blocks|key|1789231|text|一种对你来说很简单的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1789232|number_dec+=+str(number-int(number))[1:]|code-block|syntax|javascript|1789233|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|$]]

An easy approach for you:

<pre><code>number_dec = str(number-int(number))[1:]
</code></pre>

blocks|key|2002559|text|5.55+%25+1|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2002560|请记住，这不会帮助您解决浮点舍入问题。也就是说，你可能会得到：|unstyled|2002561|0.550000000001|2002562|或者比你期望的0.55稍微低一点。|2002563|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>5.55 % 1
</code></pre>

Keep in mind this won't help you with floating point rounding problems. I.e., you may get:

<pre><code>0.550000000001
</code></pre>

Or otherwise a little off the 0.55 you are expecting.

blocks|key|2002951|text|使用modf|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2002952|>>>+import+math
>>>+frac,+whole+=+math.modf(2.5)
>>>+frac
0.5
>>>+whole
2.0|code-block|syntax|javascript|2002953|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/math.html#math.modf^0|2|4|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Use <a href="https://docs.python.org/2/library/math.html#math.modf" rel="noreferrer">modf</a>:

<pre><code>&gt;&gt;&gt; import math
&gt;&gt;&gt; frac, whole = math.modf(2.5)
&gt;&gt;&gt; frac
0.5
&gt;&gt;&gt; whole
2.0
</code></pre>

blocks|key|2002541|text|下面是什么：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2002542|a+=+1.3927278749291
b+=+a+-+int(a)

b
>>+0.39272787492910011|code-block|syntax|javascript|2002543|或者，使用numpy：|2002544|import+numpy
a+=+1.3927278749291
b+=+a+-+numpy.fix(a)|2002545|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|$]]

What about:

<pre><code>a = 1.3927278749291
b = a - int(a)

b
&gt;&gt; 0.39272787492910011
</code></pre>

Or, using numpy:

<pre><code>import numpy
a = 1.3927278749291
b = a - numpy.fix(a)
</code></pre>

blocks|key|2002675|text|使用标准库中的decimal模块，可以保留原始精度并避免浮点舍入问题：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2002676|>>>+from+decimal+import+Decimal
>>>+Decimal('4.20')+%25+1
Decimal('0.20')|code-block|syntax|javascript|2002677|作为注释中的kindall+notes，您必须首先将本机float转换为字符串。|2002678|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/decimal.html|1|https://stackoverflow.com/users/416467/kindall|2|https://stackoverflow.com/questions/3886402/python-how-to-get-numbers-after-decimal-point/3886439#comment-4138283^0|7|7|7|7|0|0|0|S|5|6|7|1|E|5|2|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@$9|18|A|19|1|1A]|$9|1B|A|1C|1|1D]]|E|$]]|$1|M|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]|W|$5|P|Q|R|E|$S|X]]]]

Using the <a href="http://docs.python.org/library/decimal.html" rel="noreferrer"><code>decimal</code></a> module from the standard library, you can retain the original precision and avoid floating point rounding issues:

<pre><code>&gt;&gt;&gt; from decimal import Decimal
&gt;&gt;&gt; Decimal('4.20') % 1
Decimal('0.20')
</code></pre>

As <a href="https://stackoverflow.com/users/416467/kindall">kindall</a> <a href="https://stackoverflow.com/questions/3886402/python-how-to-get-numbers-after-decimal-point/3886439#comment-4138283">notes</a> in the comments, you'll have to convert native <code>float</code>s to strings first.

blocks|key|2002581|text|尝试Modulo：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2002582|5.55%251+=+0.54999999999999982|code-block|syntax|javascript|2002583|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|$]]

Try Modulo:

<pre><code>5.55%1 = 0.54999999999999982
</code></pre>

blocks|key|2003488|text|要使其同时适用于正数和负数:请尝试abs(x)%251。对于负数，如果不使用abs，它将出错。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2003489|5.55+%25+1|2003490|输出0.5499999999999998|2003491|-5.55+%25+1|2003492|输出0.4500000000000002|2003493|entityMap^0|H|8|10|3|0|0|8|0|0|0|9|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|X|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|11|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|O|$]]

