Change count into an array of letters, python

Question

I am struggling with an idea I want put into a python script. I'm not even sure how to ask the most appropriate question as I've been surfing through the net trying to find what I want with no luck.

Basically, I have a script that does a simple calculation:

divider = int(math.ceil(df.scale / 3000))

This is because I want ' divider' to return the value divided by 3000 and always rounded up. I want to use that value to help me return letters.

So, it goes like this:

if 1 returns then I want a to return 'A' if 2 returns, then I want to create 'A', 'B' if 3 returns, then I want to create 'A', 'B', 'C' and so on....

My end result is, that I want to save some files. 'divider' will determine how many files I want to save and then each file will recursively be named with the letter in it (i.e. FileA, FileB, FileC...)

Ok, I know my question isn't exactly well put together, but I'm struggling with the logic, so if you need some clarity, please let me know.

Niklas B. · Accepted Answer · 2012-03-20 20:18:07Z

6

Do you mean something like:

for i in range(int(math.ceil(df.scale / 3000))):
  # i will contain 0, 1, 2, ...
  # letter will contain 'A', 'B', 'C', ...
  letter = chr(ord('A') + i)

Or if you need the actual list:

[chr(ord('A') + i) for i in range(int(math.ceil(df.scale / 3000)))]

You can also use string.ascii_uppercase for a list of uppercase letters and slice it as you need:

from string import ascii_uppercase

print list(ascii_uppercase)[:int(math.ceil(df.scale / 3000))]

edited Mar 20, 2012 at 20:18

answered Mar 20, 2012 at 20:11

Niklas B.

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1 Comment

Mike Over a year ago

Niklas. Thanks for interpreting my gibberish question and coming up with exactly what I had in mind!

DSM · Accepted Answer · 2012-03-20 20:29:11Z

3

Because it's the end of the day for me, here's another approach which generalizes to more than 26 files, spreadsheet-column-name style:

import string, itertools

def name_generator(min_length=1, alphabet=string.ascii_uppercase):
    for length in itertools.count(min_length):
        for chars in itertools.product(alphabet, repeat=length):
            yield ''.join(chars)

def get_names(n):
    return list(itertools.islice(name_generator(), 0, n))

which gives

>>> get_names(1)
['A']
>>> get_names(3)
['A', 'B', 'C']
>>> get_names(30)
['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', 'AA', 'AB', 'AC', 'AD']

answered Mar 20, 2012 at 20:29

DSM

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1 Comment

Mike Over a year ago

Thanks DSM. I like the define function approach. I had thought about using one when trying to calculate my end result and I still just might do that.

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