How to count binary sequence in binary number in Python?

Question

I would like to count '01' sequence in 5760 binary bits.

First, I would like to combine several binary numbers then count # of '01' occurrences.

For example, I have 64 bits integer. Say, 6291456. Then I convert it into binary. Most significant 4 bits are not used. So I'll get 60 bits binary 000...000011000000000000000000000 Then I need to combine(just put bits together since I only need to count '01') first 60 bits + second 60 bits + ...so 96 of 60 bits are stitched together.

Finally, I want to count how many '01' appears.

s = binToString(5760 binary bits)
cnt = s.count('01');

"Is there a function to convert binary into string?" <- I'm not sure what you mean. '01000000000100000000000' is already a string. Can you give examples of the input and output you'd want for such a function? — Mark Dickinson
– Mark Dickinson, Commented Aug 10, 2017 at 19:26
@MarkDickinson Sorry I translated it into binary. I just updated the value. But, note that I will combine 96 of 60 bits into 5760 binary bits. — ejshin1
– ejshin1, Commented Aug 10, 2017 at 19:34
@MarkDickinson It's not the same problem. I'm converting integer to binary then to String. — ejshin1
– ejshin1, Commented Aug 10, 2017 at 19:35
@ejshin1: What do you mean by "then to String"? A binary representation of an integer is already a string, consisting of the characters '0' and '1'. The solutions in the linked question all return strings. — Mark Dickinson
– Mark Dickinson, Commented Aug 10, 2017 at 19:36

Shrikant Shete · Accepted Answer · 2017-08-10 20:28:32Z

1

num = 6291226
binary = format(num, 'b')
print(binary)
print(binary.count('01'))

If I use number given by you i.e 6291456 it's binary representation is 11000000000000000000000 which gives 0 occurrences of '01'.

If you always want your number to be 60 bits in length you can use

binary = format(num,'060b')

It will add leading 0 to make it of given length

edited Aug 10, 2017 at 20:28

answered Aug 10, 2017 at 19:40

Shrikant Shete

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2 Comments

ejshin1 Over a year ago

Thanks! That's what I was looking for. It worked for me.

Shrikant Shete Over a year ago

Since you have updated the question and as @johnColeman said regarding not using most 4 significant bits I have changed numbers in the answer

John Coleman · Accepted Answer · 2017-08-10 20:16:03Z

0

Say that nums is your list of 96 numbers, each of which can be stored in 64 bits. Since you want to throw away the most 4 significant bits, you are really taking the number modulo 2**60. Thus, to count the number of 01 in the resulting string, using the idea of @ShrikantShete to use the format function, you can do it all in one line:

''.join(format(n%2**60,'060b') for n in nums).count('01')

answered Aug 10, 2017 at 20:16

John Coleman

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