To make it work with both positive and negative numbers:
try <code> abs(x)%1</code>. For negative numbers, without with <code>abs</code>, it will go wrong.
<code>5.55 % 1</code>
output 0.5499999999999998
<code>-5.55 % 1</code>
output 0.4500000000000002

blocks|key|1788946|text|import+math
orig+=+5.55
whole+=+math.floor(orig)++++#+whole+=+5.0
frac+=+orig+-+whole+++++++++#+frac+=+0.55|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1788947|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>import math
orig = 5.55
whole = math.floor(orig) # whole = 5.0
frac = orig - whole # frac = 0.55
</code></pre>

blocks|key|2002993|text|与公认的答案类似，使用字符串的更简单的方法是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2002994|def+number_after_decimal(number1):
++++number+=+str(number1)
++++if+'e-'+in+number:+#+scientific+notation
++++++++number_dec+=+format(float(number),+'.%25df'%25(len(number.split(".")[1].split("e-")[0])%2Bint(number.split('e-')[1])))
++++elif+"."+in+number:+#+quick+check+if+it+is+decimal
++++++++number_dec+=+number.split(".")[1]
++++return+number_dec|code-block|syntax|javascript|2002995|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|$]]

similar to the accepted answer, even easier approach using strings would be

<pre><code>def number_after_decimal(number1):
 number = str(number1)
 if 'e-' in number: # scientific notation
 number_dec = format(float(number), '.%df'%(len(number.split(".")[1].split("e-")[0])+int(number.split('e-')[1])))
 elif "." in number: # quick check if it is decimal
 number_dec = number.split(".")[1]
 return number_dec
</code></pre>

blocks|key|1789013|text|>>>+n=5.55
>>>+if+"."+in+str(n):
...+++++print+"."%2Bstr(n).split(".")[-1]
...
.55|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1789014|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>&gt;&gt;&gt; n=5.55
&gt;&gt;&gt; if "." in str(n):
... print "."+str(n).split(".")[-1]
...
.55
</code></pre>

blocks|key|1789378|text|有时尾随零很重要|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1789379|In+[4]:+def+split_float(x):
+++...:+++++'''split+float+into+parts+before+and+after+the+decimal'''
+++...:+++++before,+after+=+str(x).split('.')
+++...:+++++return+int(before),+(int(after)*10+if+len(after)==1+else+int(after))
+++...:+
+++...:+

In+[5]:+split_float(105.10)
Out[5]:+(105,+10)

In+[6]:+split_float(105.01)
Out[6]:+(105,+1)

In+[7]:+split_float(105.12)
Out[7]:+(105,+12)|code-block|syntax|javascript|1789380|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|$]]

Sometimes trailing zeros matter

<pre><code>In [4]: def split_float(x):
 ...: '''split float into parts before and after the decimal'''
 ...: before, after = str(x).split('.')
 ...: return int(before), (int(after)*10 if len(after)==1 else int(after))
 ...: 
 ...: 

In [5]: split_float(105.10)
Out[5]: (105, 10)

In [6]: split_float(105.01)
Out[6]: (105, 1)

In [7]: split_float(105.12)
Out[7]: (105, 12)
</code></pre>

blocks|key|1789304|text|这是我尝试过的一种解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1789305|num+=+45.7234
(whole,+frac)+=+(int(num),+int(str(num)[(len(str(int(num)))%2B1):]))|code-block|syntax|javascript|1789306|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|$]]

This is a solution I tried:

<pre><code>num = 45.7234
(whole, frac) = (int(num), int(str(num)[(len(str(int(num)))+1):]))
</code></pre>

blocks|key|1789360|text|浮点数不以十进制(base10)格式存储。请通读一下python+documentation，看看为什么会这样。因此，从浮点型获取base10表示是不可取的。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1789361|现在有一些工具允许以十进制格式存储数字数据。下面是一个使用Decimal库的示例。|style|CODE|1789362|from+decimal+import+*

x+=+Decimal('0.341343214124443151466')
str(x)[-2:]+==+'66'++#+True

y+=+0.341343214124443151466
str(y)[-2:]+==+'66'++#+False|code-block|syntax|javascript|1789363|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/tutorial/floatingpoint.html^0|Q|K|0|0|T|7|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@$A|Z|B|10|F|G]]|9|@]|C|$]]|$1|H|3|I|5|J|7|11|8|@]|9|@]|C|$K|L]]|$1|M|3|-4|5|6|7|12|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

Float numbers are not stored in decimal (base10) format. Have a read through the <a href="https://docs.python.org/3/tutorial/floatingpoint.html" rel="nofollow noreferrer">python documentation</a> on this to satisfy yourself why. Therefore, to get a base10 representation from a float is not advisable.

