Python auto object generator, auto object naming

Asked Jan 13 '16 at 21:08

Active Jan 13 '16 at 21:20

Viewed 24 times

I tried to create 52 different object to present cards.

 group = ["Karo", "Sinek", "Kupa", "Maca"]
 rank = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
 cardlist = {}

 class Card:
     pass

     def __init__(self, cins, seviye):
         self.cins = cins
         self.seviye = seviye
 x = 1
 for i in group:
     kart = "kart" + str(x) # kart will have values like kart1, kart2...
     for j in rank:
         kart = Card(i,j)   # ^1
         cardlist[x] = kart
         x += 1

 print kart41 # ^2

I wanted to create objects with kart's name. I mean in the first loop kart1 would be the object name and in the second loop kart2... And if I print kart here, I can see all objects with different object id's.
But if I try to print kart41 here, it says that kart41 is not defined.

Can somebody please explain me why kart41 is not defined at 1. And how can I reach the name of object which is defined at 1?

edited Jan 13 '16 at 21:20

asked Jan 13 '16 at 21:08

MossTeMuerA

Point 1 doesn't create any new name. – Ignacio Vazquez-Abrams Jan 13 '16 at 21:09
kart = "kart" + str(x) does it in each loop. Like kart1, kart2, kart3, ... , kart52 – MossTeMuerA Jan 13 '16 at 21:15
No. It doesn't. It makes a string. – Ignacio Vazquez-Abrams Jan 13 '16 at 21:22
Hmm. Okay, thank you. How can i create a name like these(kart1, kart2...) ? – MossTeMuerA Jan 13 '16 at 21:25
See the answer at http://stackoverflow.com/questions/1373164/how-do-i-do-variable-variables-in-python, referenced above your question. You could use `locals()[kart] = Card(i, j); cardlist[x] = locals()[kart]` but you would probably be much better off using your dictionary directly and not creating kart1, kart2, etc.: `cardlist[x] = Card(i, j)` ... `print cardlist[41]`. – Matthias Fripp Jan 13 '16 at 22:03
Thank you. It's clear right now. – MossTeMuerA Jan 13 '16 at 22:08

Python auto object generator, auto object naming

0 Answers0