blocks|key|1327357|text|我不知道Python中有没有标准函数，但这对我来说很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1327358|Python+3|1327359|def+myround(x,+base=5):
++++return+base+*+round(x/base)|code-block|syntax|javascript|1327360|很容易看出为什么上面的方法是有效的。你想要确保你的数字除以5是一个整数，并且是正确的四舍五入。所以，我们首先做这件事(round(x/5))，然后因为我们除以5，所以我们也乘以5。|offset|length|style|CODE|1327361|我为该函数提供了一个base参数，默认为5，从而使该函数更加通用。|1327362|Python+2|1327363|在Python2中，需要使用float(x)来确保/执行浮点除法，并且需要最终转换为int，因为在Python2中round()返回一个浮点值。|1327364|def+myround(x,+base=5):
++++return+int(base+*+round(float(x)/base))|1327365|entityMap^0|0|0|0|1N|A|0|A|4|0|0|E|8|P|1|16|3|1L|7|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@$K|12|L|13|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@$K|15|L|16|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|17|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|18|8|@$K|19|L|1A|M|N]|$K|1B|L|1C|M|N]|$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]]|$1|W|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|X|$]]

I don't know of a standard function in Python, but this works for me:
<h3>Python 3</h3>
<pre><code>def myround(x, base=5):
 return base * round(x/base)
</code></pre>
It is easy to see why the above works. You want to make sure that your number divided by 5 is an integer, correctly rounded. So, we first do exactly that (<code>round(x/5)</code>), and then since we divided by 5, we multiply by 5 as well.
I made the function more generic by giving it a <code>base</code> parameter, defaulting to 5.
<h3>Python 2</h3>
In Python 2, <code>float(x)</code> would be needed to ensure that <code>/</code> does floating-point division, and a final conversion to <code>int</code> is needed because <code>round()</code> returns a floating-point value in Python 2.
<pre><code>def myround(x, base=5):
 return int(base * round(float(x)/base))
</code></pre>

blocks|key|1112304|text|这只是一个缩放的问题|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112305|>>>+a=[10,11,12,13,14,15,16,17,18,19,20]
>>>+for+b+in+a:
...+++++int(round(b/5.0)*5.0)
...+
10
10
10
15
15
15
15
15
20
20
20|code-block|syntax|javascript|1112306|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|$]]

It's just a matter of scaling 

<pre><code>&gt;&gt;&gt; a=[10,11,12,13,14,15,16,17,18,19,20]
&gt;&gt;&gt; for b in a:
... int(round(b/5.0)*5.0)
... 
10
10
10
15
15
15
15
15
20
20
20
</code></pre>

blocks|key|1112398|text|round(x，n)：值被舍入到最接近的10的倍数减去n的幂。所以如果n是负数...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112399|def+round5(x):
++++return+int(round(x*2,+-1))+/+2|code-block|syntax|javascript|1112400|由于10+=5*+2，所以您可以使用整数除法和乘法与2，而不是浮点除法和乘法与5.0。这并不重要，除非您喜欢位移位|1112401|def+round5(x):
++++return+int(round(x+<<+1,+-1))+>>+1|1112402|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|$]]

round(x[, n]): values are rounded to the closest multiple of 10 to the power minus n. So if n is negative...

<pre><code>def round5(x):
 return int(round(x*2, -1)) / 2
</code></pre>

Since 10 = 5 * 2, you can use integer division and multiplication with 2, rather than float division and multiplication with 5.0. Not that that matters much, unless you like bit shifting

<pre><code>def round5(x):
 return int(round(x &lt;&lt; 1, -1)) &gt;&gt; 1
</code></pre>

blocks|key|1112407|text|这样如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112408|+def+divround(value,+step):
+++++return+divmod(value,+step)[0]+*+step|code-block|syntax|javascript|1112409|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|$]]

What about this:

<pre><code> def divround(value, step):
 return divmod(value, step)[0] * step
</code></pre>

blocks|key|1112447|text|divround的修改版本:-)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112448|def+divround(value,+step,+barrage):
++++result,+rest+=+divmod(value,+step)
++++return+result*step+if+rest+<+barrage+else+(result%2B1)*step|code-block|syntax|javascript|1112449|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|$]]

Modified version of divround :-)

<pre><code>def divround(value, step, barrage):
 result, rest = divmod(value, step)
 return result*step if rest &lt; barrage else (result+1)*step
</code></pre>

blocks|key|1327627|text|移除“rest”将会起作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1327628|rounded+=+int(val)+-+int(val)+%25+5|code-block|syntax|javascript|1327629|如果该值是一个整数：|1327630|rounded+=+val+-+val+%25+5|1327631|作为函数：|1327632|def+roundint(value,+base=5):
++++return+int(value)+-+int(value)+%25+int(base)|1327633|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|$]]

