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 l
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)
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]
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]]
Just for fun - the solution as a recursive one-liner:
def tobin(x):
return tobin(x/2) + [x%2] if x > 1 else [x]
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
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)