How to slice off a certain number of bytes from a string in Python?

Question

I am trying to write a specific number of bytes of a string to a file. In C, this would be trivial: since each character is 1 byte, I would simply write however many characters from the string I want.

In Python, however, since apparently each character/string is an object, they are of varying sizes, and I have not been able to find how to slice the string at byte-level specificity.

Things I have tried:

Bytearray: (For $, read >>>, which messes up the formatting.)

$ barray = bytearray('a')
$ import sys
$ sys.getsizeof(barray[0])
24

So turning a character into a bytearray doesn't turn it into an array of bytes as I expected and it's not clear to me how to isolate individual bytes.

Slicing byte objects as described here:

$ value = b'a'
$ sys.getsizeof(value[:1])
34 

Again, a size of 34 is clearly not 1 byte.

memoryview:

$ value = b'a'  
$ mv = memoryview(value)  
$ sys.getsizeof(mv[0])  
34  
$ sys.getsizeof(mv[0][0])  
34  

ord():

$ n = ord('a')  
$ sys.getsizeof(n)  
24  
$ sys.getsizeof(n[0])  

Traceback (most recent call last):  
  File "<pyshell#29>", line 1, in <module>  
    sys.getsizeof(n[0])  
TypeError: 'int' object has no attribute '__getitem__'  

So how can I slice a string into a particular number of bytes? I don't care if slicing the string actually leads to individual characters being preserved or anything as with C; it just has to be the same each time.

Answer 1

Make sure the string is encoded into a byte array (this is the default behaviour in Python 2.7).

And then just slice the string object and write the result to file.

In [26]: s = '一二三四'

In [27]: len(s)
Out[27]: 12

In [28]: with open('test', 'wb') as f:
   ....:     f.write(s[:2])
   ....:

In [29]: !ls -lh test
-rw-r--r--  1 satoru  wheel     2B Aug 24 08:41 test