I want an array to loop when the code throws the IndexError. Here is the Example:
a = [1, 2, 3, 4, 5]
a[x] -> output
0 -> 1
1 -> 2
...
4 -> 5
after except IndexError
5 -> 1
6 -> 2
...
9 -> 5
10 -> IndexError (Should be 1)
My code works but when pos > 9 it still throws the IndexError.
pos = 5
try:
a = [1, 2, 3, 4, 5]
print(a[pos])
except IndexError:
print(a[pos - len(a)])
If you want a circular iterator, use itertools.cycle. If you just want circular behaviour when indexing, you can use modulo-based indexing.
In [20]: a = [1, 2, 3, 4, 5]
In [21]: pos = 9
In [22]: a[pos % len(a)]
Out[22]: 5
That is because when pos > 4 and pos < 10 your code throws an IndexError exception, and it then runs a[pos - len(a)] which will give the desired result.
However, when pos >= 10, the control will go to the Except block, but then the statement a[pos - len(a)] will also give an IndexError exception, becuase pos - len(a) would be greater than 4, since len(a) is a constant.
I would recommend you to implement a circular iterator about which coldspeed said in his answer, or do something like this if you want to follow your approach:
except IndexError:
print(a[pos % len(a)])
P.S. You also don't need to put this whole thing in a try-except block. ^.^
itertools.cycle. Gotta brush up me English 'mate. I just wrote how OP could implement this in his "algorithm", the try-except approach.You want to print a[pos % len(a)], not a[pos - len(a)], as anything greater than 10, minus 5, is greater than 5.
