Divide the list into 4 equal parts in python

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I have list as below

list = [' [00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 10]']

I want tto strip off the "[" and "]" from the list and then divide the list into 4 elements. i.e. 

output_list = [0x00000000 , 0x00000000 , 0x00000000 , 0x00000010]

Any help is appreciated.

Cheers

python
list
asked on Stack Overflow May 3, 2018 by akumar

2 Answers

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Step 1

l = mylist[0].strip(' []').split(' ')

Output:

['00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '00', '10']

Step 2 

parts = [l[x:x+4] for x in range(0, len(l), 4)]

Output:

[['00', '00', '00', '00'],
 ['00', '00', '00', '00'],
 ['00', '00', '00', '00'],
 ['00', '00', '00', '10']]

Step 3

output_list = ['0x%s' % ''.join(p) for p in parts]

Output:

['0x00000000', '0x00000000', '0x00000000', '0x00000010']

To get integers, just do it like this:

[int(op, 16) for op in output_list]

Output:

[0, 0, 0, 16]
answered on Stack Overflow May 3, 2018 by Ashish Acharya • edited May 3, 2018 by Ashish Acharya
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Try this

list = [' [00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 10]']

#removing [ and ] and extra spaces
string=list[0]
string=string.replace("[","")
string=string.replace("]","")
string=string.replace(" ","")


output_list=["0x"+string[i:i+8] for i in range(0,len(string),8)]
print(output_list)
#['0x00000000', '0x00000000', '0x00000000', '0x00000010']

This is a simpler approach. Just take the string from your list and first remove the square brackets. Then it's just a matter of iterating through this string and adding 0x to the beginning of each part.

If the output should be a list of ints and not strings :

output_list=[int("0x"+string[i:i+8],16) for i in range(0,len(string),8)]
#[0, 0, 0, 16]
answered on Stack Overflow May 3, 2018 by Sruthi V • edited May 3, 2018 by Sruthi V

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