blocks|key|2892314|text|if+x+%25+3+==+0:
++++print+'x+is+divisible+by+3'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2892315|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>if x % 3 == 0:
 print 'x is divisible by 3'
</code></pre>

blocks|key|2892359|text|整数没有小数。如果你的意思是"check+If+a+number+decimals+in+Python"，你可以这样做：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2892360|not+float(your_number).is_integer()|code-block|syntax|javascript|2892361|entityMap^0|O|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Integers have no decimals. If you meant "check if a number got decimals in Python", you can do:

<pre><code>not float(your_number).is_integer()
</code></pre>

blocks|key|4682536|text|编辑:正如Ollie在这篇文章下面的评论中指出的那样，++是标准库的一部分，因此不应该像我在下面所做的那样重新实现。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|4682537|此函数使用这样一个事实，即每隔一个整数，至少有一个数可以被2整除，没有余数。n或n%2B1中的任何非零分数表示都将导致n%252和(n%2B1)%252都具有余数。这样做的好处是，表示为浮点值的整数将返回True。据我所知，该函数对正数和负数以及零都能正常工作。正如在函数中提到的，对于非常接近整数的值，它会失败。|CODE|4682538|def+isInteger(n):
++++"""Return+True+if+argument+is+a+whole+number,+False+if+argument+has+a+fractional+part.

++++Note+that+for+values+very+close+to+an+integer,+this+test+breaks.+During
++++superficial+testing+the+closest+value+to+zero+that+evaluated+correctly
++++was+9.88131291682e-324.+When+dividing+this+number+by+10,+Python+2.7.1+evaluated
++++the+result+to+zero"""

++++if+n%252+==+0+or+(n%2B1)%252+==+0:
++++++++return+True
++++return+False|code-block|syntax|javascript|4682539|entityMap^0|0|R|T|T|0|12|1|14|3|1L|3|1P|7|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|H]|$9|X|A|Y|B|H]|$9|Z|A|10|B|H]|$9|11|A|12|B|H]]|D|@]|E|$]]|$1|I|3|J|5|K|7|13|8|@]|D|@]|E|$L|M]]|$1|N|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|O|$]]

Edit: As Ollie pointed out in the comment below this post, <a href="http://docs.python.org/library/stdtypes.html#float.is_integer" rel="nofollow">is_integer</a> is part of the standard library and should therefore not be reimplemented as I did below.

This function uses the fact that every other whole number will have at least one number divisible by two with no remainder. Any non-zero fractional representation in either <code>n</code> or <code>n+1</code> will cause both <code>n%2</code> and <code>(n+1)%2</code> to have a remainder. This has the benefit that whole numbers represented as float values will return True. 
The function works correctly for positive
and negative numbers and zero as far as I can determine. As mentioned in the function, it fails for values very close to an integer.

<pre><code>def isInteger(n):
 """Return True if argument is a whole number, False if argument has a fractional part.

 Note that for values very close to an integer, this test breaks. During
 superficial testing the closest value to zero that evaluated correctly
 was 9.88131291682e-324. When dividing this number by 10, Python 2.7.1 evaluated
 the result to zero"""

 if n%2 == 0 or (n+1)%2 == 0:
 return True
 return False
</code></pre>

blocks|key|2608136|text|如果x+%25+3+==+0为整数，则x+/+3将为True。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2608137|entityMap^0|2|A|H|5|O|4|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|$]]

<code>x % 3 == 0</code> will be <code>True</code> if <code>x / 3</code> is an integer.

blocks|key|990703|text|下面是另一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|990704|x+=+1/3++#+insert+your+number+here
print(x+-+int(x)+==+0)++#+True+if+x+is+a+whole+number,+False+if+it+has+decimals.|code-block|syntax|javascript|990705|这之所以有效，是因为int(x)本质上取了数字(例如，3.6453+->+3)。如果减去楼层后还剩下一些东西，它不可能是一个整数。|990706|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|$]]

Here's another method:

<pre><code>x = 1/3 # insert your number here
print(x - int(x) == 0) # True if x is a whole number, False if it has decimals.
</code></pre>

This works because int(x) essentially takes the floor of the number (ex. 3.6453 -> 3). If there's something left over once you subtract the floor, it can't have been a whole number.

