How to create a list of variables that differ from each other with only last character in python

Question

I have a lot of identical variables that differ only with the last character

b1 = Class(argument)
b2 = Class(argument)
b3 = Class(argument)
b4 = Class(argument)
b5 = Class(argument)
b6 = Class(argument)

Is there a way to create them in a loop or somehow?

Possible duplicate of http://stackoverflow.com/questions/8799446/is-it-possible-to-dynamically-create-local-variables-in-python — twasbrillig, Dec 26 '14 at 08:53

score 3 · Answer 1 · edited Dec 26 '14 at 08:51

3

You can use a dictionary for this purpose

d = {}
for i in range(1, 7):
    d['b' + str(i)] = Class(argument)

edited Dec 26 '14 at 08:51

twasbrillig

answered Dec 26 '14 at 07:39

hyades

atupal · Answer 2 · 2014-12-26T07:50:48.950

2

for i in xrange(1,7):exec("b%d=Class(argument)" % i)

Then you can check with:

from pprint import pprint
pprint(locals())

edited Dec 26 '14 at 07:50

answered Dec 26 '14 at 07:43

atupal

So I made this: `for i in range(1,7):exec("b%d=Class(argument)" % i)` - changed xrange() to range(). Thanks! – messier101 Dec 26 '14 at 08:10

score 2 · Answer 3 · answered Dec 26 '14 at 08:00

2

You could also use a list

b = []
for i in range(1,7):
    b.append(Class(argument))

or

b = [Class() for x in range(1,7)]

answered Dec 26 '14 at 08:00

baited

3 Answers3