Removing the 'rest' would work:

<pre><code>rounded = int(val) - int(val) % 5
</code></pre>

If the value is aready an integer:

<pre><code>rounded = val - val % 5
</code></pre>

As a function:

<pre><code>def roundint(value, base=5):
 return int(value) - int(value) % int(base)
</code></pre>

blocks|key|1327696|text|对于舍入为非整数值，例如0.05：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1327697|def+myround(x,+prec=2,+base=.05):
++return+round(base+*+round(float(x)/base),prec)|code-block|syntax|javascript|1327698|我发现这很有用，因为我可以在代码中进行搜索和替换，将"round(“更改为"myround(”)，而不必更改参数值。|1327699|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

For rounding to non-integer values, such as 0.05:

<pre><code>def myround(x, prec=2, base=.05):
 return round(base * round(float(x)/base),prec)
</code></pre>

I found this useful since I could just do a search and replace in my code to change "round(" to "myround(", without having to change the parameter values.

blocks|key|1112657|text|您可以通过将0.5添加到传递给int()的数字来“欺骗”int()进行四舍五入，而不是四舍五入。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1112658|entityMap^0|6|3|F|5|S|5|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

You can “trick” <code>int()</code> into rounding off instead of rounding down by adding <code>0.5</code> to the 
number you pass to <code>int()</code>.

blocks|key|1327873|text|对不起，我想对Alok+Singhai的回答发表评论，但由于缺乏声誉，它不会让我发表评论=/|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1327874|无论如何，我们可以再概括一步，然后开始：|1327875|def+myround(x,+base=5):
++++return+base+*+round(float(x)+/+base)|code-block|syntax|javascript|1327876|这允许我们使用非整数基，如.25或任何其他小数基。|offset|length|style|CODE|1327877|entityMap^0|0|0|0|D|3|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Sorry, I wanted to comment on Alok Singhai's answer, but it won't let me due to a lack of reputation =/

Anyway, we can generalize one more step and go:

<pre><code>def myround(x, base=5):
 return base * round(float(x) / base)
</code></pre>

This allows us to use non-integer bases, like <code>.25</code> or any other fractional base.

blocks|key|1327951|text|使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1327952|>>>+def+round_to_nearest(n,+m):
++++++++r+=+n+%25+m
++++++++return+n+%2B+m+-+r+if+r+%2B+r+>=+m+else+n+-+r|code-block|syntax|javascript|1327953|它不使用乘法，也不会将浮点数转换为浮点数。|1327954|舍入到最接近的10的倍数：|1327955|>>>+for+n+in+range(-21,+30,+3):+print('{:3d}++=>++{:3d}'.format(n,+round_to_nearest(n,+10)))
-21++=>++-20
-18++=>++-20
-15++=>++-10
-12++=>++-10
+-9++=>++-10
+-6++=>++-10
+-3++=>++++0
++0++=>++++0
++3++=>++++0
++6++=>+++10
++9++=>+++10
+12++=>+++10
+15++=>+++20
+18++=>+++20
+21++=>+++20
+24++=>+++20
+27++=>+++30|1327956|如你所见，它对负数和正数都有效。平局(例如-15和15)将始终向上舍入。|1327957|一个类似的例子，四舍五入到最接近的5的倍数，表明它的行为也与预期的不同的“基数”：|1327958|>>>+for+n+in+range(-21,+30,+3):+print('{:3d}++=>++{:3d}'.format(n,+round_to_nearest(n,+5)))
-21++=>++-20
-18++=>++-20
-15++=>++-15
-12++=>++-10
+-9++=>++-10
+-6++=>+++-5
+-3++=>+++-5
++0++=>++++0
++3++=>++++5
++6++=>++++5
++9++=>+++10
+12++=>+++10
+15++=>+++15
+18++=>+++20
+21++=>+++20
+24++=>+++25
+27++=>+++25|1327959|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]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Use:

<pre><code>&gt;&gt;&gt; def round_to_nearest(n, m):
 r = n % m
 return n + m - r if r + r &gt;= m else n - r
</code></pre>

It does not use multiplication and will not convert from/to floats.

Rounding to the nearest multiple of 10:

<pre><code>&gt;&gt;&gt; for n in range(-21, 30, 3): print('{:3d} =&gt; {:3d}'.format(n, round_to_nearest(n, 10)))
-21 =&gt; -20
-18 =&gt; -20
-15 =&gt; -10
-12 =&gt; -10
 -9 =&gt; -10
 -6 =&gt; -10
 -3 =&gt; 0
 0 =&gt; 0
 3 =&gt; 0
 6 =&gt; 10
 9 =&gt; 10
 12 =&gt; 10
 15 =&gt; 20
 18 =&gt; 20
 21 =&gt; 20
 24 =&gt; 20
 27 =&gt; 30
</code></pre>

As you can see, it works for both negative and positive numbers. Ties (e.g. -15 and 15) will always be rounded upwards.