blocks|key|2608106|text|假设您的意思是，如果包含数字的字符串也有小数点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2608107|Python+2.6.6+(r266:84292,+Apr+20+2011,+11:58:30)+
[GCC+4.5.2]+on+linux2
Type+"help",+"copyright",+"credits"+or+"license"+for+more+information.
>>>+number='123.4'
>>>+'.'+in+number
True
>>>+number='123'
>>>+'.'+in+number
False
>>>|code-block|syntax|javascript|2608108|要测试它是否是整数，你可以mod+1：|2608109|>>>+1.0/3+%25+1
0.33333333333333331
>>>+1/3+%25+1
0|2608110|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|$]]

assuming you mean if a string containing digits also has a decimal point:

<pre><code>Python 2.6.6 (r266:84292, Apr 20 2011, 11:58:30) 
[GCC 4.5.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
&gt;&gt;&gt; number='123.4'
&gt;&gt;&gt; '.' in number
True
&gt;&gt;&gt; number='123'
&gt;&gt;&gt; '.' in number
False
&gt;&gt;&gt;
</code></pre>

To test if it's integral you could mod 1:

<pre><code>&gt;&gt;&gt; 1.0/3 % 1
0.33333333333333331
&gt;&gt;&gt; 1/3 % 1
0
</code></pre>

blocks|key|2608533|text|在Python2中，将一个整数除以一个整数将返回一个整数(除非使用-Qnew选项调用Python，或者from+__future__+import+division位于源代码的开头；在这种情况下，/返回一个浮点数)；//指定整数除法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2608534|在Python+3中，如果使用"/"，则将int除以int将返回一个浮点数，如果使用"//“，则返回一个int。|2608535|如果你想知道一个int是否会被精确地分成另一个int，可以使用"%25“来查找余数。|2608536|entityMap^0|X|5|1F|V|2Q|1|30|2|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|V|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|K|$]]

In Python 2, dividing an int by an int returns an int (unless python was invoked with the <code>-Qnew</code> option, or a <code>from __future__ import division</code> is at the beginning of the source; in that case <code>/</code> returns a float); a <code>//</code> specifies integer division.

In Python 3, dividing an int by an int returns a float if you use "/", or an int if you use "//".

If you want to know whether an int will divide into another int exactly, use "%" to look for a remainder.

blocks|key|4682563|text|转换1.0+=>+1和转换1.x+=>+1.x|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4682564|如果浮点数包含小数部分，如1.5，则此代码将返回1.5；如果是35.00，则返回35：|4682565|a+=+ReadFrom()++++
if+float(a).is_integer():+#+it+is+an+integer+number+like+23.00+so+return+23
+++++++return+int(a)
+else:+#+for+numbers+with+decimal+part+like+:+1.5+return+1.5
+++++++return+float(a)|code-block|syntax|javascript|4682566|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|$]]

<h1>convert 1.0 => 1 &amp; convert 1.x => 1.x</h1>

This code if float numbers has decimal part like 1.5 will return 1.5
&amp; if it is 35.00 it return 35:

<pre><code>a = ReadFrom() 
if float(a).is_integer(): # it is an integer number like 23.00 so return 23
 return int(a)
 else: # for numbers with decimal part like : 1.5 return 1.5
 return float(a)
</code></pre>

blocks|key|990658|text|假设您的x变量是一个整数，最好在除法之前进行确定。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|990659|尝试对浮点数进行相等测试或比较是危险的：http://www.lahey.com/float.htm|offset|length|990660|在进行除法以查看一个整数是否可被另一个整数除以之前，已经使用模数提供的答案是安全的。在进行除法并处理可能的浮点值之后，数字不再是精确的整数或不是整数。|990661|entityMap|0|LINK|mutability|MUTABLE|url|http://www.lahey.com/float.htm^0|0|K|U|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@$D|R|E|S|1|T]]|A|$]]|$1|F|3|G|5|6|7|U|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]]]

It is best to make your determination before doing the division, assuming that your x variable is an integer.

Trying to do equality tests or comparisons on floating point numbers is dangerous: <a href="http://www.lahey.com/float.htm" rel="nofollow">http://www.lahey.com/float.htm</a>

The answer already provided using modulus before doing the division to see if one integer is divsible by the other integer is safe. After you do a division and are dealing with possibly floating point values, then numbers are no longer exactly integers or not.

I would like to determine if a numeric value in Python is a whole number. For example, given: 

<pre><code>y = x / 3
</code></pre>

I want to distinguish between values of <code>x</code> which are evenly divisible by 3 those which are not.

Determining whether an value is a whole number in Python

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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我想确定Python中的数值是否为整数。例如，给定：y = x / 3我想区分x的值，它们可以被3整除，而不是3。

问在Python中确定值是否为整数
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