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I have one column in pandas data frame with hex values, for example:

Data
1A
2B
BB
FF
A7
78
CB

I want to convert hex values in binary, then from binary to take first 3 bits and finally convert 3 bits value in decimal.

Data column in binary will be:

Data
00011010
00101011
10111011
11111111
10100111
01111000
11001011

the first 3 bits:

Data
010
011
011
111
111
000
011

and finally the desired value in decimal:

Data
2
3
3
7
7
0
3

How to do this? I tried with bin() function, but it doesn't work with pandas data frames.

We can do this by a chain of actions:

1. first we convert the hexadecimal number to an int with .apply(int, base=16);


2. next we convert this to binary data, with .apply(bin);


3. next we chunk off the first two characters with .str[2:];


4. then we obtain the last three characters with .str[-3:]; and


5. finally we again interpret these as ints, with .apply(int, base=2).

So:

>>> df.Data.apply(int, base=16).apply(bin).str[2:].str[-3:].apply(int, base=2)
0 2
1 3
2 3
3 7
4 7
5 0
6 3
Name: Data, dtype: int64

We can however use another strategy here:

1. we first convert the hexadecimal number to an int; and


2. then we apply a bitwise and with 0b111.

for example:

>>> df.Data.apply(int, base=16) & 0b111
0 2
1 3
2 3
3 7
4 7
5 0
6 3
Name: Data, dtype: int64

The second attempt is not only simpler, but faster as well, approximately by 66%:

>>> timeit(first_strategy, number=10000)
6.962630775000434
>>> timeit(second_strategy, number=10000)
2.330652763019316

for a dataframe that repeats the sample data 100 times, we get:

>>> timeit(first_strategy, number=10000)
17.603060900000855
>>> timeit(second_strategy, number=10000)
5.901462858979357

this is again 66% faster.

You can use:

df.Data.apply(lambda v: int(format(int(v, 16), '08b')[-3:], 2))

Which gives you:

0 2
1 3
2 3
3 7
4 7
5 0
6 3
Name: Data, dtype: int64

Those steps are:

• Take your original data and convert it to decimal using int(number, 16) (base 16 is hex) (int('1A', 16) == 26)


• Take that number and format it as a binary string format(number, '08b') gives you an character string of 0/1's zero filled on the left (format(26, '08b') == '00011010')


• Take the last 3 characters of that string [-3:] ('010') and convert it to decimal with a base 2, int(binary_string[-3:], 2) gives you: 2