Finding multiple occurrences of a string within a

Question

How do I find multiple occurrences of a string within a string in Python? Consider this:

>>> text = "Allowed Hello Hollow"
>>> text.find("ll")
1
>>> 

So the first occurrence of ll is at 1 as expected. How do I find the next occurrence of it?

Same question is valid for a list. Consider:

>>> x = ['ll', 'ok', 'll']

How do I find all the ll with their indexes?

Answer 1

Using regular expressions, you can use re.finditer to find all (non-overlapping) occurences:

>>> import re
>>> text = 'Allowed Hello Hollow'
>>> for m in re.finditer('ll', text):
         print('ll found', m.start(), m.end())

ll found 1 3
ll found 10 12
ll found 16 18

Alternatively, if you don't want the overhead of regular expressions, you can also repeatedly use str.find to get the next index:

>>> text = 'Allowed Hello Hollow'
>>> index = 0
>>> while index < len(text):
        index = text.find('ll', index)
        if index == -1:
            break
        print('ll found at', index)
        index += 2 # +2 because len('ll') == 2

ll found at  1
ll found at  10
ll found at  16

This also works for lists and other sequences.

Answer 2

>>> for n,c in enumerate(text):
...   try:
...     if c+text[n+1] == "ll": print n
...   except: pass
...
1
10
16

Answer 3

For your list example:

In [1]: x = ['ll','ok','ll']

In [2]: for idx, value in enumerate(x):
   ...:     if value == 'll':
   ...:         print idx, value       
0 ll
2 ll

If you wanted all the items in a list that contained 'll', you could also do that.

In [3]: x = ['Allowed','Hello','World','Hollow']

In [4]: for idx, value in enumerate(x):
   ...:     if 'll' in value:
   ...:         print idx, value
   ...:         
   ...:         
0 Allowed
1 Hello
3 Hollow

Answer 4

Brand new to programming in general and working through an online tutorial. I was asked to do this as well, but only using the methods I had learned so far (basically strings and loops). Not sure if this adds any value here, and I know this isn't how you would do it, but I got it to work with this:

needle = input()
haystack = input()
counter = 0
n=-1
for i in range (n+1,len(haystack)+1):
   for j in range(n+1,len(haystack)+1):
      n=-1
      if needle != haystack[i:j]:
         n = n+1
         continue
      if needle == haystack[i:j]:
         counter = counter + 1
print (counter)

Answer 5

You can split to get relative positions then sum consecutive numbers in a list and add (string length * occurence order) at the same time to get the wanted string indexes.

>>> key = 'll'
>>> text = "Allowed Hello Hollow"
>>> x = [len(i) for i in text.split(key)[:-1]]
>>> [sum(x[:i+1]) + i*len(key) for i in range(len(x))]
[1, 10, 16]
>>> 

Answer 6

A simple iterative code which returns a list of indices where the substring occurs.

        def allindices(string, sub):
           l=[]
           i = string.find(sub)
           while i >= 0:
              l.append(i)
              i = string.find(sub, i + 1)
           return l