Python 3: finding a set of letters in a string using loops

i'm using python 3 and i would like to know how to count the amount of times a set of 3 letters comes up in a sentence:

con = ""

count = 0

for i in mystery_string:

    if i == "c":

        con = "c" 

    if i == "a" and con == "c":

        con = "ca"   

    if i == "t" and con == "ca":

        con = "cat"

    if con == "cat":

        count +=1

        con = ""

print (count)

this is the code i have so far, however it doesn't seem to work for every case
can someone help me

the thing i also need to explain is that i cannot use the builtin function count()

edited Nov 24 '18 at 18:28

asked Nov 24 '18 at 18:05

Theo Boston

115

1

Can you also provide a sample input and expected output?

– slider
Nov 24 '18 at 18:09

5

Possible duplicate of Count the number occurrences of a character in a string

– Muzol
Nov 24 '18 at 18:13

add a comment |

i'm using python 3 and i would like to know how to count the amount of times a set of 3 letters comes up in a sentence:

con = ""

count = 0

for i in mystery_string:

    if i == "c":

        con = "c" 

    if i == "a" and con == "c":

        con = "ca"   

    if i == "t" and con == "ca":

        con = "cat"

    if con == "cat":

        count +=1

        con = ""

print (count)

this is the code i have so far, however it doesn't seem to work for every case
can someone help me

the thing i also need to explain is that i cannot use the builtin function count()

edited Nov 24 '18 at 18:28

asked Nov 24 '18 at 18:05

Theo Boston

115

1

Can you also provide a sample input and expected output?

– slider
Nov 24 '18 at 18:09

5

Possible duplicate of Count the number occurrences of a character in a string

– Muzol
Nov 24 '18 at 18:13

add a comment |

i'm using python 3 and i would like to know how to count the amount of times a set of 3 letters comes up in a sentence:

con = ""

count = 0

for i in mystery_string:

    if i == "c":

        con = "c" 

    if i == "a" and con == "c":

        con = "ca"   

    if i == "t" and con == "ca":

        con = "cat"

    if con == "cat":

        count +=1

        con = ""

print (count)

this is the code i have so far, however it doesn't seem to work for every case
can someone help me

the thing i also need to explain is that i cannot use the builtin function count()

edited Nov 24 '18 at 18:28

asked Nov 24 '18 at 18:05

Theo Boston

115

i'm using python 3 and i would like to know how to count the amount of times a set of 3 letters comes up in a sentence:

con = ""

count = 0

for i in mystery_string:

    if i == "c":

        con = "c" 

    if i == "a" and con == "c":

        con = "ca"   

    if i == "t" and con == "ca":

        con = "cat"

    if con == "cat":

        count +=1

        con = ""

print (count)

this is the code i have so far, however it doesn't seem to work for every case
can someone help me

the thing i also need to explain is that i cannot use the builtin function count()

python

edited Nov 24 '18 at 18:28

asked Nov 24 '18 at 18:05

Theo Boston

115

edited Nov 24 '18 at 18:28

asked Nov 24 '18 at 18:05

Theo Boston

115

edited Nov 24 '18 at 18:28

asked Nov 24 '18 at 18:05

Theo Boston

115

asked Nov 24 '18 at 18:05

Theo Boston

115

asked Nov 24 '18 at 18:05

Theo Boston

115

1

Can you also provide a sample input and expected output?

– slider
Nov 24 '18 at 18:09

5

Possible duplicate of Count the number occurrences of a character in a string

– Muzol
Nov 24 '18 at 18:13

add a comment |

1

Can you also provide a sample input and expected output?

– slider
Nov 24 '18 at 18:09

5

Possible duplicate of Count the number occurrences of a character in a string

– Muzol
Nov 24 '18 at 18:13

Can you also provide a sample input and expected output?

– slider
Nov 24 '18 at 18:09

Possible duplicate of Count the number occurrences of a character in a string

– Muzol
Nov 24 '18 at 18:13

add a comment |

2 Answers
2

active

oldest

votes

you can use slicing:

word='cat'

count = 0

for i in range(len(mystery_string)-len(word)+1):

    if mystery_string[i:i+len(word)]==word:

        count +=1

print (count)

answered Nov 24 '18 at 18:14

prophet-five

1089

this solution supports overlapping, contrary to count() function, for example the word 'pop' will be counted once in the string 'popop' if you use the count function

– prophet-five
Nov 24 '18 at 18:17

Thanks @slider, I guess some people think they know better

– prophet-five
Nov 24 '18 at 18:36

add a comment |

just do the following

mystery_string.count('cat')

answered Nov 24 '18 at 18:10

Biswadip Mandal

1809

1

It is a good solution for this specific word, however the count() method does not support overlapping

– prophet-five
Nov 24 '18 at 18:38

1

you mean examples like if I am counting 'aba' in 'gtfababacd' ?

– Biswadip Mandal
Nov 24 '18 at 18:41

yes, exactly, this will return 1 instead of 2.

