Python decode and encode

Question

I cannot figure out how to feed the list of decimal codepoints for them to be decoded (or encoded?) via a specific codepage.

str_0 = 'abc'
lst_decimal = list(str_0.encode("utf-8"))
for i in lst_decimal:
 print (str(int(i)).decode("cp437"))

Error: 'str' object has no attribute 'decode'

Answer 1

str_0.encode("utf-8") returns a bytes object, which has a decode method, but when you turn it into a list, it becomes a list on ints. Just keep it as a bytes object:

my_bytes = str_0.encode("utf-8") 
print (my_bytes.decode("cp437")) #prints abc

Even more simply:

print(str_0.encode("utf-8").decode("cp437"))

As an added benefit, there is no need for using decode in a loop -- the entire bytes object can be decoded at once.

If you wanted to keep your original lst_decimal and do what you were trying to do, your loop could look like:

for i in lst_decimal:
 print(bytes([i]).decode("cp437"))

list() turns a bytes object into a list of ints, and bytes goes backwards. Note, however that simply bytes(i) returns a list of i bytes, each initialized to 0.

• Dmitry Starostin: I stumbled upon print (struct.pack('B', int(i).decode('cp866'))), this did what I wanted. But is there a way to it with the method you suggested? 'i' the original decimal value I want to have a possibility to change.
• John Coleman: @DmitryStarostin What I gave at the end should work. In your comment here, i is already an integer so int(i) is superfluous. bytes([i]).decode() works -- but note the square brackets! There is an int.to_bytes() method, but it isn't very convenient (you need specify number of bytes and byte order). One complicating fact is that when you convert bytes to a list of ints, those ints are in range 0-255. With Unicode, an individual character might correspond to 2 or even more bytes. ord() gives you the actual codepoint of a Unicode character.