A similar example that rounds to the nearest multiple of 5, demonstrating that it also behaves as expected for a different "base":

<pre><code>&gt;&gt;&gt; for n in range(-21, 30, 3): print('{:3d} =&gt; {:3d}'.format(n, round_to_nearest(n, 5)))
-21 =&gt; -20
-18 =&gt; -20
-15 =&gt; -15
-12 =&gt; -10
 -9 =&gt; -10
 -6 =&gt; -5
 -3 =&gt; -5
 0 =&gt; 0
 3 =&gt; 5
 6 =&gt; 5
 9 =&gt; 10
 12 =&gt; 10
 15 =&gt; 15
 18 =&gt; 20
 21 =&gt; 20
 24 =&gt; 25
 27 =&gt; 25
</code></pre>

blocks|key|1327967|text|def+round_to_next5(n):
++++return+n+%2B+(5+-+n)+%25+5|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1327968|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_to_next5(n):
 return n + (5 - n) % 5
</code></pre>

blocks|key|1112846|text|这是我的C代码。如果我理解正确的话，它应该是这样的；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112847|#include+<stdio.h>

int+main(){
int+number;

printf("Enter+number:+\n");
scanf("%25d"+,+&number);

if(number%255+==+0)
++++printf("It+is+multiple+of+5\n");
else{
++++while(number%255+!=+0)
++++++++number%2B%2B;
++printf("%25d\n",number);
++}
}|code-block|syntax|javascript|1112848|并且这也舍入到最接近的5的倍数，而不是仅仅向上舍入；|1112849|#include+<stdio.h>

int+main(){
int+number;

printf("Enter+number:+\n");
scanf("%25d"+,+&number);

if(number%255+==+0)
++++printf("It+is+multiple+of+5\n");
else{
++++while(number%255+!=+0)
++++++++if+(number%255+<+3)
++++++++++++number--;
++++++++else
++++++++number%2B%2B;
++printf("nearest+multiple+of+5+is:+%25d\n",number);
++}
}|1112850|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|$]]

Here is my C code. If I understand it correctly, it should supposed to be something like this;

<pre><code>#include &lt;stdio.h&gt;

int main(){
int number;

printf("Enter number: \n");
scanf("%d" , &amp;number);

if(number%5 == 0)
 printf("It is multiple of 5\n");
else{
 while(number%5 != 0)
 number++;
 printf("%d\n",number);
 }
}
</code></pre>

and this also rounds to nearest multiple of 5 instead of just rounding up;

<pre><code>#include &lt;stdio.h&gt;

int main(){
int number;

printf("Enter number: \n");
scanf("%d" , &amp;number);

if(number%5 == 0)
 printf("It is multiple of 5\n");
else{
 while(number%5 != 0)
 if (number%5 &lt; 3)
 number--;
 else
 number++;
 printf("nearest multiple of 5 is: %d\n",number);
 }
}
</code></pre>

blocks|key|1112876|text|对于整数和Python+3：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112877|def+divround_down(value,+step):
++++return+value//step*step


def+divround_up(value,+step):
++++return+(value%2Bstep-1)//step*step|code-block|syntax|javascript|1112878|制作：|1112879|>>>+[divround_down(x,5)+for+x+in+range(20)]
[0,+0,+0,+0,+0,+5,+5,+5,+5,+5,+10,+10,+10,+10,+10,+15,+15,+15,+15,+15]
>>>+[divround_up(x,5)+for+x+in+range(20)]
[0,+5,+5,+5,+5,+5,+10,+10,+10,+10,+10,+15,+15,+15,+15,+15,+20,+20,+20,+20]|1112880|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|$]]

For integers and with Python 3:

<pre><code>def divround_down(value, step):
 return value//step*step


def divround_up(value, step):
 return (value+step-1)//step*step
</code></pre>

Producing:

<pre><code>&gt;&gt;&gt; [divround_down(x,5) for x in range(20)]
[0, 0, 0, 0, 0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15]
&gt;&gt;&gt; [divround_up(x,5) for x in range(20)]
[0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15, 20, 20, 20, 20]
</code></pre>

blocks|key|1112928|text|另一种方法(没有显式的乘法或除法运算符)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112929|def+rnd(x,+b=5):
++++return+round(x+%2B+min(-(x+%25+b),+b+-+(x+%25+b),+key=abs))|code-block|syntax|javascript|1112930|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|$]]

Another way to do this (without explicit multiplication or division operators):

<pre><code>def rnd(x, b=5):
 return round(x + min(-(x % b), b - (x % b), key=abs))
</code></pre>

blocks|key|1112964|text|我猜还没有人真正写下这段代码，但你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1112965|round(12,+-1)+-->+10
round(18,+-1)+-->+20|code-block|syntax|javascript|1112966|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|$]]