Now there are tools which allow storage of numeric data in decimal format. Below is an example using the <code>Decimal</code> library.

<pre><code>from decimal import *

x = Decimal('0.341343214124443151466')
str(x)[-2:] == '66' # True

y = 0.341343214124443151466
str(y)[-2:] == '66' # False
</code></pre>

blocks|key|1789570|text|使用modf的另一个示例|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1789571|from+math+import+modf
number+=+1.0124584

#+[0]+decimal,+[1]+integer
result+=+modf(number)
print(result[0])
#+output+=+0124584
print(result[1])
#+output+=+1|code-block|syntax|javascript|1789572|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 example using modf

<pre><code>from math import modf
number = 1.0124584

# [0] decimal, [1] integer
result = modf(number)
print(result[0])
# output = 0124584
print(result[1])
# output = 1
</code></pre>

blocks|key|1788976|text|使用floor并从原始数字中减去结果：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1788977|>>+import+math+#gives+you+floor.
>>+t+=+5.55+#Give+a+variable+5.55
>>+x+=+math.floor(t)+#floor+returns+t+rounded+down+to+5..
>>+z+=+t+-+x+#z+=+5.55+-+5+=+0.55|code-block|syntax|javascript|1788978|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|$]]

Use floor and subtract the result from the original number:

<pre><code>&gt;&gt; import math #gives you floor.
&gt;&gt; t = 5.55 #Give a variable 5.55
&gt;&gt; x = math.floor(t) #floor returns t rounded down to 5..
&gt;&gt; z = t - x #z = 5.55 - 5 = 0.55
</code></pre>

blocks|key|2002973|text|示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2002974|import+math
x+=+5.55
print((math.floor(x*100)%25100))|code-block|syntax|javascript|2002975|这将在小数点后给出两个数字，即该示例中的55。如果您需要一个数字，则将上述计算减少10，或根据小数点后想要多少个数字而增加。|2002976|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|$]]

Example:

<pre><code>import math
x = 5.55
print((math.floor(x*100)%100))
</code></pre>

This is will give you two numbers after the decimal point, 55 from that example. If you need one number you reduce by 10 the above calculations or increase depending on how many numbers you want after the decimal.

blocks|key|2003020|text|import+math

x+=+1245342664.6
print(+(math.floor(x*1000)%251000)+//100+)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2003021|它肯定是有效的|unstyled|2003022|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>import math

x = 1245342664.6
print( (math.floor(x*1000)%1000) //100 )
</code></pre>

It definitely worked

blocks|key|2003226|text|我发现当使用模1来得到分数时，带有非常大的小数部分的非常大的数字可能会导致问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2003227|import+decimal

>>>+d+=+decimal.Context(decimal.MAX_PREC).create_decimal(
...+'143000000000000000000000000000000000000000000000000000000000000000000000000000.1231200000000000000002013210000000'
...+)
...
>>>+d+%25+1
Traceback+(most+recent+call+last):
++File+"<stdin>",+line+1,+in+<module>
decimal.InvalidOperation:+[<class+'decimal.DivisionImpossible'>]|code-block|syntax|javascript|2003228|相反，我抓取了整数部分，并先减去它，以帮助简化剩下的部分。|2003229|>>>+d+-+d.to_integral()
Decimal('0.1231200000000000000002013210')|2003230|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|$]]

I've found that really large numbers with really large fractional parts can cause problems when using modulo 1 to get the fraction.

<pre class="lang-py prettyprint-override"><code>import decimal

&gt;&gt;&gt; d = decimal.Context(decimal.MAX_PREC).create_decimal(
... '143000000000000000000000000000000000000000000000000000000000000000000000000000.1231200000000000000002013210000000'
... )
...
&gt;&gt;&gt; d % 1
Traceback (most recent call last):
 File "&lt;stdin&gt;", line 1, in &lt;module&gt;
decimal.InvalidOperation: [&lt;class 'decimal.DivisionImpossible'&gt;]
</code></pre>

I instead grabbed the integral part and subtracted it first to help simplify the rest of it.

