blocks|key|1018889|text|您使用的是Python+2.x，其中整数除法将被截断，而不是变成浮点数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1018890|>>>+1+/+2
0|code-block|syntax|javascript|1018891|您应该将其中一个设置为float|offset|length|style|CODE|1018892|>>>+float(10+-+20)+/+(100+-+10)
-0.1111111111111111|1018893|或者from+__future__+import+division，它强制/采用Python3.x总是返回浮点数的行为。|1018894|>>>+from+__future__+import+division
>>>+(10+-+20)+/+(100+-+10)
-0.1111111111111111|1018895|entityMap^0|0|0|B|5|0|0|2|V|11|1|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]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|T|$]]

You're using Python 2.x, where integer divisions will truncate instead of becoming a floating point number.

<pre><code>&gt;&gt;&gt; 1 / 2
0
</code></pre>

You should make one of them a <code>float</code>:

<pre><code>&gt;&gt;&gt; float(10 - 20) / (100 - 10)
-0.1111111111111111
</code></pre>

or <code>from __future__ import division</code>, which the forces <code>/</code> to adopt Python 3.x's behavior that always returns a float.

<pre><code>&gt;&gt;&gt; from __future__ import division
&gt;&gt;&gt; (10 - 20) / (100 - 10)
-0.1111111111111111
</code></pre>

blocks|key|1233520|text|你是putting+Integers+in+so+Python+is+giving+you+an+integer+back|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1233521|>>>+10+/+90
0|code-block|syntax|javascript|1233522|如果之后将其强制转换为浮点数，则四舍五入已经完成，换句话说，0整数将始终变为0浮点数。|1233523|如果你在除法的两端使用浮点数，那么Python会给出你所期望的答案。|1233524|>>>+10+/+90.0
0.1111111111111111|1233525|所以在你的例子中：|1233526|>>>+float(20-10)+/+(100-10)
0.1111111111111111
>>>+(20-10)+/+float(100-10)
0.1111111111111111|1233527|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/reference/expressions.html#binary-arithmetic-operations^0|2|1N|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|15|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|16|8|@]|9|@]|C|$]]|$1|M|3|N|5|F|7|17|8|@]|9|@]|C|$G|H]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|C|$]]|$1|Q|3|R|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]]]

You're <a href="http://docs.python.org/reference/expressions.html#binary-arithmetic-operations" rel="noreferrer">putting Integers in so Python is giving you an integer back</a>:

<pre><code>&gt;&gt;&gt; 10 / 90
0
</code></pre>

If if you cast this to a float afterwards the rounding will have already been done, in other words, 0 integer will always become 0 float.

If you use floats on either side of the division then Python will give you the answer you expect.

<pre><code>&gt;&gt;&gt; 10 / 90.0
0.1111111111111111
</code></pre>

So in your case:

<pre><code>&gt;&gt;&gt; float(20-10) / (100-10)
0.1111111111111111
&gt;&gt;&gt; (20-10) / float(100-10)
0.1111111111111111
</code></pre>

blocks|key|1018741|text|在进行除法运算之前，需要将其更改为浮点数。这就是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1018742|float(20+-+10)+/+(100+-+10)|code-block|syntax|javascript|1018743|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|$]]

You need to change it to a float BEFORE you do the division. That is:

<pre><code>float(20 - 10) / (100 - 10)
</code></pre>

blocks|key|1019112|text|在Python2.7中，如果输入是整数，则/运算符是整数除法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1019113|>>>20/15
1

>>>20.0/15.0
1.33333333333

>>>20.0/15
1.33333333333|code-block|syntax|javascript|1019114|在Python3.3中，即使输入是整数，/运算符也是浮点数除法。|1019115|>>>+20/15
1.33333333333

>>>20.0/15
1.33333333333|1019116|对于Python3中的整数除法，我们将使用//运算符。|1019117|在Python2.7和Python3.3中，//运算符都是整数除法运算符。|1019118|在Python+2.7和Python+3.3中：|1019119|>>>20//15
1|1019120|现在，请看比较|1019121|>>>a+=+7.0/4.0
>>>b+=+7/4
>>>print+a+==+b|1019122|对于上面的程序，Python+2.7中的输出将为False，Python+3.3中的输出将为True。|1019123|在Python2.7中，a=+1.75，b=1。|1019124|在Python3.3中，a=+1.75和b=+1.75只是因为/是一个浮点数除法。|1019125|entityMap^0|L|1|0|0|K|1|0|0|L|2|0|M|2|0|0|0|0|0|0|0|V|1|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1G|8|@$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1J|8|@$9|1K|A|1L|B|C]]|D|@]|E|$]]|$1|S|3|T|5|6|7|1M|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1N|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1O|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|1P|8|@]|D|@]|E|$I|J]]|$1|10|3|11|5|6|7|1Q|8|@]|D|@]|E|$]]|$1|12|3|13|5|6|7|1R|8|@]|D|@]|E|$]]|$1|14|3|15|5|6|7|1S|8|@$9|1T|A|1U|B|C]]|D|@]|E|$]]|$1|16|3|-4|5|6|7|1V|8|@]|D|@]|E|$]]]|17|$]]

In Python 2.7, the <code>/</code> operator is an integer division if inputs are integers:

<pre><code>&gt;&gt;&gt;20/15
1

&gt;&gt;&gt;20.0/15.0
1.33333333333

&gt;&gt;&gt;20.0/15
1.33333333333
</code></pre>

In Python 3.3, the <code>/</code> operator is a float division even if the inputs are integer.

