entityMap|blocks|key|fdq0c|text|简单|type|unstyled|depth|inlineStyleRanges|entityRanges|data|fqlvc|print+int(x)|code-block|syntax|js|1srt5|也同样有效。^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|J|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|K|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|L|9|@]|A|@]|B|$]]]]

Simple

<pre><code>print int(x)
</code></pre>

will work as well.

entityMap|blocks|key|7p7n1|text|其中之一应该是有效的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|45nci|import+math
math.trunc(1.5)
>+1
math.trunc(-1.5)
>+-1
math.floor(1.5)
>+1
math.floor(-1.5)
>+-2|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

One of these should work:

<pre><code>import math
math.trunc(1.5)
&gt; 1
math.trunc(-1.5)
&gt; -1
math.floor(1.5)
&gt; 1
math.floor(-1.5)
&gt; -2
</code></pre>

entityMap|blocks|key|f6isb|text|x//1|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|js|e2jvl|The+//运算符返回除法的底数。因为除以1不会改变你的数字，所以这等同于floor，但不需要导入。备注：|unstyled|offset|length|style|CODE|2vrsf|这将返回一个浮点数|ordered-list-item|44p36|这轮到-∞^0|0|4|2|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|Q|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|R|9|@$H|S|I|T|J|K]]|A|@]|B|$]]|$2|L|4|M|6|N|8|U|9|@]|A|@]|B|$]]|$2|O|4|P|6|N|8|V|9|@]|A|@]|B|$]]]]

<pre><code>x//1
</code></pre>

The <code>//</code> operator returns the floor of the division. Since dividing by 1 doesn't change your number, this is equivalent to floor but no import is needed. 
Notes:

<ol>
<li>This returns a float</li>
<li>This rounds towards -∞</li>
</ol>

entityMap|blocks|key|76fgh|text|要获得浮点结果，只需使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4ea7u|round(x-0.5)|code-block|syntax|js|7aca3|它也适用于负数。^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|J|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|K|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|L|9|@]|A|@]|B|$]]]]

To get floating point result simply use:

<pre><code>round(x-0.5)
</code></pre>

It works for negative numbers as well.

entityMap|0|type|LINK|mutability|MUTABLE|data|url|http://docs.python.org/2/library/math.html|blocks|key|br3o2|text|我认为你需要一个floor函数：|unstyled|depth|inlineStyleRanges|entityRanges|al9e9|math.floor(x)|offset|length^0|0|0|D|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|M|G|@]|H|@]|6|$]]|$A|I|C|J|2|E|F|N|G|@]|H|@$K|O|L|P|A|Q]]|6|$]]]]

I think you need a floor function :

<a href="http://docs.python.org/2/library/math.html">math.floor(x)</a>

entityMap|blocks|key|bv5bh|text|很多人说要用int(x)，这在大多数情况下工作得很好，但有一个小问题。如果OP的结果是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|advb6|x+=+1.9999999999999999|code-block|syntax|js|9r6ek|它将四舍五入为|al93a|x+=+2|e6si7|在16号9号之后它就会转了。如果你确定你永远不会遇到这样的事情，这没什么大不了的。但这是需要牢记的事情。^0|6|6|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|R|9|@$A|S|B|T|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|U|9|@]|E|@]|F|$J|K]]|$2|L|4|M|6|7|8|V|9|@]|E|@]|F|$]]|$2|N|4|O|6|I|8|W|9|@]|E|@]|F|$J|K]]|$2|P|4|Q|6|7|8|X|9|@]|E|@]|F|$]]]]

a lot of people say to use <code>int(x)</code>, and this works ok for most cases, but there is a little problem. If OP's result is:

<pre><code>x = 1.9999999999999999
</code></pre>

it will round to

<pre><code>x = 2
</code></pre>

after the 16th 9 it will round. This is not a big deal if you are sure you will never come across such thing. But it's something to keep in mind.

entityMap|blocks|key|6nf25|text|如果你不想导入数学，你可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4svg2|int(round(x))|offset|length|style|CODE|7shpo|这里有一段文档：|5747l|>>>+help(round)
Help+on+built-in+function+round+in+module+__builtin__:

round(...)
++++round(number[,+ndigits])+->+floating+point+number

++++Round+a+number+to+a+given+precision+in+decimal+digits+(default+0+digits).
++++This+always+returns+a+floating+point+number.++Precision+may+be+negative.|code-block|syntax|javascript^0|0|0|D|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|P|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|Q|9|@$E|R|F|S|G|H]]|A|@]|B|$]]|$2|I|4|J|6|7|8|T|9|@]|A|@]|B|$]]|$2|K|4|L|6|M|8|U|9|@]|A|@]|B|$N|O]]]]

