blocks|key|2752823|text|如果您只需要控制格式的精度|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2752824|pi+=+3.14159265
format(pi,+'.3f')+#print+3.142+#+3+precision+after+the+decimal+point
format(pi,+'.1f')+#print+3.1
format(pi,+'.10f')+#print+3.1415926500,+more+precision+than+the+original|code-block|syntax|javascript|2752825|如果需要控制浮点算法的精度|2752826|import+decimal
decimal.getcontext().prec=4+#4+precision+in+total
pi+=+decimal.Decimal(3.14159265)
pi**2+#print+Decimal('9.870')+whereas+'3.142+squared'+would+be+off|2752827|-编辑--|2752828|没有“四舍五入”，从而截断数字。|2752829|import+decimal
from+decimal+import+ROUND_DOWN
decimal.getcontext().prec=4
pi*1+#print+Decimal('3.142')

decimal.getcontext().rounding+=+ROUND_DOWN
pi*1+#print+Decimal('3.141')|2752830|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

If you just need to control the precision in format

<pre><code>pi = 3.14159265
format(pi, '.3f') #print 3.142 # 3 precision after the decimal point
format(pi, '.1f') #print 3.1
format(pi, '.10f') #print 3.1415926500, more precision than the original
</code></pre>

If you need to control the precision in floating point arithmetic

<pre><code>import decimal
decimal.getcontext().prec=4 #4 precision in total
pi = decimal.Decimal(3.14159265)
pi**2 #print Decimal('9.870') whereas '3.142 squared' would be off
</code></pre>

--edit--

Without "rounding", thus truncating the number

<pre><code>import decimal
from decimal import ROUND_DOWN
decimal.getcontext().prec=4
pi*1 #print Decimal('3.142')

decimal.getcontext().rounding = ROUND_DOWN
pi*1 #print Decimal('3.141')
</code></pre>

blocks|key|3247445|text|一个超级简单的解决方案是使用字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3247446|x+=+float+(str+(w)[:-1])
y+=+float+(str+(w)[:-2])
z+=+float+(str+(w)[:-3])|code-block|syntax|javascript|3247447|任何浮点库解决方案都需要你回避一些四舍五入，并且使用10的地板/幂来挑选小数可以得到一些毛茸茸的与上面的相比。|3247448|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|$]]

A super simple solution is to use strings

<pre><code>x = float (str (w)[:-1])
y = float (str (w)[:-2])
z = float (str (w)[:-3])
</code></pre>

Any of the floating point library solutions would require you dodge some rounding, and using floor/powers of 10 to pick out the decimals can get a little hairy by comparison to the above.

blocks|key|85005|text|你可以：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|85006|def+truncate(f,+n):
++++return+math.floor(f+*+10+**+n)+/+10+**+n|code-block|syntax|javascript|85007|测试：|85008|>>>+f=1.923328437452
>>>+[truncate(f,+n)+for+n+in+range(7)]
[1.0,+1.9,+1.92,+1.923,+1.9233,+1.92332,+1.923328]|85009|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|$]]

You can do:

<pre><code>def truncate(f, n):
 return math.floor(f * 10 ** n) / 10 ** n
</code></pre>

testing:

<pre><code>&gt;&gt;&gt; f=1.923328437452
&gt;&gt;&gt; [truncate(f, n) for n in range(7)]
[1.0, 1.9, 1.92, 1.923, 1.9233, 1.92332, 1.923328]
</code></pre>

blocks|key|3247485|text|整数比浮动/双倍(比字符串更快)操作更快。在本例中，我试图为这两种方法争取时间：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3247486|++timeit.timeit(stmt+=+"float(str(math.pi)[:12])",+setup+=+"import+math",+number+=+1000000)|code-block|syntax|javascript|3247487|~1.1929605630000424|3247488|适用于：|3247489|timeit.timeit(stmt+=+"math.floor(math.pi+*+10+**+10)+/+10+**+10",+setup+=+"import+math",+number+=+1000000)|3247490|~0.3455968870000561|3247491|因此，使用math.floor而不是对其进行字符串操作是安全的。|3247492|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Integers are faster to manipulate than floats/doubles which are faster than strings. In this case, I tried to get time with both approach :

<pre><code> timeit.timeit(stmt = "float(str(math.pi)[:12])", setup = "import math", number = 1000000)
</code></pre>

~1.1929605630000424

for :

<pre><code>timeit.timeit(stmt = "math.floor(math.pi * 10 ** 10) / 10 ** 10", setup = "import math", number = 1000000)
</code></pre>

~0.3455968870000561

So it's safe to use math.floor rather than string operation on it.

blocks|key|85024|text|此外，还包括：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|85025|>>>+f+=+1.678
>>>+n+=+2
>>>+int(f+*+10+**+n)+/+10+**+n
1.67|code-block|syntax|javascript|85026|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|$]]

also this:
<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; f = 1.678
&gt;&gt;&gt; n = 2
&gt;&gt;&gt; int(f * 10 ** n) / 10 ** n
1.67
</code></pre>

blocks|key|85032|text|我认为最简单的答案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|85033|from+math+import+trunc

w+=+1.678
x+=+trunc(w+*+100)+/+100
y+=+trunc(w+*+10)+/+10
z+=+trunc(w)|code-block|syntax|javascript|85034|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|$]]

I think the easiest answer is :
<pre><code>from math import trunc

w = 1.678
x = trunc(w * 100) / 100
y = trunc(w * 10) / 10
z = trunc(w)
</code></pre>

blocks|key|3247537|text|获取整数的最简单方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3247538|series_col.round(2).apply(lambda+:float(str(X).split(“.”，1))|3247539|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Easiest way to get integer:
series_col.round(2).apply(lambda x: float(str(x).split(&quot;.&quot;,1)[0]))

How can I take a float variable, and control how far out the float goes without round()? For example. 

<pre><code>w = float(1.678)
</code></pre>

I want to take x and make the following variables out of it. 

<pre><code>x = 1.67
y = 1.6
z = 1
</code></pre>

If I use the respective round methods:

<pre><code>x = round(w, 2) # With round I get 1.68 
y = round(y, 1) # With round I get 1.7
z = round(z, 0) # With round I get 2.0
</code></pre>

It's going to round and alter the numbers to the point where there no use to me. I understand this is the point of round and its working properly. How would I go about getting the information that I need in the x,y,z variables and still be able to use them in other equations in a float format?

Python setting Decimal Place range without rounding?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我如何才能获得一个浮动变量，并控制浮点数在不带圆形()的情况下走多远？例如。w = float(1.678)我想取x，然后用它做下面的变量。x = 1.67y = 1.6z = 1如果我使用各自的圆方法：x = round(w, 2) # With round I get 1.68 y = round(y, 1) # ...

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