Problem when appending a variable with a numbering scheme to a list

Question

a1 = 0
a2 = 1

x = [] #here I have declared an empty list
for i in range(2):
    x.append('a'+str(i+1)) #to append the variable with a numbering scheme
print (x)

This is a sample python code. A similar situation that I am facing in a programming task .

Here the output is ['a1','a2'] instead I need the output as [0,1]. Can someone help me with this ?

Possible duplicate of [How do I create a variable number of variables?](https://stackoverflow.com/questions/1373164/how-do-i-create-a-variable-number-of-variables) — Georgy, Jun 11 '19 at 13:57
@Error-SyntacticalRemorse in this example, `i` is correct but he wants to use the values of `a1` and `a2`. Your comment won't work in more elaborate examples — DSC, Jun 11 '19 at 14:02
The marked duplicate should be considered than, and you are most likely doing something wrong. You probably want to use a `dict`. — Error - Syntactical Remorse, Jun 11 '19 at 14:04

score 2 · Answer 1 · answered Jun 11 '19 at 14:01

2

Use a dictionary for this:

d = {'a1': 0, 'a2':1}
x = [] #here I have declared an empty list
for i in range(2):
    x.append(d['a'+str(i+1)]) #to append the variable with a numbering scheme
print (x)

answered Jun 11 '19 at 14:01

DSC

1,153
7
21

Adam.Er8 · Accepted Answer · 2019-06-11T14:03:52.080

1

If you MUST use variables already existing in your scope, you can use locals() to get all local variables as a dict

a1 = 0
a2 = 1

x = [] #here I have declared an empty list
for i in range(2):
    x.append(locals()['a'+str(i+1)]) #to append the variable with a numbering scheme
print (x)

edited Jun 11 '19 at 14:03

answered Jun 11 '19 at 14:02

Adam.Er8

12,675
3
26
38

Good for indicating that this should be avoided at all costs – DSC Jun 11 '19 at 14:03
2

Even then it would probably be better (less bad?) to use `locals` or `globals` instead of `eval`. (Oh, you just did while I typed this...) – tobias_k Jun 11 '19 at 14:04
Hi , Thankyou so much for the help . I dont have much experience in programming. Also the solution that you gave me works like I expected. I would like to know what consequences will I have in this kind of programming since you have mentioned "very bad programming practice" – Aadithya Sathya Jun 11 '19 at 14:09
1

eval/exec are generally abused to do a task where they aren't needed, very slow, opens up security issues,.... – DSC Jun 11 '19 at 14:12
@AadithyaSathya my first version was using `eval`, which is not recommended because it executes code (it interprents the parameter as python code), which can lead to severe security issues and bugs. – Adam.Er8 Jun 11 '19 at 14:13

score 0 · Answer 3 · answered Jun 11 '19 at 14:05

0

vars = {
    'a1': 0,
    'a2': 1,
}

x = []

for var in vars.keys():
    x.append(vars[var])

print(x)

answered Jun 11 '19 at 14:05

Gokhan Gerdan

1,222
6
19

please add an explanation to your answer – d_kennetz Jun 11 '19 at 17:27

Problem when appending a variable with a numbering scheme to a list

3 Answers3