quantize(Decimal('0.0'),+rounding=ROUND_HALF_UP)
print(res)+
#+prints+0.3|code-block|syntax|javascript|439593|更多参考：|439594|https://gist.github.com/jackiekazil/6201722|439595|entityMap|0|LINK|mutability|MUTABLE|url^0|7|5|0|4|G|0|0|0|0|0|17|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|15|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|16|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|17|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|18|8|@]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|S|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|R]]]]

The python <code>round</code> function could be rounding the way not you expected.

You can be more specific about the rounding method by using <code>Decimal.quantize</code>

eg.

<pre><code>from decimal import Decimal, ROUND_HALF_UP
res = Decimal('0.25').quantize(Decimal('0.0'), rounding=ROUND_HALF_UP)
print(res) 
# prints 0.3
</code></pre>

More reference:

<a href="https://gist.github.com/jackiekazil/6201722" rel="nofollow noreferrer">https://gist.github.com/jackiekazil/6201722</a>

blocks|key|654100|text|这里有一个我在其他答案中看不到的简单方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|654101|向上舍入到小数点后第二位：|654102|>>>+n+=+0.022499999999999999
>>>+
>>>+-(-n//.01)+*+.01
0.03
>>>+|code-block|syntax|javascript|654103|其他值：|654104|>>>+n+=+0.1111111111111000
>>>+
>>>+-(-n//.01)+*+.01
0.12
>>>+|654105|对于浮点数，偶尔会有一些不精确的值，如果您正在显示值，则可以对其进行校正，例如：|654106|>>>+n+=+10.1111111111111000
>>>+
>>>+-(-n//0.01)+*+0.01
10.120000000000001
>>>+
>>>+f"{-(-n//0.01)+*+0.01:.2f}"
'10.12'
>>>+|654107|一个简单的四舍五入函数，带有指定精度的参数：|654108|>>>+roundup+=+lambda+n,+p:+-(-n//10**-p)+*+10**-p
>>>+
>>>+#+Or+if+you+want+to+ensure+truncation+using+the+f-string+method:
>>>+roundup+=+lambda+n,+p:+float(f"{-(-n//10**-p)+*+10**-p:.{p}f}")
>>>+
>>>+roundup(0.111111111,+2)
0.12
>>>+roundup(0.111111111,+3)
0.112|654109|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Here's a simple way to do it that I don't see in the other answers. 

To round up to the second decimal place:

<pre><code>&gt;&gt;&gt; n = 0.022499999999999999
&gt;&gt;&gt; 
&gt;&gt;&gt; -(-n//.01) * .01
0.03
&gt;&gt;&gt; 
</code></pre>

Other value:

<pre><code>&gt;&gt;&gt; n = 0.1111111111111000
&gt;&gt;&gt; 
&gt;&gt;&gt; -(-n//.01) * .01
0.12
&gt;&gt;&gt; 
</code></pre>

With floats there's the occasional value with some minute imprecision, which can be corrected for if you're displaying the values for instance:

<pre><code>&gt;&gt;&gt; n = 10.1111111111111000
&gt;&gt;&gt; 
&gt;&gt;&gt; -(-n//0.01) * 0.01
10.120000000000001
&gt;&gt;&gt; 
&gt;&gt;&gt; f"{-(-n//0.01) * 0.01:.2f}"
'10.12'
&gt;&gt;&gt; 
</code></pre>

A simple roundup function with a parameter to specify precision:

<pre><code>&gt;&gt;&gt; roundup = lambda n, p: -(-n//10**-p) * 10**-p
&gt;&gt;&gt; 
&gt;&gt;&gt; # Or if you want to ensure truncation using the f-string method:
&gt;&gt;&gt; roundup = lambda n, p: float(f"{-(-n//10**-p) * 10**-p:.{p}f}")
&gt;&gt;&gt; 
&gt;&gt;&gt; roundup(0.111111111, 2)
0.12
&gt;&gt;&gt; roundup(0.111111111, 3)
0.112
</code></pre>

How can I round up a number to the second decimal place in python? For example:

<pre><code>0.022499999999999999
</code></pre>

Should round up to <code>0.03</code>

<pre><code>0.1111111111111000
</code></pre>

Should round up to <code>0.12</code>

If there is any value in the third decimal place, I want it to always round up leaving me 2 values behind the decimal point.

Round up to Second Decimal Place in Python

在python中，如何将一个数字四舍五入到小数点后第二位？例如：0.022499999999999999应四舍五入为0.030.1111111111111000应四舍五入为0.12如果在小数点的第三位有任何值，我希望它总是四舍五入，在小数点后留下两个值。

问向上舍入到Python中的第二位小数
EN

回答 8

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问向上舍入到Python中的第二位小数EN

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