count number of ones in a given integer
Question:
How do you count the number of ones in a given integer’s binary representation.
Say you are given a number 20, which is 10100 in binary, so number of ones is 2.
Answers:
If the input number is ‘number’
number =20
len(bin(number)[2:].replace('0',''))
Another solution is
from collections import Counter
Counter(list(bin(number))[2:])['1']
Use the awesome collections module.
>>> from collections import Counter
>>> binary = bin(20)[2:]
>>> Counter(binary)
Counter({'0': 3, '1': 2})
Or you can use the built-in function count():
>>> binary = bin(20)[2:]
>>> binary.count('1')
2
Or even:
>>> sum(1 for i in bin(20)[2:] if i == '1')
2
But that last solution is slower than using count()
>>> num = 20
>>> bin(num)[2:].count('1')
2
What you’re looking for is called the Hamming weight, and there are a lot of algorithms to do it. Here’s another straightforward one:
def ones(n):
w = 0
while (n):
w += 1
n &= n - 1
return w
The usual way to make this blinding fast is to use lookup tables:
table = [bin(i)[2:].count('1') for i in range(256)]
def pop_count(n):
cnt = 0
while n > 0:
cnt += table[n & 255]
n >>= 8
return cnt
In Python, any solution using bin and list.count will be faster, but this is nice if you want to write it in assembler.
I am a new coder and I found this one logic simple. Might be easier for newbies to understand.
def onesInDecimal(n):
count = 0
while(n!=0):
if (n%2!=0):
count = count+1
n = n-1
n = n/2
else:
n = n/2
return count
You can do this using bit shifting >> and bitwise and & to inspect the least significant bit, like this:
def count_ones(x):
result = 0
while x > 0:
result += x & 1
x = x >> 1
return result
This works by shifting the bits right until the value becomes zero, counting the number of times the least significant bit is 1 along the way.
The int type has a new method int.bit_count() since python 3.10a, returning the number of ones in the binary expansion of a given integer, also known as the population count as follows:
n = 20
bin(n)
'0b10100'
n.bit_count() returns 2 as it has 2 ones in the binary representation.
For a special case when you need to check quickly whether the binary form of the integer x has only a single 1 (and thus is a power of 2), you can use this check:
if x == -(x | (-x)):
...
The expression -(x | (-x)) is the number that you get if you replace all 1s except the last one (the least significant bit) in the binary representation of x with 0.
Example:
12 = 1100 in binary
-12 = …110100 in binary (with an infinite number of leading 1s)
12 | (-12) = …111100 in binary (with an infinite number of leading 1s)
-(12 | (-12)) = 100 in binary
How do you count the number of ones in a given integer’s binary representation.
Say you are given a number 20, which is 10100 in binary, so number of ones is 2.
If the input number is ‘number’
number =20
len(bin(number)[2:].replace('0',''))
Another solution is
from collections import Counter
Counter(list(bin(number))[2:])['1']
Use the awesome collections module.
>>> from collections import Counter
>>> binary = bin(20)[2:]
>>> Counter(binary)
Counter({'0': 3, '1': 2})
Or you can use the built-in function count():
>>> binary = bin(20)[2:]
>>> binary.count('1')
2
Or even:
>>> sum(1 for i in bin(20)[2:] if i == '1')
2
But that last solution is slower than using count()
>>> num = 20
>>> bin(num)[2:].count('1')
2
What you’re looking for is called the Hamming weight, and there are a lot of algorithms to do it. Here’s another straightforward one:
def ones(n):
w = 0
while (n):
w += 1
n &= n - 1
return w
The usual way to make this blinding fast is to use lookup tables:
table = [bin(i)[2:].count('1') for i in range(256)]
def pop_count(n):
cnt = 0
while n > 0:
cnt += table[n & 255]
n >>= 8
return cnt
In Python, any solution using bin and list.count will be faster, but this is nice if you want to write it in assembler.
I am a new coder and I found this one logic simple. Might be easier for newbies to understand.
def onesInDecimal(n):
count = 0
while(n!=0):
if (n%2!=0):
count = count+1
n = n-1
n = n/2
else:
n = n/2
return count
You can do this using bit shifting >> and bitwise and & to inspect the least significant bit, like this:
def count_ones(x):
result = 0
while x > 0:
result += x & 1
x = x >> 1
return result
This works by shifting the bits right until the value becomes zero, counting the number of times the least significant bit is 1 along the way.
The int type has a new method int.bit_count() since python 3.10a, returning the number of ones in the binary expansion of a given integer, also known as the population count as follows:
n = 20
bin(n)
'0b10100'
n.bit_count() returns 2 as it has 2 ones in the binary representation.
For a special case when you need to check quickly whether the binary form of the integer x has only a single 1 (and thus is a power of 2), you can use this check:
if x == -(x | (-x)):
...
The expression -(x | (-x)) is the number that you get if you replace all 1s except the last one (the least significant bit) in the binary representation of x with 0.
Example:
12 = 1100 in binary
-12 = …110100 in binary (with an infinite number of leading 1s)
12 | (-12) = …111100 in binary (with an infinite number of leading 1s)
-(12 | (-12)) = 100 in binary