Python Reverse Find in String

Question

I have a string and an arbitrary index into the string. I want find the first occurrence of a substring before the index.

An example: I want to find the index of the 2nd I by using the index and str.rfind()

s = "Hello, I am 12! I like plankton but I don't like Baseball."
index = 34 #points to the 't' in 'but'
index_of_2nd_I = s.rfind('I', index)
#returns = 36 and not 16 

Now I would expect rfind() to return the index of the 2nd I (16) but it returns 36. after looking it up in the docs I found out rfind does not stand for reverse find.

I'm totally new to Python so is there a built in solution to reverse find? Like reversing the string with some python [::-1] magic and using find, etc? Or will I have to reverse iterate char by char through the string?

Answer 1

Your call tell rfind to start looking at index 34. You want to use the rfind overload that takes a string, a start and an end. Tell it to start at the beginning of the string (0) and stop looking at index:

>>> s = "Hello, I am 12! I like plankton but I don't like Baseball."
>>> index = 34 #points to the 't' in 'but'
>>> index_of_2nd_I = s.rfind('I', 0, index)
>>>
>>> index_of_2nd_I
16

Answer 2

I became curious how to implement looking n times for string from end by rpartition and did this nth rpartition loop:

orig = s = "Hello, I am 12! I like plankton but I don't like Baseball."
found = tail = ''
nthlast = 2
lookfor = 'I'
for i in range(nthlast):
    tail = found+tail
    s,found,end = s.rpartition(lookfor)
    if not found:
        print "Only %i (less than %i) %r in \n%r" % (i, nthlast, lookfor, orig)
        break
    tail = end + tail
else:
    print(s,found,tail)

Answer 3

To find the 2nd 'I':
s.find('I', s.find('I')+1) # returns 16

Find the last 'I':
s.rfind('I') # returns 36

Find the indeces of all 'I' occurrences:
locs = [idx for idx,ltr in enumerate(s) if ltr == 'I'] # [7, 16, 36]