– prophet-five
Nov 24 '18 at 18:44

cool. sounds about right. Thanks for mentioning

– Biswadip Mandal
Nov 24 '18 at 18:45

sure, np, I recently encountered this problem myself :)

– prophet-five
Nov 24 '18 at 18:47

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

you can use slicing:

word='cat'

count = 0

for i in range(len(mystery_string)-len(word)+1):

    if mystery_string[i:i+len(word)]==word:

        count +=1

print (count)

answered Nov 24 '18 at 18:14

prophet-five

1089

this solution supports overlapping, contrary to count() function, for example the word 'pop' will be counted once in the string 'popop' if you use the count function

– prophet-five
Nov 24 '18 at 18:17

Thanks @slider, I guess some people think they know better

– prophet-five
Nov 24 '18 at 18:36

add a comment |

you can use slicing:

word='cat'

count = 0

for i in range(len(mystery_string)-len(word)+1):

    if mystery_string[i:i+len(word)]==word:

        count +=1

print (count)

answered Nov 24 '18 at 18:14

prophet-five

1089

this solution supports overlapping, contrary to count() function, for example the word 'pop' will be counted once in the string 'popop' if you use the count function

– prophet-five
Nov 24 '18 at 18:17

Thanks @slider, I guess some people think they know better

– prophet-five
Nov 24 '18 at 18:36

add a comment |

you can use slicing:

word='cat'

count = 0

for i in range(len(mystery_string)-len(word)+1):

    if mystery_string[i:i+len(word)]==word:

        count +=1

print (count)

answered Nov 24 '18 at 18:14

prophet-five

1089

you can use slicing:

word='cat'

count = 0

for i in range(len(mystery_string)-len(word)+1):

    if mystery_string[i:i+len(word)]==word:

        count +=1

print (count)

answered Nov 24 '18 at 18:14

prophet-five

1089

answered Nov 24 '18 at 18:14

prophet-five

1089

answered Nov 24 '18 at 18:14

prophet-five

1089

answered Nov 24 '18 at 18:14

prophet-five

1089

this solution supports overlapping, contrary to count() function, for example the word 'pop' will be counted once in the string 'popop' if you use the count function

– prophet-five
Nov 24 '18 at 18:17

Thanks @slider, I guess some people think they know better

– prophet-five
Nov 24 '18 at 18:36

add a comment |

this solution supports overlapping, contrary to count() function, for example the word 'pop' will be counted once in the string 'popop' if you use the count function

– prophet-five
Nov 24 '18 at 18:17

Thanks @slider, I guess some people think they know better

– prophet-five
Nov 24 '18 at 18:36

this solution supports overlapping, contrary to count() function, for example the word 'pop' will be counted once in the string 'popop' if you use the count function

– prophet-five
Nov 24 '18 at 18:17

Thanks @slider, I guess some people think they know better

– prophet-five
Nov 24 '18 at 18:36

add a comment |

just do the following

mystery_string.count('cat')

answered Nov 24 '18 at 18:10

Biswadip Mandal

1809

1

It is a good solution for this specific word, however the count() method does not support overlapping

– prophet-five
Nov 24 '18 at 18:38

1

you mean examples like if I am counting 'aba' in 'gtfababacd' ?

– Biswadip Mandal
Nov 24 '18 at 18:41

yes, exactly, this will return 1 instead of 2.

– prophet-five
Nov 24 '18 at 18:44

cool. sounds about right. Thanks for mentioning

– Biswadip Mandal
Nov 24 '18 at 18:45

sure, np, I recently encountered this problem myself :)

– prophet-five
Nov 24 '18 at 18:47

add a comment |

just do the following

mystery_string.count('cat')

answered Nov 24 '18 at 18:10

Biswadip Mandal

1809

1

It is a good solution for this specific word, however the count() method does not support overlapping

– prophet-five
Nov 24 '18 at 18:38

1

you mean examples like if I am counting 'aba' in 'gtfababacd' ?

– Biswadip Mandal
Nov 24 '18 at 18:41

yes, exactly, this will return 1 instead of 2.

– prophet-five
Nov 24 '18 at 18:44

cool. sounds about right. Thanks for mentioning

– Biswadip Mandal
Nov 24 '18 at 18:45

sure, np, I recently encountered this problem myself :)

– prophet-five
Nov 24 '18 at 18:47

add a comment |

just do the following

mystery_string.count('cat')

answered Nov 24 '18 at 18:10

Biswadip Mandal

1809

just do the following

mystery_string.count('cat')

answered Nov 24 '18 at 18:10

Biswadip Mandal

1809

answered Nov 24 '18 at 18:10

Biswadip Mandal

1809

answered Nov 24 '18 at 18:10

Biswadip Mandal

1809

answered Nov 24 '18 at 18:10

Biswadip Mandal

1809

1

It is a good solution for this specific word, however the count() method does not support overlapping

– prophet-five
Nov 24 '18 at 18:38

1

you mean examples like if I am counting 'aba' in 'gtfababacd' ?

– Biswadip Mandal
Nov 24 '18 at 18:41

yes, exactly, this will return 1 instead of 2.

– prophet-five
Nov 24 '18 at 18:44

cool. sounds about right. Thanks for mentioning

– Biswadip Mandal
Nov 24 '18 at 18:45

sure, np, I recently encountered this problem myself :)

– prophet-five
Nov 24 '18 at 18:47

add a comment |

1

It is a good solution for this specific word, however the count() method does not support overlapping

– prophet-five
Nov 24 '18 at 18:38

1

you mean examples like if I am counting 'aba' in 'gtfababacd' ?

– Biswadip Mandal
Nov 24 '18 at 18:41

yes, exactly, this will return 1 instead of 2.

– prophet-five
Nov 24 '18 at 18:44

cool. sounds about right. Thanks for mentioning

– Biswadip Mandal
Nov 24 '18 at 18:45

sure, np, I recently encountered this problem myself :)

– prophet-five
Nov 24 '18 at 18:47

It is a good solution for this specific word, however the count() method does not support overlapping

– prophet-five
Nov 24 '18 at 18:38

you mean examples like if I am counting 'aba' in 'gtfababacd' ?

– Biswadip Mandal
Nov 24 '18 at 18:41

yes, exactly, this will return 1 instead of 2.

– prophet-five
Nov 24 '18 at 18:44

cool. sounds about right. Thanks for mentioning

– Biswadip Mandal
Nov 24 '18 at 18:45

sure, np, I recently encountered this problem myself :)

– prophet-five
Nov 24 '18 at 18:47

add a comment |

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