<pre><code>&gt;&gt;&gt; 20/15
1.33333333333

&gt;&gt;&gt;20.0/15
1.33333333333
</code></pre>

For integer division in Python 3, we will use the <code>//</code> operator.

The <code>//</code> operator is an integer division operator in both Python 2.7 and Python 3.3.

In Python 2.7 and Python 3.3:

<pre><code>&gt;&gt;&gt;20//15
1
</code></pre>

Now, see the comparison

<pre><code>&gt;&gt;&gt;a = 7.0/4.0
&gt;&gt;&gt;b = 7/4
&gt;&gt;&gt;print a == b
</code></pre>

For the above program, the output will be False in Python 2.7 and True in Python 3.3.

In Python 2.7 a = 1.75 and b = 1.

In Python 3.3 a = 1.75 and b = 1.75, just because <code>/</code> is a float division.

blocks|key|1233756|text|这与您使用的python版本有关。基本上它采用了C语言的行为:如果你将两个整数相除，结果将被向下舍入为一个整数。还要记住，Python从左到右执行操作，这在您进行类型转换时起到了一定的作用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1233757|例子:当我做算术运算时，这是一个总是在我脑海中跳出的问题(我应该转换成浮点数和哪个数字)，下面给出一个来自这方面的例子：|1233758|>>>+a+=+1/2/3/4/5/4/3
>>>+a
0|code-block|syntax|javascript|1233759|当我们除以整数时，毫不奇怪，它的四舍五入较低。|1233760|>>>+a+=+1/2/3/4/5/4/float(3)
>>>+a
0.0|1233761|如果我们将最后一个整数类型转换为浮点数，我们仍然会得到零，因为当我们的数字除以浮点数时，由于整数除法已经变成了0。|1233762|>>>+a+=+1/2/3/float(4)/5/4/3
>>>+a
0.0|1233763|与上面的场景相同，但将浮动类型转换稍微向左移动一点。|1233764|>>>+a+=+float(1)/2/3/4/5/4/3
>>>+a
0.0006944444444444445|1233765|最后，当我们将第一个整数类型转换为浮点数时，结果是所需的，因为从第一个除法开始，即最左边的除法，我们使用浮点数。|1233766|Extra+1:如果您试图回答这个问题以改进算术计算，您应该检查this|offset|length|1233767|额外2:请注意以下场景：|1233768|>>>+a+=+float(1/2/3/4/5/4/3)
>>>+a
0.0|1233769|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/3264345^0|0|0|0|0|0|0|0|0|0|0|W|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1D|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1E|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1F|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|1G|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|1H|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|1I|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|1J|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|1K|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1L|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1M|8|@]|9|@$Y|1N|Z|1O|1|1P]]|A|$]]|$1|10|3|11|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|12|3|13|5|F|7|1R|8|@]|9|@]|A|$G|H]]|$1|14|3|-4|5|6|7|1S|8|@]|9|@]|A|$]]]|15|$16|$5|17|18|19|A|$1A|1B]]]]

It has to do with the version of python that you use. Basically it adopts the C behavior: if you divide two integers, the results will be rounded down to an integer. Also keep in mind that Python does the operations from left to right, which plays a role when you typecast.

Example:
Since this is a question that always pops in my head when I am doing arithmetic operations (should I convert to float and which number), an example from that aspect is presented:

<pre><code>&gt;&gt;&gt; a = 1/2/3/4/5/4/3
&gt;&gt;&gt; a
0
</code></pre>

When we divide integers, not surprisingly it gets lower rounded.

<pre><code>&gt;&gt;&gt; a = 1/2/3/4/5/4/float(3)
&gt;&gt;&gt; a
0.0
</code></pre>

If we typecast the last integer to float, we will still get zero, since by the time our number gets divided by the float has already become 0 because of the integer division.

<pre><code>&gt;&gt;&gt; a = 1/2/3/float(4)/5/4/3
&gt;&gt;&gt; a
0.0
</code></pre>

Same scenario as above but shifting the float typecast a little closer to the left side.

