CODEWITHSUNDEEP
pythonloopsiterationinfinite-loop
Ruben_phy
Ruben_phy

Reputation: 73

How can I loop through an unknown number of buckets in python?

I want to obtain a sequence like this

a,b,c,.....,x,y,z,aa,ab,ac,....,az,ba,bb,....,bz,....,aaa,....,azz....,zzz,...

and so on, until I achieve a given length.

My problem is that I don't know how to code an infinite number of loops.

abc = 'abcdefghijklmnopqrstuvwxyz'

def next_plate(i):
    for letter in abc:
        i += 1
        yield letter, i

num_plates = 10000
i = 0
all_plates = {}

for plate,i in next_plate(i):
    if i > num_plates:
        break
    all_plates[i] = plate

if i < num_plates:
    for plate_1, _ in next_plate(i):
        for plate_2,i in next_plate(i):
            if i > num_plates:
                break
            all_plates[i] = plate_1 + plate_2

if i < num_plates:
    for plate_1, _ in next_plate(i):
        for plate_2, _ in next_plate(i):
            for plate_3,i in next_plate(i):
                if i > num_plates:
                    break
                all_plates[i] = plate_1 + plate_2 + plate_3

Could someone help me with this?

Upvotes: 2

Views: 283

Answers (2)

John Coleman
John Coleman

Reputation: 52008

Here is an itertools based generator:

import itertools, string

def plates():
    n = 1
    while True:
        for plate in itertools.product(string.ascii_lowercase,repeat = n):
            yield ''.join(plate)
        n += 1

#test:

p = plates()
test = [next(p) for _ in range(10000)]
print(test[:30])
print(test[-30:])

output:

['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']
['nsm', 'nsn', 'nso', 'nsp', 'nsq', 'nsr', 'nss', 'nst', 'nsu', 'nsv', 'nsw', 'nsx', 'nsy', 'nsz', 'nta', 'ntb', 'ntc', 'ntd', 'nte', 'ntf', 'ntg', 'nth', 'nti', 'ntj', 'ntk', 'ntl', 'ntm', 'ntn', 'nto', 'ntp']

Upvotes: 2

Icelain17l
Icelain17l

Reputation: 11

You need to use the while loop and set it to true(adding the condition to break when your condition is met). I don't have my computer now so the answer is very brief.

Upvotes: 0

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