blocks|key|582046|text|有点无聊的方法..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|582047|>>>+bin8+=+lambda+x+:+''.join(reversed(+[str((x+>>+i)+&+1)+for+i+in+range(8)]+)+)
>>>+bin8(6)
'00000110'
>>>+bin8(-3)
'11111101'|code-block|syntax|javascript|582048|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|$]]

A bit twiddling method...

<pre><code>&gt;&gt;&gt; bin8 = lambda x : ''.join(reversed( [str((x &gt;&gt; i) &amp; 1) for i in range(8)] ) )
&gt;&gt;&gt; bin8(6)
'00000110'
&gt;&gt;&gt; bin8(-3)
'11111101'
</code></pre>

blocks|key|366225|text|只需使用格式函数即可|type|unstyled|depth|inlineStyleRanges|entityRanges|data|366226|format(6,+"08b")|code-block|syntax|javascript|366227|一般的形式是|366228|format(<the_integer>,+"<0><width_of_string><format_specifier>")|366229|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|$]]

Just use the format function

<pre><code>format(6, "08b")
</code></pre>

The general form is 

<pre><code>format(&lt;the_integer&gt;, "&lt;0&gt;&lt;width_of_string&gt;&lt;format_specifier&gt;")
</code></pre>

blocks|key|581984|text|eumiro的答案更好，但我只是为了多样化而张贴这篇文章：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|581985|>>>+"%2508d"+%25+int(bin(6)[2:])
00000110|code-block|syntax|javascript|581986|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|$]]

eumiro's answer is better, however I'm just posting this for variety:

<pre><code>&gt;&gt;&gt; "%08d" % int(bin(6)[2:])
00000110
</code></pre>

blocks|key|582026|text|。。或者，如果您不确定它是否应始终为8位数，则可以将其作为参数传递：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|582027|>>>+'%250*d'+%25+(8,+int(bin(6)[2:]))
'00000110'|code-block|syntax|javascript|582028|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|$]]

.. or if you're not sure it should always be 8 digits, you can pass it as a parameter:

<pre><code>&gt;&gt;&gt; '%0*d' % (8, int(bin(6)[2:]))
'00000110'
</code></pre>

blocks|key|366263|text|走老校门总是管用的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|366264|def+intoBinary(number):
binarynumber=""
if+(number!=0):
++++while+(number>=1):
++++++++if+(number+%252==0):
++++++++++++binarynumber=binarynumber%2B"0"
++++++++++++number=number/2
++++++++else:
++++++++++++binarynumber=binarynumber%2B"1"
++++++++++++number=(number-1)/2

else:
++++binarynumber="0"

return+"".join(reversed(binarynumber))|code-block|syntax|javascript|366265|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|$]]

Going Old School always works

<pre><code>def intoBinary(number):
binarynumber=""
if (number!=0):
 while (number&gt;=1):
 if (number %2==0):
 binarynumber=binarynumber+"0"
 number=number/2
 else:
 binarynumber=binarynumber+"1"
 number=(number-1)/2

else:
 binarynumber="0"

return "".join(reversed(binarynumber))
</code></pre>

blocks|key|366307|text|假设你想从一个不总是常量的变量中解析出用于表示的位数，一个不错的方法是使用numpy.binary。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|366308|在将二进制应用于电源集时可能会很有用|366309|import+numpy+as+np
np.binary_repr(6,+width=8)|code-block|syntax|javascript|366310|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|$]]

Assuming you want to parse the number of digits used to represent from a variable which is not always constant, a good way will be to use numpy.binary.

could be useful when you apply binary to power sets

<pre><code>import numpy as np
np.binary_repr(6, width=8)
</code></pre>

blocks|key|582163|text|最好的方法是指定格式。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|582164|format(a,+'b')|code-block|syntax|javascript|582165|以字符串格式返回a的二进制值。|582166|要将二进制字符串转换回整数，请使用int()函数。|582167|int('110',+2)|582168|返回二进制字符串的整数值。|582169|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The best way is to specify the format.

<pre><code>format(a, 'b')
</code></pre>

returns the binary value of a in string format.

