A cleaner way to generate a list of running IDs

Question

I just made a function to generate a list of running ids between a given range. IDs begin with an alphabet and follow with 5 numbers (e.g. A00002). The function below works, but I was wondering if there was a cleaner way to do this. Thanks!

def running_ids(start, end):
list = []
start = int(start[1:])
end = int(end[1:])
steps = end - start

def zeros(n):
    zeros = 5 - len(str(n))
    return zeros

while start <= end:
    string = "A" + "0"*zeros(start) + str(start)
    list.append(string)
    start += 1

return list

print running_ids('A00001', 'A00005')

['A00001', 'A00002', 'A00003', 'A00004', 'A00005']

Answer 1

Use a generator. This way you can generate the numbers as needed and not store them all at once. It also maintains the state of your counter, useful if you start building large projects and you forget to add one to your index. It's a very powerful way of programming in Python:

def running_id():
    n = 1
    while True:
        yield 'A{0:05d}'.format(n)
        n += 1

C = running_id()
for n in xrange(5):
    print next(C)

Giving:

A00001
A00002
A00003
A00004
A00005

Answer 2

You could just use simple builtin string formatting:

>>> 'A%05d'%1
'A00001'
>>> 'A{0:05d}'.format(1)
'A00001'

Answer 3

You can use the builtin format method

print "A" + format(1, "05d")   # A00001
print "A" + format(100, "05d") # A00100

Or you can use str.zfill method like this

print "A" + str(1).zfill(5)    # A00001
print "A" + str(100).zfill(5)  # A00100

Answer 4

def running_ids(start, end):
    t = start[0]
    low = int(start[1:])
    high = int(end[1:]) + 1
    res = []
    for x in range(low, high):
        res.append(t + '{0:05d}'.format(x))
    return res

print(running_ids('A00001', 'A00005'))