Find First Occurrence of Substring Before Index

Question

Is there an analogy to myStr.find(subStr, startInd) in order to get the index of the first occurrence of subStr before my startInd . Like taking step -1 instead of 1 from startInd ?

Edit

Here an example:

myStr = "(I am a (cool) str)"

startInd = 9  # print(myStr[startInd]) -> "c"

print(myStr.find(")", startInd))  # -> 13
print(myStr.findBefore("(", startInd))  # -> 8

Edit II

The following code solves my problem but it is not really convenient. Wanted to ask if there is a simple method to fullfill that task

startInd = 9

myStr = "(I am a (cool) str)"

print(myStr.find(")", startInd))  # -> 13
print(len(myStr[::-1]) - myStr[::-1].find("(", len(myStr[::-1]) - startInd - 1) - 1)  # -> 8

Answer 1

str.find takes an optional end parameter:

str.find(sub[, start[, end]])

Return the lowest index in the string where substring sub is found within the slice s[start:end]. Optional arguments start and end are interpreted as in slice notation.

So, if you want subStr to end before endIndex , you can use myStr.find(subStr, 0, endIndex) :

>>> 'hello world'.find('ello', 0, 5)
1
>>> 'hello world'.find('ello', 0, 4)  # "ello" ends at index 5, so it's not found
-1
>>> 'hello world'[0:4]
'hell'

If you want subStr to start anywhere before endIndex , you have to use myStr.find(subStr, 0, endIndex + len(subStr)) instead:

>>> 'hello world'.find('ello', 0, 1 + len('ello'))
1
>>> 'hello world'.find('ello', 0, 0 + len('ello'))  # it starts at index 1, so it's not found
-1