Field calculate sequential letters

[RIMNTIR]

Hello,

I am trying to figure out how I can use the field calculator with python to calculate sequential letters. For example, I would like "A,B,C,D....Z, AA,AB, AC...".  I did find a code that works just fine in a regular python environment but I am having a hard time incorporating it on the field calculator. 

excel - How to make a continuous alphabetic list python (from a-z then from aa, ab, ac etc) - Stack ... 

I tried using this as the code block:

"""

from string import ascii_lowercase
  import itertools

def iter_all_strings():
   size = 1
   while True:
      for s in itertools.product(ascii_lowercase, repeat=size):
            yield "".join(s)
        size +=1

"""

and  "iter_all_strings()" as the expression in the field calculator but it just gives me null values.

Any help would be much appreciated, thank you. 

[DanPatterson_Retired]

how big is the file? aka does it cut off at ZZ?  If it is short, it might just be easier to generate the list, then slice it based on a unique ID field (perhaps OBJECTID -1 )

[Rhys-Donoghue]

This is fairly straightforward.  Sorry, I don't have the time to write the code for you but if you check the end of this video I did, you can see how I used Python for creating the letters: https://www.youtube.com/watch?v=QA0iFlP0V3Y - I needed A1 to AA1 to AAA1, etc. If you're really stuck and still can't work it out, let me know and I can create a working example for you.

[RandyBurton]

Using a modified number base function, try this in the code block:

i = -1

def letterString():
    global i
    i += 1
    return toStr(i,26)

def toStr(n,base):
    convertString = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
    if n < base:
        return convertString[n]
    else:
        return toStr(n//base-1,base) + convertString[n%base]‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍

and this for the expression

letterString()‍‍

Basically, you will need to set up a global variable in your code block and increment it each time your function is called.

[RandyBurton · Accepted solution]

And using your code, this would go in the code block:

from string import ascii_lowercase
import itertools

def iter_all_strings():
    size = 1
    while True:
        for s in itertools.product(ascii_lowercase, repeat=size):
            yield "".join(s)
        size +=1
        
s = iter_all_strings()
‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍

The expression would be:

s.next()‍‍

The generator needs to be set up in the code block.

[RIMNTIR]

Randy, 

Thank you very much, problem solved!

[JoshuaBixby]

Expression:  next(values)

Expression Type:  PYTHON_9.3

Code Block:

from itertools import count, product
from string import ascii_lowercase

cnt = count(1)
values = ("".join(v)
          for c in cnt
          for v in product(ascii_lowercase, repeat=c))‍‍‍‍‍‍‍

[TedKowal]

from string import ascii_uppercase
import itertools

def iter_all_strings():
    size = 1
    while True:
        for s in itertools.product(ascii_uppercase, repeat=size):
            yield "".join(s)
        size +=1

for s in itertools.islice(iter_all_strings(), 54):
    print s‍‍‍‍‍‍‍‍‍‍‍‍

Just save your stopping point in line 11 (number of iterations you want  54... I believe is "BB"

[RandyBurton]

A clean, efficient solution.  Thanks for sharing bixb0012‌.