blocks|key|5886210|text|def+trans(x):
++++if+x+==+0:+return+[0]
++++bit+=+[]
++++while+x:
++++++++bit.append(x+%25+2)
++++++++x+>>=+1
++++return+bit[::-1]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5886211|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 trans(x):
 if x == 0: return [0]
 bit = []
 while x:
 bit.append(x % 2)
 x &gt;&gt;= 1
 return bit[::-1]
</code></pre>

blocks|key|5886255|text|这样就可以了。如果有内置函数，就没有必要使用自己的函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5886256|def+binary(x):
++++return+[int(i)+for+i+in+bin(x)[2:]]|code-block|syntax|javascript|5886257|bin()函数将转换为二进制字符串。去掉0b，你就准备好了。|offset|length|style|CODE|5886258|entityMap^0|0|0|0|5|K|2|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]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

This will do it. No sense in rolling your own function if there's a builtin.

<pre><code>def binary(x):
 return [int(i) for i in bin(x)[2:]]
</code></pre>

The <code>bin()</code> function converts to a string in binary. Strip of the <code>0b</code> and you're set.

blocks|key|2358339|text|只是为了好玩-作为递归一行程序的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2358340|def+tobin(x):
++++return+tobin(x/2)+%2B+[x%252]+if+x+>+1+else+[x]|code-block|syntax|javascript|2358341|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|$]]

Just for fun - the solution as a recursive one-liner:

<pre><code>def tobin(x):
 return tobin(x/2) + [x%2] if x &gt; 1 else [x]
</code></pre>

blocks|key|2358345|text|我可以提出以下建议：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2358346|def+tobin(x,s):
++++return+[(x>>k)&1+for+k+in+range(0,s)]|code-block|syntax|javascript|2358347|这可能是最快的方法，而且对我来说似乎很清楚。当性能很重要时，bin+way太慢了。|2358348|干杯|2358349|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

may I propose this:

<pre><code>def tobin(x,s):
 return [(x&gt;&gt;k)&amp;1 for k in range(0,s)]
</code></pre>

it is probably the fastest way and it seems pretty clear to me.
bin way is too slow when performance matters.

cheers

blocks|key|2358351|text|#+dec2bin.py
#+FB+-+201012057
import+math

def+dec2bin(f):
++++if+f+>=+1:
++++++++g+=+int(math.log(f,+2))
++++else:
++++++++g+=+-1
++++h+=+g+%2B+1
++++ig+=+math.pow(2,+g)
++++st+=+""++++
++++while+f+>+0+or+ig+>=+1:+
++++++++if+f+<+1:
++++++++++++if+len(st[h:])+>=+10:+#+10+fractional+digits+max
+++++++++++++++++++break
++++++++if+f+>=+ig:
++++++++++++st+%2B=+"1"
++++++++++++f+-=+ig
++++++++else:
++++++++++++st+%2B=+"0"
++++++++ig+/=+2
++++st+=+st[:h]+%2B+"."+%2B+st[h:]
++++return+st

#+MAIN
while+True:
++++f+=+float(raw_input("Enter+decimal+number+>0:+"))
++++if+f+<=+0:+break
++++print+"Binary+#:+",+dec2bin(f)
++++print+"bin(int(f)):+",+bin(int(f))+#+for+comparison|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2358352|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># dec2bin.py
# FB - 201012057
import math

def dec2bin(f):
 if f &gt;= 1:
 g = int(math.log(f, 2))
 else:
 g = -1
 h = g + 1
 ig = math.pow(2, g)
 st = "" 
 while f &gt; 0 or ig &gt;= 1: 
 if f &lt; 1:
 if len(st[h:]) &gt;= 10: # 10 fractional digits max
 break
 if f &gt;= ig:
 st += "1"
 f -= ig
 else:
 st += "0"
 ig /= 2
 st = st[:h] + "." + st[h:]
 return st

# MAIN
while True:
 f = float(raw_input("Enter decimal number &gt;0: "))
 if f &lt;= 0: break
 print "Binary #: ", dec2bin(f)
 print "bin(int(f)): ", bin(int(f)) # for comparison
</code></pre>

