Is there an easy way to get the number of repeating character in a word?

up vote
6
down vote

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I'm trying to get how many any character repeats in a word. The repetitions must be sequential.

For example, the method with input "loooooveee" should return 6 (4 times 'o', 2 times 'e').

I'm trying to implement string level functions and I can do it this way but, is there an easy way to do this? Regex, or some other sort of things?

So far I tried this:

def measure_normalized_emphasis(text):
 char = text[-1]
 emphasis_size = 0
 for i in range(1, len(text)):
 if text[-i] == char:
 emphasis_size += 1
 else:
 char = text[i - 1]

 return emphasis_size

And it returns 8 with "loooooveee".

edited 2 hours ago

Jan

23.7k52347

asked 2 hours ago

emremrah

478

add a commentÂ |Â

up vote
6
down vote

favorite

I'm trying to get how many any character repeats in a word. The repetitions must be sequential.

For example, the method with input "loooooveee" should return 6 (4 times 'o', 2 times 'e').

I'm trying to implement string level functions and I can do it this way but, is there an easy way to do this? Regex, or some other sort of things?

So far I tried this:

def measure_normalized_emphasis(text):
 char = text[-1]
 emphasis_size = 0
 for i in range(1, len(text)):
 if text[-i] == char:
 emphasis_size += 1
 else:
 char = text[i - 1]

 return emphasis_size

And it returns 8 with "loooooveee".

edited 2 hours ago

Jan

23.7k52347

asked 2 hours ago

emremrah

478

add a commentÂ |Â

up vote
6
down vote

favorite

I'm trying to get how many any character repeats in a word. The repetitions must be sequential.

For example, the method with input "loooooveee" should return 6 (4 times 'o', 2 times 'e').

I'm trying to implement string level functions and I can do it this way but, is there an easy way to do this? Regex, or some other sort of things?

So far I tried this:

def measure_normalized_emphasis(text):
 char = text[-1]
 emphasis_size = 0
 for i in range(1, len(text)):
 if text[-i] == char:
 emphasis_size += 1
 else:
 char = text[i - 1]

 return emphasis_size

And it returns 8 with "loooooveee".

edited 2 hours ago

Jan

23.7k52347

asked 2 hours ago

emremrah

478

I'm trying to get how many any character repeats in a word. The repetitions must be sequential.

For example, the method with input "loooooveee" should return 6 (4 times 'o', 2 times 'e').

I'm trying to implement string level functions and I can do it this way but, is there an easy way to do this? Regex, or some other sort of things?

So far I tried this:

def measure_normalized_emphasis(text):
 char = text[-1]
 emphasis_size = 0
 for i in range(1, len(text)):
 if text[-i] == char:
 emphasis_size += 1
 else:
 char = text[i - 1]

 return emphasis_size

And it returns 8 with "loooooveee".

python regex string counter

edited 2 hours ago

Jan

23.7k52347

asked 2 hours ago

emremrah

478

edited 2 hours ago

Jan

23.7k52347

asked 2 hours ago

emremrah

478

edited 2 hours ago

Jan

23.7k52347

edited 2 hours ago

Jan

23.7k52347

edited 2 hours ago

Jan

23.7k52347

asked 2 hours ago

emremrah

478

asked 2 hours ago

emremrah

478

asked 2 hours ago

emremrah

478

add a commentÂ |Â

6 Answers
6

active

oldest

votes

up vote
10
down vote

accepted

Original question: `order of repetition does not matter`

You can subtract the number of unique letters by the number of total letters. set applied to a string will return a unique collection of letters.

x = "loooooveee"
res = len(x) - len(set(x)) # 6

Or you can use collections.Counter, subtract 1 from each value, then sum:

from collections import Counter

c = Counter("loooooveee")

res = sum(i-1 for i in c.values()) # 6

New question: `repetitions must be sequential`

You can use itertools.groupby to group sequential identical characters:

from itertools import groupby

g = groupby("aooooaooaoo")

res = sum(sum(1 for _ in j) - 1 for i, j in g) # 5

edited 1 hour ago

answered 2 hours ago

jpp

79.1k184593

add a commentÂ |Â

up vote
2
down vote

You could use a regular expression if you want:

import re

rx = re.compile(r'(w)1+')

repeating = sum([m.span()[1] - m.span()[0] - 1 for m in rx.finditer("loooooveee")])
print(repeating)

This correctly yields 6 and makes use of the .span() function.