To convert a binary string back to integer, use 
int() function.

<pre><code>int('110', 2)
</code></pre>

returns integer value of binary string.

blocks|key|366267|text|('0'+*+7+%2B+bin(6)[2:])[-8:]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|366268|或|unstyled|366269|right_side+=+bin(6)[2:]
'0'+*+(+8+-+len(+right_side+))+%2B+right_side|366270|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>('0' * 7 + bin(6)[2:])[-8:]
</code></pre>

or

<pre><code>right_side = bin(6)[2:]
'0' * ( 8 - len( right_side )) + right_side
</code></pre>

blocks|key|366442|text|您可以只使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|366443|"{0:b}".format(n)|code-block|syntax|javascript|366444|在我看来，这是最简单的方法！|366445|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|$]]

You can use just:
<pre><code>&quot;{0:b}&quot;.format(n)
</code></pre>
In my opinion this is the easiest way!

blocks|key|366498|text|即使是更简单的方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|366499|my_num+=+6
print(f'{my_num:b}')|code-block|syntax|javascript|366500|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|$]]

even an easier way
<pre><code>my_num = 6
print(f'{my_num:b}')
</code></pre>

blocks|key|366379|text|def+int_to_bin(num,+fill):
++++bin_result+=+''

++++def+int_to_binary(number):
++++++++nonlocal+bin_result
++++++++if+number+>+1:
++++++++++++int_to_binary(number+//+2)
++++++++bin_result+=+bin_result+%2B+str(number+%25+2)

++++int_to_binary(num)
++++return+bin_result.zfill(fill)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|366380|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 int_to_bin(num, fill):
 bin_result = ''

 def int_to_binary(number):
 nonlocal bin_result
 if number &gt; 1:
 int_to_binary(number // 2)
 bin_result = bin_result + str(number % 2)

 int_to_binary(num)
 return bin_result.zfill(fill)
</code></pre>

blocks|key|366542|text|python包Binary+Fractions有一个完整的二进制文件和二进制片段的实现。您可以按如下方式执行操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|366543|from+binary_fractions+import+Binary
b+=+Binary(6)+#+creates+a+binary+fraction+string
b.lfill(8)+#+fills+to+length+8|code-block|syntax|javascript|366544|这个包还有许多其他方法可以完全精确地操作二进制字符串。|366545|entityMap|0|LINK|mutability|MUTABLE|url|https://pypi.org/project/binary-fractions/^0|7|G|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

The python package <a href="https://pypi.org/project/binary-fractions/" rel="nofollow noreferrer">Binary Fractions</a> has a full implementation of binaries as well as binary fractions. You can do your operation as follows:
<pre><code>from binary_fractions import Binary
b = Binary(6) # creates a binary fraction string
b.lfill(8) # fills to length 8
</code></pre>
This package has many other methods for manipulating binary strings with full precision.

blocks|key|366557|text|使用递归的简单代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|366558|+def+bin(n,number=('')):
+++if+n==0:
+++++return(number)
+++else:
+++++number=str(n%252)%2Bnumber
+++++n=n//2
+++++return+bin(n,number)|code-block|syntax|javascript|366559|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|$]]

Simple code with recursion:
<pre><code> def bin(n,number=('')):
 if n==0:
 return(number)
 else:
 number=str(n%2)+number
 n=n//2
 return bin(n,number)
</code></pre>

In order to convert an integer to a binary, I have used this code :

<pre><code>&gt;&gt;&gt; bin(6) 
'0b110'
</code></pre>

and when to erase the '0b', I use this :

<pre><code>&gt;&gt;&gt; bin(6)[2:] 
'110'
</code></pre>

What can I do if I want to show <code>6</code> as <code>00000110</code> instead of <code>110</code>?

Converting integer to binary in python

为了将整数转换为二进制，我使用了以下代码：>>> bin(6)  '0b110'当要删除“0b”时，我使用以下命令：>>> bin(6)[2:]  '110'如果我想将6显示为00000110而不是110，该怎么办

问在python中将整数转换为二进制
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问在python中将整数转换为二进制
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