blocks|key|1796507|text|这是我为大学设计的代码。单击Here+for+a+youtube+video+of+the+code.！https://www.youtube.com/watch?v=SGTZzJ5H-CE|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1796508|__author__+=+'Derek'
print('Int+to+binary')
intStr+=+input('Give+me+an+int:+')
myInt+=+int(intStr)
binStr+=+''
while+myInt+>+0:
++++binStr+=+str(myInt+%25+2)+%2B+binStr
++++myInt+//=+2
print('The+binary+of',+intStr,+'is',+binStr)
print('\nBinary+to+int')
binStr+=+input('Give+me+a+binary+string:+')
temp+=+binStr
newInt+=+0
power+=+0
while+len(temp)+>+0:+++#+While+the+length+of+the+array+if+greater+than+zero+keep+looping+through
++++bit+=+int(temp[-1])+++#+bit+is+were+you+temporally+store+the+converted+binary+number+before+adding+it+to+the+total
++++newInt+=+newInt+%2B+bit+*+2+**+power++#+newInt+is+the+total,++Each+time+it+loops+it+adds+bit+to+newInt.
++++temp+=+temp[:-1]++#+this+moves+you+to+the+next+item+in+the+string.
++++power+%2B=+1++#+adds+one+to+the+power+each+time.
print("The+binary+number+"+%2B+binStr,+'as+an+integer+is',+newInt)|code-block|syntax|javascript|1796509|entityMap|0|LINK|mutability|MUTABLE|url|https://www.youtube.com/watch?v=SGTZzJ5H-CE|1^0|E|11|0|1G|17|1|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@$A|S|B|T|1|U]|$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|Z|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|P]]]]

Here is the code for one that I made for college.
Click <a href="https://www.youtube.com/watch?v=SGTZzJ5H-CE" rel="nofollow" title="YouTube video if the code">Here for a youtube video of the code.</a>! 
<a href="https://www.youtube.com/watch?v=SGTZzJ5H-CE" rel="nofollow">https://www.youtube.com/watch?v=SGTZzJ5H-CE</a>

<pre><code>__author__ = 'Derek'
print('Int to binary')
intStr = input('Give me an int: ')
myInt = int(intStr)
binStr = ''
while myInt &gt; 0:
 binStr = str(myInt % 2) + binStr
 myInt //= 2
print('The binary of', intStr, 'is', binStr)
print('\nBinary to int')
binStr = input('Give me a binary string: ')
temp = binStr
newInt = 0
power = 0
while len(temp) &gt; 0: # While the length of the array if greater than zero keep looping through
 bit = int(temp[-1]) # bit is were you temporally store the converted binary number before adding it to the total
 newInt = newInt + bit * 2 ** power # newInt is the total, Each time it loops it adds bit to newInt.
 temp = temp[:-1] # this moves you to the next item in the string.
 power += 1 # adds one to the power each time.
print("The binary number " + binStr, 'as an integer is', newInt)
</code></pre>

blocks|key|2358372|text|用长度填充|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2358373|在大多数情况下，您希望二进制数是一个特定的长度。例如，您希望1是长度为0，0，0,0,0,0,0,0,0,1的8位二进制数。我自己使用：|2358374|def+convert_to_binary(num,+length=8):
++++binary_string_list+=+list(format(num,+'0{}b'.format(length)))
++++return+[int(digit)+for+digit+in+binary_string_list]|code-block|syntax|javascript|2358375|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|$]]

<h2>Padded with length</h2>

In most cases you want your binary number to be a specific length. For example you want 1 to be 8 binary digits long [0,0,0,0,0,0,0,1]. I use this myself:

<pre><code>def convert_to_binary(num, length=8):
 binary_string_list = list(format(num, '0{}b'.format(length)))
 return [int(digit) for digit in binary_string_list]
</code></pre>

blocks|key|4470923|text|您可以首先使用format函数来获取与current函数类似的二进制字符串。例如，下面的代码片段创建了一个对应于整数58的8位二进制串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4470924|>>>u+=+format(58,+"08b")
'00111010'|code-block|syntax|javascript|4470925|现在迭代字符串以将每个位转换为int，以获得所需的编码为整数的位列表。|4470926|>>>[int(d)+for+d+in+u]
[0,+0,+1,+1,+1,+0,+1,+0]|4470927|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 first use the format function to get a binary string like your current function. For e.g the following snippet creates a binary string of 8 bits corresponding to integer 58.

<pre><code>&gt;&gt;&gt;u = format(58, "08b")
'00111010'
</code></pre>

Now iterate the string to convert each bit to an int to get your desired list of bits encoded as integers.

<pre><code>&gt;&gt;&gt;[int(d) for d in u]
[0, 0, 1, 1, 1, 0, 1, 0]
</code></pre>

blocks|key|5886384|text|不是真正最有效的，但至少它提供了一种简单的概念性方法来理解它……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5886385|1)将所有数字重复除以2，直到得到1为止|5886386|2)以相反的顺序，创建这个数字数组的位，如果它是偶数，则附加一个0，如果它是奇数，则附加一个1。|5886387|下面是它的字面实现：|5886388|def+intToBin(n):
++++nums+=+[n]
++++while+n+>+1:
++++++++n+=+n+//+2
++++++++nums.append(n)

++++bits+=+[]
++++for+i+in+nums:
++++++++bits.append(str(0+if+i%252+==+0+else+1))
++++bits.reverse()
++++print+''.join(bits)|code-block|syntax|javascript|5886389|下面是一个更好地利用内存的版本：|5886390|def+intToBin(n):
++++bits+=+[]

++++bits.append(str(0+if+n%252+==+0+else+1))
++++while+n+>+1:
++++++++n+=+n+//+2
++++++++bits.append(str(0+if+n%252+==+0+else+1))

++++bits.reverse()
++++return+''.join(bits)|5886391|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|V|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Not really the most efficient but at least it provides a simple conceptual way of understanding it...

