Is there an operator to specify an index "not less than 0"?

Question

To "peek" at characters in a string preceding the current index, i, is there a one-liner for avoiding negative indices? I want n chars before the current char index of the string OR beginning of the string to current char.

My intuition says there might be something simple I can put in the same line as the list operation instead of another line to make an index. I'd rather avoid any libraries, and I'm aware a simple if check would work... Hoping there's a magic operator I missed.


>>> s = 'ABCDEFG'
>>> s[0:5]
'ABCDE'
>>> s[-5:5]
'CDE'

I think you'll need an `if` if you want to always start at `0`. — Code-Apprentice, Aug 25 '20 at 23:12

score 3 · Answer 1 · answered Aug 25 '20 at 23:15

3

There is no operator for doing this, but you can achieve the same effect with max:

>>> s = 'abcdefg'
>>> i = 3
>>> s[max(i - 6, 0):i + 2]
'abcde'

answered Aug 25 '20 at 23:15

gmds

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i think this is the answer, but abhilash got it just a few second earlier; i'll give them the answer. thanks for helping – 123 Aug 25 '20 at 23:19

score 2 · Accepted Answer · answered Aug 25 '20 at 23:15

2

How about this?

s = 'ABCDEFG'
i = 5
n = 3

print(s[max(0, i-n): i])

Output:

CDE

answered Aug 25 '20 at 23:15

Abhilash

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totally. thanks! it's unlikely `max()` on two ints would be a bottleneck right? – 123 Aug 25 '20 at 23:21
1

true it'd run in constant time, so not an issue. :) – Abhilash Aug 25 '20 at 23:21

Is there an operator to specify an index "not less than 0"?

2 Answers2