<pre><code>&gt;&gt;&gt; a = float(1)/2/3/4/5/4/3
&gt;&gt;&gt; a
0.0006944444444444445
</code></pre>

Finally, when we typecast the first integer to float, the result is the desired one, since beginning from the first division, i.e. the leftmost one, we use floats.

Extra 1: If you are trying to answer that to improve arithmetic evaluation, you should check <a href="https://stackoverflow.com/questions/3264345">this</a> 

Extra 2: Please be careful of the following scenario:

<pre><code>&gt;&gt;&gt; a = float(1/2/3/4/5/4/3)
&gt;&gt;&gt; a
0.0
</code></pre>

blocks|key|1233663|text|通过放置'.‘来指定浮点数。之后的数字也会使其缺省为浮点型。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1233664|>>>+1+/+2
0

>>>+1.+/+2.
0.5|code-block|syntax|javascript|1233665|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|$]]

Specifying a float by placing a '.' after the number will also cause it to default to float. 

<pre><code>&gt;&gt;&gt; 1 / 2
0

&gt;&gt;&gt; 1. / 2.
0.5
</code></pre>

blocks|key|1018928|text|让其中至少有一个是浮点型，那么它将是浮点型除法，而不是整型：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1018929|>>>+(20.0-10)+/+(100-10)
0.1111111111111111|code-block|syntax|javascript|1018930|将结果强制转换为float为时已晚。|1018931|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|$]]

Make at least one of them float, then it will be float division, not integer:

<pre><code>&gt;&gt;&gt; (20.0-10) / (100-10)
0.1111111111111111
</code></pre>

Casting the result to float is too late.

blocks|key|1019199|text|在python中，cv2没有更新除法计算。因此，您必须在程序的第一行包含from+__future__+import+division。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1019200|entityMap^0|9|3|10|V|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

In python <code>cv2</code> not updated the division calculation. so, you must include <code>from __future__ import division</code> in first line of the program.

blocks|key|1233415|text|无论哪种方式，都是整数除法。10/90+=+0。在第二种情况下，您只是将0转换为一个浮点数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1233416|尝试将"/“的一个操作数转换为浮点数：|1233417|float(20-10)+/+(100-10)|code-block|syntax|javascript|1233418|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Either way, it's integer division. 10/90 = 0. In the second case, you're merely casting 0 to a float.

Try casting one of the operands of "/" to be a float:

<pre><code>float(20-10) / (100-10)
</code></pre>

blocks|key|1018793|text|在你的第二个例子中除法已经发生后，你将强制转换为float。试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1018794|float(20-10)+/+float(100-10)|code-block|syntax|javascript|1018795|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|$]]

You're casting to float after the division has already happened in your second example. Try this:

<pre><code>float(20-10) / float(100-10)
</code></pre>

blocks|key|1019156|text|就我个人而言，我更喜欢在一开始插入一个1.+*。因此，表达式如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1019157|1.+*+(20-10)+/+(100-10)|code-block|syntax|javascript|1019158|因为我总是对一些公式做除法运算，比如：|1019159|accuracy+=+1.+*+(len(y_val)+-+sum(y_val))+/+len(y_val)|1019160|因此，不可能简单地添加像20.0这样的.0。在我的例子中，用float()包装可能会失去一点可读性。|1019161|entityMap^0|J|4|0|0|0|0|C|4|J|2|U|7|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

Personally I preferred to insert a <code>1. *</code> at the very beginning. So the expression become something like this:

<pre><code>1. * (20-10) / (100-10)
</code></pre>

As I always do a division for some formula like:

<pre><code>accuracy = 1. * (len(y_val) - sum(y_val)) / len(y_val)
</code></pre>

so it is impossible to simply add a <code>.0</code> like <code>20.0</code>. And in my case, wrapping with a <code>float()</code> may lose a little bit readability.

I was trying to normalize a set of numbers from -100 to 0 to a range of 10-100 and was having problems only to notice that even with no variables at all, this does not evaluate the way I would expect it to:

<pre><code>&gt;&gt;&gt; (20-10) / (100-10)
0
</code></pre>

Float division doesn't work either:

<pre><code>&gt;&gt;&gt; float((20-10) / (100-10))
0.0
</code></pre>

If either side of the division is cast to a float it will work:

<pre><code>&gt;&gt;&gt; (20-10) / float((100-10))
0.1111111111111111
</code></pre>

Each side in the first example is evaluating as an int which means the final answer will be cast to an int. Since 0.111 is less than .5, it rounds to 0. It is not transparent in my opinion, but I guess that's the way it is.

What is the explanation?

Python division

Python

我试图将一组从-100到0的数字归一化到10-100的范围，并且遇到了问题，但我注意到即使根本没有变量，这也不会以我期望的方式进行评估：>>> (20-10) / (100-10)0浮点数除法也不起作用：>>> float((20-10) / (100-10))0.0如果除法的任一侧被强制转换为浮点数，它将工作：>>> (20-10) / float((100-10))0.111111111111

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