1) Floor divide all the numbers by two repeatedly until you reach 1

2) Going in reverse order, create bits of this array of numbers, if it is even, append a 0 if it is odd append a 1.

Here's the literal implementation of that:

<pre><code>def intToBin(n):
 nums = [n]
 while n &gt; 1:
 n = n // 2
 nums.append(n)

 bits = []
 for i in nums:
 bits.append(str(0 if i%2 == 0 else 1))
 bits.reverse()
 print ''.join(bits)
</code></pre>

Here's a version that better utilizes memory:

<pre><code>def intToBin(n):
 bits = []

 bits.append(str(0 if n%2 == 0 else 1))
 while n &gt; 1:
 n = n // 2
 bits.append(str(0 if n%2 == 0 else 1))

 bits.reverse()
 return ''.join(bits)
</code></pre>

blocks|key|2358433|text|你可以使用numpy包，并得到非常快速的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2358434|python+-m+timeit+-s+"import+numpy+as+np;+x=np.array([8],+dtype=np.uint8)"+"np.unpackbits(x)"
1000000+loops,+best+of+3:+0.65+usec+per+loop

python+-m+timeit+"[int(x)+for+x+in+list('{0:0b}'.format(8))]"
100000+loops,+best+of+3:+3.68+usec+per+loop|code-block|syntax|javascript|2358435|Np.view只处理uint8类型的输入，但您仍然可以使用unpackbits：|2358436|python+-m+timeit+-s+"import+numpy+as+np;+x=np.array([124567],+dtype=np.uint64).view(np.uint8)"+"np.unpackbits(x)"
1000000+loops,+best+of+3:+0.697+usec+per+loop|2358437|entityMap|0|LINK|mutability|MUTABLE|url|http://numpy.org/^0|5|5|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

You can use <a href="http://numpy.org" rel="nofollow">numpy</a> package and get very fast solution:

<pre><code>python -m timeit -s "import numpy as np; x=np.array([8], dtype=np.uint8)" "np.unpackbits(x)"
1000000 loops, best of 3: 0.65 usec per loop

python -m timeit "[int(x) for x in list('{0:0b}'.format(8))]"
100000 loops, best of 3: 3.68 usec per loop
</code></pre>

unpackbits handles inputs of uint8 type only, but you can still use np.view:

<pre><code>python -m timeit -s "import numpy as np; x=np.array([124567], dtype=np.uint64).view(np.uint8)" "np.unpackbits(x)"
1000000 loops, best of 3: 0.697 usec per loop
</code></pre>

blocks|key|4471007|text|不是pythonic式的way...but仍然可以工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4471008|def+get_binary_list_from_decimal(integer,+bits):
++++'''Return+a+list+of+0's+and+1's+representing+a+decimal+type+integer.

++++Keyword+arguments:
++++integer+--+decimal+type+number.
++++bits+--+number+of+bits+to+represent+the+integer.

++++Usage+example:
++++#Convert+3+to+a+binary+list
++++get_binary_list_from_decimal(3,+4)
++++#Return+will+be+[0,+0,+1,+1]
++++'''
++++#Validate+bits+parameter.
++++if+2**bits+<=+integer:
++++++++raise+ValueError("Error:+Number+of+bits+is+not+sufficient+to+\
++++++++++++++++++++++++++represent+the+integer.+Increase+bits+parameter.")

++++#Initialise+binary+list
++++binary_list+=+[]
++++remainder+=+integer
++++for+i+in+range(bits-1,+-1,+-1):
++++++++#If+current+bit+value+is+less+than+or+equal+to+the+remainder+of+
++++++++#the+integer+then+bit+value+is+1.
++++++++if+2**i+<=+remainder:
++++++++++++binary_list.append(1)
++++++++++++#Subtract+the+current+bit+value+from+the+integer.
++++++++++++remainder+=+remainder+-+2**i
++++++++else:
++++++++++++binary_list.append(0)

++++return+binary_list|code-block|syntax|javascript|4471009|如何使用的示例：|4471010|get_binary_list_from_decimal(1,+3)
#Return+will+be+[0,+0,+1]|4471011|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|$]]

Not the pythonic way...but still works:

<pre><code>def get_binary_list_from_decimal(integer, bits):
 '''Return a list of 0's and 1's representing a decimal type integer.

