在python中添加二进制数需要帮助
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Need help in adding binary numbers in python
提问by user3246901
If I have 2 numbers in binary form as a string, and I want to add them I will do it digit by digit, from the right most end. So 001 + 010 = 011 But suppose I have to do 001+001, how should I create a code to figure out how to take carry over responses?
如果我有 2 个二进制形式的数字作为字符串，并且我想将它们相加，我将从最右边开始逐位进行。所以 001 + 010 = 011 但是假设我必须做 001+001，我应该如何创建一个代码来弄清楚如何接受结转响应？
回答by Mostly Harmless
binand intare very useful here:
bin并且int在这里非常有用：
a = '001'
b = '011'
c = bin(int(a,2) + int(b,2))
# 0b100
intallows you to specify what base the first argument is in when converting from a string (in this case two), and binconverts a number back to a binary string.
int允许您在从字符串（在本例中为两个）bin转换时指定第一个参数的基数，并将数字转换回二进制字符串。
回答by mshsayem
Can be simple if you parse the strings by int(shown in the other answer). Here is a kindergarten-school-mathway:
如果您通过int（显示在另一个答案中）解析字符串，则可能很简单。这是幼儿园-学校-数学的方法：
>>> def add(x,y):
maxlen = max(len(x), len(y))
#Normalize lengths
x = x.zfill(maxlen)
y = y.zfill(maxlen)
result = ''
carry = 0
for i in range(maxlen-1, -1, -1):
r = carry
r += 1 if x[i] == '1' else 0
r += 1 if y[i] == '1' else 0
# r can be 0,1,2,3 (carry + x[i] + y[i])
# and among these, for r==1 and r==3 you will have result bit = 1
# for r==2 and r==3 you will have carry = 1
result = ('1' if r % 2 == 1 else '0') + result
carry = 0 if r < 2 else 1
if carry !=0 : result = '1' + result
return result.zfill(maxlen)
>>> add('1','111')
'1000'
>>> add('111','111')
'1110'
>>> add('111','1000')
'1111'
回答by Algorithmatic
Here's an easy to understand version
这是一个易于理解的版本
def binAdd(s1, s2):
if not s1 or not s2:
return ''
maxlen = max(len(s1), len(s2))
s1 = s1.zfill(maxlen)
s2 = s2.zfill(maxlen)
result = ''
carry = 0
i = maxlen - 1
while(i >= 0):
s = int(s1[i]) + int(s2[i])
if s == 2: #1+1
if carry == 0:
carry = 1
result = "%s%s" % (result, '0')
else:
result = "%s%s" % (result, '1')
elif s == 1: # 1+0
if carry == 1:
result = "%s%s" % (result, '0')
else:
result = "%s%s" % (result, '1')
else: # 0+0
if carry == 1:
result = "%s%s" % (result, '1')
carry = 0
else:
result = "%s%s" % (result, '0')
i = i - 1;
if carry>0:
result = "%s%s" % (result, '1')
return result[::-1]
回答by SalchiPapa
This accepts an arbitrary number or arguments:
这接受任意数字或参数：
def bin_add(*args): return bin(sum(int(x, 2) for x in args))[2:]
>>> bin_add('1', '10', '100')
'111'
回答by Arturo Morales Rangel
you can use this function I did:
你可以使用我做的这个功能：
def addBinary(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
#a = int('10110', 2) #(0*2** 0)+(1*2**1)+(1*2**2)+(0*2**3)+(1*2**4) = 22
#b = int('1011', 2) #(1*2** 0)+(1*2**1)+(0*2**2)+(1*2**3) = 11
sum = int(a, 2) + int(b, 2)
if sum == 0: return "0"
out = []
while sum > 0:
res = int(sum) % 2
out.insert(0, str(res))
sum = sum/2
return ''.join(out)
回答by Sanya
Not an optimal solution but a working one without use of any inbuilt functions.
不是最佳解决方案，而是不使用任何内置功能的有效解决方案。
# two approaches
# first - binary to decimal conversion, add and then decimal to binary conversion
# second - binary addition normally
# binary addition - optimal approach
# rules
# 1 + 0 = 1
# 1 + 1 = 0 (carry - 1)
# 1 + 1 + 1(carry) = 1 (carry -1)
aa = a
bb = b
len_a = len(aa)
len_b = len(bb)
min_len = min(len_a, len_b)
carry = 0
arr = []
while min_len > 0:
last_digit_aa = int(aa[len(aa)-1])
last_digit_bb = int(bb[len(bb)-1])
add_digits = last_digit_aa + last_digit_bb + carry
carry = 0
if add_digits == 2:
add_digits = 0
carry = 1
if add_digits == 3:
add_digits = 1
carry = 1
arr.append(add_digits) # will rev this at the very end for output
aa = aa[:-1]
bb = bb[:-1]
min_len -= 1
a_len_after = len(aa)
b_len_after = len(bb)
if a_len_after > 0:
while a_len_after > 0:
while carry == 1:
if len(aa) > 0:
sum_digit = int(aa[len(aa) - 1]) + carry
if sum_digit == 2:
sum_digit = 0
carry = 1
arr.append(sum_digit)
aa = aa[:-1]
else:
carry = 0
arr.append(sum_digit)
aa = aa[:-1]
else:
arr.append(carry)
carry = 0
if carry == 0 and len(aa) > 0:
arr.append(aa[len(aa) - 1])
aa = aa[:-1]
a_len_after -= 1
if b_len_after > 0:
while b_len_after > 0:
while carry == 1:
if len(bb) > 0:
sum_digit = int(bb[len(bb) - 1]) + carry
if sum_digit == 2:
sum_digit = 0
carry = 1
arr.append(sum_digit)
bb = bb[:-1]
else:
carry = 0
arr.append(sum_digit)
bb = bb[:-1]
else:
arr.append(carry)
carry = 0
if carry == 0 and len(bb) > 0:
arr.append(bb[len(bb) - 1])
bb = bb[:-1]
b_len_after -= 1
if carry == 1:
arr.append(carry)
out_arr = reversed(arr)
out_str = "".join(str(x) for x in out_arr)
return out_str
