How to concatenate bytes together

Question

Here is the code snippet:-

join_values = []
new_byteArray = [128, 0, 0, 0, 86, 70, 124, -96]
for values in byteArray:
 values = long(values)
 store_values = struct.pack('!q', values)
 join_values.append(store_values)
print join_values

This produces correct result which is:-

['\x00\x00\x00\x00\x00\x00\x00\x80',
 '\x00\x00\x00\x00\x00\x00\x00\x00', 
 '\x00\x00\x00\x00\x00\x00\x00\x00', 
 '\x00\x00\x00\x00\x00\x00\x00\x00', 
 '\x00\x00\x00\x00\x00\x00\x00V', 
 '\x00\x00\x00\x00\x00\x00\x00F', 
 '\x00\x00\x00\x00\x00\x00\x00|', 
 '\xff\xff\xff\xff\xff\xff\xff\xa0']

This result is correct but isn't there any way I can do a combine pack or do something so that the last out of 8 bytes concatenate together so that I can get the output as:-

\x80\x00\x00x00VF|\xa0

• David Zemens: is there a typo in your desired output? Shouldn't it be \x80\x00\x00\x00\x00V\x00F\x00|\xa0

Answer 1

Well, you can get the last byte from a string s by using s[-1].

python2.7 has a feature called list comprehension which can be used to perform a transformation on each element of a list.

Finally, join the strings together.

Putting that all together would look something like

my_transformed_list = [s[-1] for s in join_values]
concatenated = ''.join(my_transformed_list)

---

Another way to solve this would be with functional programming. The act of performing a transformation on every element of a list is called a map:

my_transformed_list2 = map(lambda s: s[-1], j)
concatenated2 = ''.join(my_transformed_list2)