为了将整数转换为二进制,我使用了以下代码:
>>> bin(6)
'0b110'
当删除'0b'时,我使用这个:
>>> bin(6)[2:]
'110'
如果我想显示6为00000110而不是110,我能做什么?
为了将整数转换为二进制,我使用了以下代码:
>>> bin(6)
'0b110'
当删除'0b'时,我使用这个:
>>> bin(6)[2:]
'110'
如果我想显示6为00000110而不是110,我能做什么?
当前回答
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 + str(number % 2)
int_to_binary(num)
return bin_result.zfill(fill)
其他回答
('0' * 7 + bin(6)[2:])[-8:]
or
right_side = bin(6)[2:]
'0' * ( 8 - len( right_side )) + right_side
只需使用format函数
format(6, "08b")
一般形式是
format(<the_integer>, "<0><width_of_string><format_specifier>")
走老路总是管用的
def intoBinary(number):
binarynumber=""
if (number!=0):
while (number>=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))
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 + str(number % 2)
int_to_binary(num)
return bin_result.zfill(fill)
简单的递归代码:
def bin(n,number=('')):
if n==0:
return(number)
else:
number=str(n%2)+number
n=n//2
return bin(n,number)