How to find a string in a text file

Question

I need to write a program that prints the number of times the string 'bob' occurs in s. For example, if s = 'azcbobobegghakl', then the program should print: 2
thought to use: mystring.find('bob') but Im not sure about this...

clear you question first. may be this link might help http://stackoverflow.com/questions/8899905/count-number-of-occurrences-of-a-given-substring-in-a-string Thanks — Sagar, Sep 17 '16 at 21:40
thanks, the link actually led me to the answer: s = 'xbxbxbbobobxvbcvbgb' sb = 'bob' results = 0 sub_len = len(sb) for i in range(len(s)): if s[i:i+sub_len] == sb: results += 1 print("Number of times 'bob' occurs is: ' " + str(results)) — troy, Sep 18 '16 at 09:14
Also, please don't post code on the comments. If you have an answer, it can go below *as an answer* — OneCricketeer, Sep 18 '16 at 13:17

score 0 · Answer 1 · edited May 23 '17 at 10:34

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str.find will return the position at which the string is found. For example:

"abc".find ("b")  # Returns 1, because it is found at index 1 in the string array.

The find the number of occurrences in the string use:

"ababac".count ("ba")  # Returns 2, because there are 2 "ba" in the string.

Read more on str.count here: count string occurences

edited May 23 '17 at 10:34

Community

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answered Sep 18 '16 at 13:23

Andrei Tumbar

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score 0 · Answer 2 · answered Sep 19 '16 at 14:24

this is the answer I managed to get to. probably one of many:

s = 'xbxbxbbobobxvbcvbgb' sb = 'bob' results = 0 sub_len = len(sb) for i in range(len(s)): if s[i:i+sub_len] == sb: results += 1 print("Number of times 'bob' occurs is: ' " + str(results))

How to find a string in a text file

2 Answers2