<pre class="lang-py prettyprint-override"><code>&gt;&gt;&gt; d - d.to_integral()
Decimal('0.1231200000000000000002013210')
</code></pre>

blocks|key|1789642|text|只有当你想要得到第一个小数的时候才会这样|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1789643|print(int(float(input())+*+10)+%25+10)|code-block|syntax|javascript|1789644|或者你可以试试这个|1789645|num+=+float(input())
b+=+num+-+int(num)+
c+=+b+*+10
print(int(c))|1789646|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|$]]

This is only if you want toget the first decimal
<pre><code>print(int(float(input()) * 10) % 10)
</code></pre>
Or you can try this
<pre><code>num = float(input())
b = num - int(num) 
c = b * 10
print(int(c))
</code></pre>

blocks|key|2003441|text|def+fractional_part(numerator,+denominator):
++++#+Operate+with+numerator+and+denominator+to+
#+keep+just+the+fractional+part+of+the+quotient
if++denominator+==+0:
++++++return+0
++else:
+++++++return+(numerator/+denominator)-(numerator+//+denominator)++
+

print(fractional_part(5,+5))+#+Should+be+0
print(fractional_part(5,+4))+#+Should+be+0.25
print(fractional_part(5,+3))+#+Should+be+0.66...
print(fractional_part(5,+2))+#+Should+be+0.5
print(fractional_part(5,+0))+#+Should+be+0
print(fractional_part(0,+5))+#+Should+be+0|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2003442|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 fractional_part(numerator, denominator):
 # Operate with numerator and denominator to 
# keep just the fractional part of the quotient
if denominator == 0:
 return 0
 else:
 return (numerator/ denominator)-(numerator // denominator) 
 

print(fractional_part(5, 5)) # Should be 0
print(fractional_part(5, 4)) # Should be 0.25
print(fractional_part(5, 3)) # Should be 0.66...
print(fractional_part(5, 2)) # Should be 0.5
print(fractional_part(5, 0)) # Should be 0
print(fractional_part(0, 5)) # Should be 0
</code></pre>

blocks|key|1789771|text|更简单如果输入是字符串，我们可以使用split()|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1789772|decimal+=+input("Input+decimal+number:+")+#123.456

#+split+123.456+by+dot+=+['123',+'456']
after_coma+=+decimal.split('.')[1]+

#+because+only+index+1+is+taken+then+'456'
print(after_coma)+#+'456'|code-block|syntax|javascript|1789773|如果想要生成一个数字，请键入print(int(after_coma))+#+456|1789774|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|$]]

Easier if the input is a string, we can use split()
<pre><code>decimal = input(&quot;Input decimal number: &quot;) #123.456

# split 123.456 by dot = ['123', '456']
after_coma = decimal.split('.')[1] 

# because only index 1 is taken then '456'
print(after_coma) # '456'
</code></pre>
if you want to make a number type
print(int(after_coma)) # 456

blocks|key|1789313|text|下面是什么：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1789314|a+=+1.234
b+=+a+-+int(a)
length+=+len(str(a))

round(b,+length-2)|code-block|syntax|javascript|1789315|输出：|1789316|print(b)|offset|length|style|CODE|1789317|0.23399999999999999|1789318|round(b,+length-2)|1789319|0.234|1789320|由于舍入被发送到十进制字符串的长度('0.234')，我们可以只减2来不计算'0.'，并计算出所需的小数点数量。这在大多数情况下应该是有效的，除非你有很多小数位，并且计算b时的舍入误差会影响到舍入的第二个参数。|1789321|entityMap^0|0|0|0|0|8|0|0|J|0|0|I|0|0|5|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|11|8|@$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|$]]|$1|Q|3|R|5|6|7|17|8|@$K|18|L|19|M|N]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1A|8|@$K|1B|L|1C|M|N]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1D|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|X|$]]

What about:

<pre><code>a = 1.234
b = a - int(a)
length = len(str(a))

round(b, length-2)
</code></pre>

Output: 
<code>print(b)</code> 
<code>0.23399999999999999</code> 
<code>round(b, length-2)</code> 
<code>0.234</code> 