No one actually wrote this yet I guess but you can do:
<pre><code>round(12, -1) --&gt; 10
round(18, -1) --&gt; 20
</code></pre>

blocks|key|1113024|text|def+round_up_to_base(x,+base=10):
++++return+x+%2B+(base+-+x)+%25+base

def+round_down_to_base(x,+base=10):
++++return+x+-+(x+%25+base)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1113025|这给了我们|unstyled|1113026|对于base=5|offset|length|style|CODE|1113027|>>>+[i+for+i+in+range(20)]
[0,+1,++2,++3,++4,++5,++6,++7,++8,++9,++10,+11,+12,+13,+14,+15,+16,+17,+18,+19]|1113028|>>>+[round_down_to_base(x=i,+base=5)+for+i+in+range(20)]
[0,+0,++0,++0,++0,++5,++5,++5,++5,++5,++10,+10,+10,+10,+10,+15,+15,+15,+15,+15]

>>>+[round_up_to_base(x=i,+base=5)+for+i+in+range(20)]
[0,+5,++5,++5,++5,++5,++10,+10,+10,+10,+10,+15,+15,+15,+15,+15,+20,+20,+20,+20]|1113029|对于base=10|1113030|>>>+[i+for+i+in+range(20)]
[0,+1,++2,++3,++4,++5,++6,++7,++8,++9,++10,+11,+12,+13,+14,+15,+16,+17,+18,+19]|1113031|>>>+[round_down_to_base(x=i,+base=10)+for+i+in+range(20)]
[0,+0,++0,++0,++0,++0,++0,++0,++0,++0,++10,+10,+10,+10,+10,+10,+10,+10,+10,+10]

>>>+[round_up_to_base(x=i,+base=10)+for+i+in+range(20)]
[0,+10,+10,+10,+10,+10,+10,+10,+10,+10,+10,+20,+20,+20,+20,+20,+20,+20,+20,+20]|1113032|在Python+3.7.9中测试|1113033|entityMap^0|0|0|2|6|0|0|0|2|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|11|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|12|8|@$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$B|C]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$B|C]]|$1|Q|3|R|5|F|7|17|8|@$I|18|J|19|K|L]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1A|8|@]|9|@]|A|$B|C]]|$1|U|3|V|5|6|7|1B|8|@]|9|@]|A|$B|C]]|$1|W|3|X|5|F|7|1C|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|F|7|1D|8|@]|9|@]|A|$]]]|Z|$]]

<pre class="lang-py prettyprint-override"><code>def round_up_to_base(x, base=10):
 return x + (base - x) % base

def round_down_to_base(x, base=10):
 return x - (x % base)
</code></pre>
which gives
for <code>base=5</code>:
<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; [i for i in range(20)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
</code></pre>
<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; [round_down_to_base(x=i, base=5) for i in range(20)]
[0, 0, 0, 0, 0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15]

&gt;&gt;&gt; [round_up_to_base(x=i, base=5) for i in range(20)]
[0, 5, 5, 5, 5, 5, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15, 20, 20, 20, 20]
</code></pre>
for <code>base=10</code>:
<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; [i for i in range(20)]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
</code></pre>
<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; [round_down_to_base(x=i, base=10) for i in range(20)]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10]

&gt;&gt;&gt; [round_up_to_base(x=i, base=10) for i in range(20)]
[0, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 20, 20, 20, 20, 20, 20, 20, 20, 20]
</code></pre>
tested in Python 3.7.9

blocks|key|1328263|text|我需要向下舍入到前面的5位。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1328264|示例16向下舍入为15或19向下舍入为15|1328265|下面是使用的代码|1328266|++++def+myround(x,segment):
++++++++preRound+=+x+/+segment
++++++++roundNum+=+int(preRound)
++++++++segVal+=+segment+*+roundNum
++++++++return+segVal|code-block|syntax|javascript|1328267|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I needed to round down to the preceding 5.
Example 16 rounds down to 15 or 19 rounds down to 15
Here's the code used
<pre><code> def myround(x,segment):
 preRound = x / segment
 roundNum = int(preRound)
 segVal = segment * roundNum
 return segVal
</code></pre>

Is there a built-in function that can round like the following?
<pre class="lang-none prettyprint-override"><code>10 -&gt; 10
12 -&gt; 10
13 -&gt; 15
14 -&gt; 15
16 -&gt; 15
18 -&gt; 20
</code></pre>

Round to 5 (or other number) in Python

Python

有没有一个内置的函数可以像下面这样舍入？10 -> 1012 -> 1013 -> 1514 -> 1516 -> 1518 -> 20

问在Python中四舍五入为5(或其他数字
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