If you don't want to import math, you could use:

<code>int(round(x))</code>

Here's a piece of documentation:

<pre><code>&gt;&gt;&gt; help(round)
Help on built-in function round in module __builtin__:

round(...)
 round(number[, ndigits]) -&gt; floating point number

 Round a number to a given precision in decimal digits (default 0 digits).
 This always returns a floating point number. Precision may be negative.
</code></pre>

entityMap|blocks|key|a9loo|text|如果你使用numpy，你可以使用下面的解决方案，它也适用于负数(它也适用于数组)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|f5600|import+numpy+as+np
def+round_down(num):
++++if+num+<+0:
++++++++return+-np.ceil(abs(num))
++++else:
++++++++return+np.int32(num)
round_down+=+np.vectorize(round_down)|code-block|syntax|javascript|1apok|round_down([-1.1,+-1.5,+-1.6,+0,+1.1,+1.5,+1.6])
>+array([-2.,+-2.,+-2.,++0.,++1.,++1.,++1.])|260rd|我认为如果您只使用math模块，而不是numpy模块。|offset|length|style|CODE^0|0|0|0|9|4|J|5^^$0|$]|1|@$2|3|4|5|6|7|8|P|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|Q|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|E|8|R|9|@]|A|@]|B|$F|G]]|$2|J|4|K|6|7|8|S|9|@$L|T|M|U|N|O]|$L|V|M|W|N|O]]|A|@]|B|$]]]]

If you working with numpy, you can use the following solution which also works with negative numbers (it's also working on arrays)

<pre><code>import numpy as np
def round_down(num):
 if num &lt; 0:
 return -np.ceil(abs(num))
 else:
 return np.int32(num)
round_down = np.vectorize(round_down)
</code></pre>

<hr>

<pre><code>round_down([-1.1, -1.5, -1.6, 0, 1.1, 1.5, 1.6])
&gt; array([-2., -2., -2., 0., 1., 1., 1.])
</code></pre>

I think it will also work if you just use the <code>math</code> module instead of <code>numpy</code> module.

entityMap|blocks|key|8ehdh|text|只需进行四舍五入(x-0.5)，这将始终返回浮点数的下一个向下舍入的Integer值。您还可以通过do+round(x%2B0.5)轻松地进行舍入|type|unstyled|depth|inlineStyleRanges|entityRanges|data^0^^$0|$]|1|@$2|3|4|5|6|7|8|C|9|@]|A|@]|B|$]]]]

Just make round(x-0.5) this will always return the next rounded down Integer value of your Float. You can also easily round up by do round(x+0.5)

entityMap|blocks|key|1i51a|text|不知道你是否解决了这个问题，我只是偶然发现了这个问题。如果你想去掉小数点，你可以使用int(x)，它会去掉所有的小数点。不需要使用round(x)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data^0^^$0|$]|1|@$2|3|4|5|6|7|8|C|9|@]|A|@]|B|$]]]]

Don't know if you solved this, but I just stumble upon this question. If you want to get rid of decimal points, you could use int(x) and it will eliminate all decimal digits. Theres no need to use round(x).

entityMap|blocks|key|clddu|text|这可能很简单，但你不能把它四舍五入然后减1吗？例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|91b7i|number=1.5
round(number)-1
>+1|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

It may be very simple, but couldn't you just round it up then minus 1?
For example:

<pre><code>number=1.5
round(number)-1
&gt; 1
</code></pre>

entityMap|blocks|key|1v8op|text|我使用这个代码，你从数字中减去0.5，当你舍入它时，它是原始的数字向下舍入。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|62ugp|圆形(a-0.5)|blockquote^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|F|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|G|9|@]|A|@]|B|$]]]]

I used this code where you subtract 0.5 from the number and when you round it, it is the original number rounded down.

<blockquote>
 round(a-0.5)
</blockquote>

I want to take a floating-point number and round it down to the nearest integer. However, if it's not a whole, I always want to round down the variable, regardless of how close it is to the next integer up. Is there a way to do this?

Round a floating-point number down to the nearest integer?

 我想取一个浮点数，并将其向下舍入为最接近的整数。但是，如果它不是一个整体，我始终希望将变量向下舍入，而不管它与下一个整数向上的距离有多近。有没有办法做到这一点？

问将浮点数向下舍入为最接近的整数？
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