Since the round is sent to a the length of the string of decimals ('0.234'), we can just minus 2 to not count the '0.', and figure out the desired number of decimal points. This should work most times, unless you have lots of decimal places and the rounding error when calculating b interferes with the second parameter of round.

blocks|key|2003092|text|你可能想试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2003093|your_num+=+5.55
n+=+len(str(int(your_num)))
float('0'+%2B+str(your_num)[n:])|code-block|syntax|javascript|2003094|它将返回0.55。|offset|length|style|CODE|2003095|entityMap^0|0|0|4|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You may want to try this:

<pre><code>your_num = 5.55
n = len(str(int(your_num)))
float('0' + str(your_num)[n:])
</code></pre>

It will return <code>0.55</code>.

blocks|key|2003103|text|number=5.55
decimal=(number-int(number))
decimal_1=round(decimal,2)
print(decimal)
print(decimal_1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2003104|输出:+0.55|unstyled|2003105|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>number=5.55
decimal=(number-int(number))
decimal_1=round(decimal,2)
print(decimal)
print(decimal_1)
</code></pre>

output: 0.55

blocks|key|2003117|text|在python+3中，我经常做些什么来获取小数点后的数字：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2003118|a=1.22
dec=str(a).split('.')
dec=+int(dec[1])|code-block|syntax|javascript|2003119|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|$]]

See what I often do to obtain numbers after the decimal point in python
 3:

<pre><code>a=1.22
dec=str(a).split('.')
dec= int(dec[1])
</code></pre>

blocks|key|1789479|text|如果您使用的是pandas：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1789480|df['decimals']+=+df['original_number'].mod(1)|code-block|syntax|javascript|1789481|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|$]]

If you are using pandas:

<pre><code>df['decimals'] = df['original_number'].mod(1)
</code></pre>

blocks|key|2003176|text|另一种选择是将re模块与re.findall或re.search一起使用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2003177|import+re


def+get_decimcal(n:+float)+->+float:
++++return+float(re.search(r'\.\d%2B',+str(n)).group(0))


def+get_decimcal_2(n:+float)+->+float:
++++return+float(re.findall(r'\.\d%2B',+str(n))[0])


def+get_int(n:+float)+->+int:
++++return+int(n)


print(get_decimcal(5.55))
print(get_decimcal_2(5.55))
print(get_int(5.55))|code-block|syntax|javascript|2003178|输出|2003179|0.55
0.55
5|2003180|2003181|如果您希望简化/修改/探索该表达式，可以在regex101.com的右上角面板中对其进行解释。如果您愿意，您还可以在this+link中观看它如何与一些样本输入进行匹配。|2003182|2003183|来源|2003184|How+to+get+rid+of+additional+floating+numbers+in+python+subtraction?|2003185|entityMap|0|LINK|mutability|MUTABLE|url|https://regex101.com/r/HcCnEL/1/|1|https://regex101.com/r/HcCnEL/1/debugger|2|https://stackoverflow.com/questions/58687240/how-to-get-rid-of-additional-floating-numbers-in-python-subtraction/58687387#58687387^0|7|2|C|A|N|9|0|0|0|0|0|L|C|0|1M|9|1|0|0|0|0|1W|2|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1G|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1J|8|@]|D|@$9|1K|A|1L|1|1M]|$9|1N|A|1O|1|1P]]|E|$]]|$1|R|3|-4|5|6|7|1Q|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1R|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1S|8|@]|D|@$9|1T|A|1U|1|1V]]|E|$]]|$1|W|3|-4|5|6|7|1W|8|@]|D|@]|E|$]]]|X|$Y|$5|Z|10|11|E|$12|13]]|14|$5|Z|10|11|E|$12|15]]|16|$5|Z|10|11|E|$12|17]]]]

Another option would be to use the <code>re</code> module with <code>re.findall</code> or <code>re.search</code>:

<pre><code>import re


def get_decimcal(n: float) -&gt; float:
 return float(re.search(r'\.\d+', str(n)).group(0))


def get_decimcal_2(n: float) -&gt; float:
 return float(re.findall(r'\.\d+', str(n))[0])


def get_int(n: float) -&gt; int:
 return int(n)


print(get_decimcal(5.55))
print(get_decimcal_2(5.55))
print(get_int(5.55))
</code></pre>

<h3>Output</h3>

<pre><code>0.55
0.55
5
</code></pre>

<hr>

If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of <a href="https://regex101.com/r/HcCnEL/1/" rel="nofollow noreferrer">regex101.com</a>. If you'd like, you can also watch in <a href="https://regex101.com/r/HcCnEL/1/debugger" rel="nofollow noreferrer">this link</a>, how it would match against some sample inputs.