 Keyword arguments:
 integer -- decimal type number.
 bits -- number of bits to represent the integer.

 Usage example:
 #Convert 3 to a binary list
 get_binary_list_from_decimal(3, 4)
 #Return will be [0, 0, 1, 1]
 '''
 #Validate bits parameter.
 if 2**bits &lt;= integer:
 raise ValueError("Error: Number of bits is not sufficient to \
 represent the integer. Increase bits parameter.")

 #Initialise binary list
 binary_list = []
 remainder = integer
 for i in range(bits-1, -1, -1):
 #If current bit value is less than or equal to the remainder of 
 #the integer then bit value is 1.
 if 2**i &lt;= remainder:
 binary_list.append(1)
 #Subtract the current bit value from the integer.
 remainder = remainder - 2**i
 else:
 binary_list.append(0)

 return binary_list
</code></pre>

Example of how to use it:

<pre><code>get_binary_list_from_decimal(1, 3)
#Return will be [0, 0, 1]
</code></pre>

blocks|key|5886447|text|def+nToKBit(n,+K=64):
+++output+=+[0]*K

+++def+loop(n,+i):
+++++++if+n+==+0:+
+++++++++++return+output
+++++++output[-i]+=+n+&+1
+++++++return+loop(n+>>+1,+i%2B1)

+++return+loop(n,+1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5886448|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 nToKBit(n, K=64):
 output = [0]*K

 def loop(n, i):
 if n == 0: 
 return output
 output[-i] = n &amp; 1
 return loop(n &gt;&gt; 1, i+1)

 return loop(n, 1)
</code></pre>

blocks|key|1799799|text|将十进制转换为二进制取决于如何使用%25和//|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1799800|def+getbin(num):
++++if+(num==0):
++++++++k=[0]+
++++++++return+k+
++++else:
++++++++s+=+[]
++++++++while(num):
++++++++++++s.append(num%252)
++++++++++++num=num//2
++++++++return+s|code-block|syntax|javascript|1799801|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|$]]

Converting decimal to binary is a matter of how you are going to use the % and // 

<pre><code>def getbin(num):
 if (num==0):
 k=[0] 
 return k 
 else:
 s = []
 while(num):
 s.append(num%2)
 num=num//2
 return s
</code></pre>

blocks|key|2358482|text|只需共享一个处理int数组的函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2358483|def+to_binary_string(x):
++++length+=+len(bin(max(x))[2:])

++++for+i+in+x:
++++++++b+=+bin(i)[2:].zfill(length)

++++++++yield+[int(n)+for+n+in+b]|code-block|syntax|javascript|2358484|测试：|2358485|x1+=+to_binary_string([1,+2,+3])
x2+=+to_binary_string([1,+2,+3,+4])

print(list(x1))+#+[[0,+1],+[1,+0],+[1,+1]]
print(list(x2))+#+[[0,+0,+1],+[0,+1,+0],+[0,+1,+1],+[1,+0,+0]]|2358486|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 sharing a function that processes an array of ints:

<pre><code>def to_binary_string(x):
 length = len(bin(max(x))[2:])

 for i in x:
 b = bin(i)[2:].zfill(length)

 yield [int(n) for n in b]
</code></pre>

Test:

<pre><code>x1 = to_binary_string([1, 2, 3])
x2 = to_binary_string([1, 2, 3, 4])

print(list(x1)) # [[0, 1], [1, 0], [1, 1]]
print(list(x2)) # [[0, 0, 1], [0, 1, 0], [0, 1, 1], [1, 0, 0]]
</code></pre>

blocks|key|1799810|text|将整数转换为固定长度的位列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1799811|[int(x)+for+x+in+list('{0:0{width}b}'.format(8,+width=5))]|code-block|syntax|javascript|1799812|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|$]]

Convert integer to list of bits with a fixed length :
<pre><code>[int(x) for x in list('{0:0{width}b}'.format(8, width=5))]
</code></pre>

This function receives as a parameter an integer and should return a list representing the same value expressed in binary as a list of bits, where the first element in the list is the most significant (leftmost) bit.

My function currently outputs <code>'1011'</code> for the number 11, I need <code>[1,0,1,1]</code> instead.

For example,

<pre><code>&gt;&gt;&gt; convert_to_binary(11)
[1,0,1,1]
</code></pre>

Convert an integer to binary without using the built-in bin function

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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此函数接收一个整数作为参数，并应返回一个列表，该列表以二进制形式表示与位列表相同的值，其中列表中的第一个元素是最高有效位(最左侧)。我的函数当前输出数字11的'1011'，我需要[1,0,1,1]。例如,>>> convert_to_binary(11)[1,0,1,1]

问在不使用内置bin函数的情况下将整数转换为二进制
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问在不使用内置bin函数的情况下将整数转换为二进制EN