Strategies for searching through strings in python?

Question

What are efficient ways to search for substrings in strings? - Are there specific functions built in python we can use? - Can we convert them to lists then access elements in the list? - Can we use for loops to search through individual elements? - Is there one generally accepted method by Python programmers?

It would be helpful to understand the different strategies to help me solve the following problem:

Return the number of times that the string "code" appears anywhere in the given string, except we'll accept any letter for the 'd', so "cope" and "cooe" count.

count_code('aaacodebbb') → 1
count_code('codexxcode') → 2
count_code('cozexxcope') → 2

Ashwini Chaudhary · Accepted Answer · 2014-01-23 05:56:44Z

2

You can use regex:

>>> import re
def count(s):
    return sum(1 for m in re.finditer(r'co.e', s))
... 
>>> count('aaacodebbb')
1
>>> count('codexxcode')
2
>>> count('cozexxcope')
2

Here . matches any character, if you only want to match alphabets then use r'co[a-zA-Z]e'.

answered Jan 23, 2014 at 5:56

Ashwini Chaudhary

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6 Comments

KishB87 Over a year ago

How could this be done without using a external library?

SethMMorton Over a year ago

re is part of the python standard library, so it is no more external than sys or os.

Ben Southgate Over a year ago

as per @robertking s answer -> 'codexxcode'.count("code") no regex or modules at all!

Tarzan Over a year ago

I'm curious that why do you use "sum(...)" instead of len(...) ?

Ashwini Chaudhary Over a year ago

@Tarzan What's the point of creating a list in memory when only count is required as output.

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Kamehameha · Accepted Answer · 2014-01-23 06:00:59Z

0

Using regular expression would work for this-

import re
regex = re.compile(r'co[a-z]e')
li = regex.findall("cozexxcope")           #Output: ['coze', 'cope']

Count can be found be the len(li).

answered Jan 23, 2014 at 6:00

Kamehameha

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Comments

Zycarsus · Accepted Answer · 2016-06-28 20:34:09Z

0

    def count_code(str):
      count = 0 #set count to 0 initially
        #For loops number of times that there are chars
      for i in range(len(str)-1): 
         #looks for "co", [any char] and "e"
        if str[i:i+2] == "co" and str[i+3:i+4] == "e":  
            #add up the number of times this happens and return it
          count+= 1 
    return count

answered Jun 28, 2016 at 20:34

Zycarsus

1

1 Comment

nbryans Over a year ago

In addition to providing code, you should explain to OP what/how this code works and why you took this approach.

Rafael Sartori · Accepted Answer · 2021-01-07 03:59:17Z

0

This was my solution:

def count_code(str):
  count = 0
  for i in range(0, len(str)-3):
    if str[i] == "c" and str[i+1] == "o" and str[i+3] == "e":
      count = count + 1
  return count

answered Jan 7, 2021 at 3:59

Rafael Sartori

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