<hr>

<h3>Source</h3>

<a href="https://stackoverflow.com/questions/58687240/how-to-get-rid-of-additional-floating-numbers-in-python-subtraction/58687387#58687387">How to get rid of additional floating numbers in python subtraction?</a>

blocks|key|2003307|text|您可以使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2003308|number+=+5.55
int(str(number).split('.')[1])|code-block|syntax|javascript|2003309|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 can use this:
<pre class="lang-py prettyprint-override"><code>number = 5.55
int(str(number).split('.')[1])
</code></pre>

blocks|key|1789654|text|def+fractional_part(numerator,+denominator):
++++if+denominator+==+0:
++++++++return+0
++++else:
++++++++return+numerator+/+denominator+-+numerator+//+denominator

print(fractional_part(5,+5))+#+Should+be+0
print(fractional_part(5,+4))+#+Should+be+0.25
print(fractional_part(5,+3))+#+Should+be+0.66...
print(fractional_part(5,+2))+#+Should+be+0.5
print(fractional_part(5,+0))+#+Should+be+0
print(fractional_part(0,+5))+#+Should+be+0|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1789655|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 fractional_part(numerator, denominator):
 if denominator == 0:
 return 0
 else:
 return numerator / denominator - numerator // denominator

print(fractional_part(5, 5)) # Should be 0
print(fractional_part(5, 4)) # Should be 0.25
print(fractional_part(5, 3)) # Should be 0.66...
print(fractional_part(5, 2)) # Should be 0.5
print(fractional_part(5, 0)) # Should be 0
print(fractional_part(0, 5)) # Should be 0
</code></pre>

blocks|key|1789805|text|在聚会的后期，给新访客留下了评论。如果你知道你只需要两位小数，那么你可以使用入站舍入|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1789806|round(1.15+%25+1,+2)|code-block|syntax|javascript|1789807|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|$]]

Late in the party, leaving comment for new visitors.
If you know you need only 2 decimal places then you can use inbound round
<pre><code>round(1.15 % 1, 2)
</code></pre>

blocks|key|2003263|text|一种解决方案是使用模数和舍入。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2003264|import+math

num+=+math.fabs(float(5.55))
rem+=+num+%25+1

rnd_by+=+++len(str(num))+-+len(str(int(num)))+-+1

print(str(round(rem,rnd_by)))|code-block|syntax|javascript|2003265|你的输出是0.55|2003266|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 solution is using modulo and rounding.

<pre><code>import math

num = math.fabs(float(5.55))
rem = num % 1

rnd_by = len(str(num)) - len(str(int(num))) - 1

print(str(round(rem,rnd_by)))
</code></pre>

Your output will be 0.55

blocks|key|2002938|text|另一个疯狂的解决方案是(不转换为字符串)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2002939|number+=+123.456
temp+=+1

while+(number*temp)%2510+!=+0:
++++temp+=+temp+*10
++++print+temp
++++print+number

temp+=+temp+/10
number+=+number*temp
number_final+=+number%25temp
print+number_final|code-block|syntax|javascript|2002940|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 crazy solution is (without converting in a string):

<pre><code>number = 123.456
temp = 1

while (number*temp)%10 != 0:
 temp = temp *10
 print temp
 print number

temp = temp /10
number = number*temp
number_final = number%temp
print number_final
</code></pre>

How do I get the numbers after a decimal point?

For example, if I have <code>5.55</code>, how do i get <code>.55</code>?

How to get numbers after decimal point?

如何获得小数点后的数字？例如，如果我有5.55，如何获取.55

问如何获取小数点后的数字？
EN

回答 32

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问如何获取小数点后的数字？EN

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问如何获取小数点后的数字？
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