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 '1011' for the number 11, I need [1,0,1,1] instead.
For example,
>>> convert_to_binary(11)
[1,0,1,1]
def trans(x):
if x == 0: return [0]
bit = []
while x:
bit.append(x % 2)
x >>= 1
return bit[::-1]
Just for fun - the solution as a recursive one-liner:
def tobin(x):
return tobin(x/2) + [x%2] if x > 1 else [x]
may I propose this:
def tobin(x,s):
return [(x>>k)&1 for k in range(0,s)]
it is probably the fastest way and it seems pretty clear to me. bin way is too slow when performance matters.
cheers
This will do it. No sense in rolling your own function if there's a builtin.
def binary(x):
return [int(i) for i in bin(x)[2:]]
The bin() function converts to a string in binary. Strip of the 0b and you're set.
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.
>>>u = format(58, "08b")
'00111010'
Now iterate the string to convert each bit to an int to get your desired list of bits encoded as integers.
>>>[int(d) for d in u]
[0, 0, 1, 1, 1, 0, 1, 0]
Here is the code for one that I made for college. Click Here for a youtube video of the code.! https://www.youtube.com/watch?v=SGTZzJ5H-CE
__author__ = 'Derek'
print('Int to binary')
intStr = input('Give me an int: ')
myInt = int(intStr)
binStr = ''
while myInt > 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) > 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)
Padded with length
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:
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]
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:
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%2 == 0 else 1))
bits.reverse()
print ''.join(bits)
Here's a version that better utilizes memory:
def intToBin(n):
bits = []
bits.append(str(0 if n%2 == 0 else 1))
while n > 1:
n = n // 2
bits.append(str(0 if n%2 == 0 else 1))
bits.reverse()
return ''.join(bits)
You can use numpy package and get very fast solution:
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
unpackbits handles inputs of uint8 type only, but you can still use np.view:
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
Not the pythonic way...but still works:
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
Example of how to use it:
get_binary_list_from_decimal(1, 3)
#Return will be [0, 0, 1]
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+1)
return loop(n, 1)
Converting decimal to binary is a matter of how you are going to use the % and //
def getbin(num):
if (num==0):
k=[0]
return k
else:
s = []
while(num):
s.append(num%2)
num=num//2
return s
Just sharing a function that processes an array of ints:
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]
Test:
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]]
# dec2bin.py
# FB - 201012057
import math
def dec2bin(f):
if f >= 1:
g = int(math.log(f, 2))
else:
g = -1
h = g + 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 += "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 >0: "))
if f <= 0: break
print "Binary #: ", dec2bin(f)
print "bin(int(f)): ", bin(int(f)) # for comparison
来源：https://stackoverflow.com/questions/13522773/convert-an-integer-to-binary-without-using-the-built-in-bin-function