The expression is

(w)1+

which captures a word character (one of a-zA-Z0-9_) and tries to repeat it as often as possible.

See a demo on regex101.com for the repeating pattern.

If you want to match any character (that is, not only word characters), change your expression to:

(.)1+

See another demo on regex101.com.

edited 2 hours ago

answered 2 hours ago

Jan

23.7k52347

Very clean solution! Thank you so much.
â€“Â emremrah
1 hour ago

@emremrah: You're welcome.
â€“Â Jan
1 hour ago

add a commentÂ |Â

up vote
0
down vote

try this:

word=input('something:')

sum = 0

chars=set(list(word)) #get the set of unique characters

for item in chars: #iterate over the set and output the count for each item
 if word.count(char)>1:
 sum+=word.count(char)
 print('|'.format(item,str(word.count(char)))

print('Total:'+str(sum))

EDIT:

added total count of repetitions

edited 2 hours ago

answered 2 hours ago

vencaslac

520114

add a commentÂ |Â

up vote
0
down vote

i'm not giving you a better solution, many have done it.
I'll just correct the one you gave.

def mne(text):
char = text[0]
emphasis_size = 0
for i in range(1,len(text)):
 print(i, text[i], char, emphasis_size)
 if text[i] == char:
 emphasis_size += 1
 else:
 char = text[i]

return emphasis_size

is giving me:

>>>1 o l 0
 2 o o 0
 3 o o 1
 4 o o 2
 5 o o 3
 6 v o 4
 7 e v 4
 8 e e 4
 9 e e 5
 6

Which is what you wanted. no need for going backward, no need of [i-1]. just go forward and use too indices in the list (i and i-1)

answered 2 hours ago

Alexis

789113

add a commentÂ |Â

up vote
0
down vote

Since it doesn't matter where the repetition is occurring or which characters are being repeated, you can make use of the set data structure provided in Python. It will discard the duplicate occurrences of any character or an object.

Therefore, the solution would look something like this:

def measure_normalized_emphasis(text):
 return len(text) - len(set(text))

This will give you the exact result.

Also, make sure to look out for some edge cases, which you should as it is a good practice.

answered 2 hours ago

Dhruv Joshi

161

New contributor

I think you missed the "The repetitions must be sequential" part of the question.
â€“Â Muhammad Ahmad
2 hours ago

@MuhammadAhmad Ahh. The question got edited while I was answering. However, it seems that jpp has already given a clear explanation.
â€“Â Dhruv Joshi
1 hour ago

add a commentÂ |Â

up vote
0
down vote

I think your code is comparing the wrong things

You start by finding the last character:

char = text[-1]

Then you compare this to itself:

for i in range(1, len(text)):
 if text[-i] == char: #<-- surely this is test[-1] to begin with?

Why not just run through the characters:

def measure_normalized_emphasis(text):
 char = text[0]
 emphasis_size = 0
 for i in range(1, len(text)):
 if text[i] == char:
 emphasis_size += 1
 else:
 char = text[i]

 return emphasis_size

This seems to work.

answered 2 hours ago

doctorlove

14.3k22850

add a commentÂ |Â

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

up vote
10
down vote

accepted

Original question: `order of repetition does not matter`

You can subtract the number of unique letters by the number of total letters. set applied to a string will return a unique collection of letters.

x = "loooooveee"
res = len(x) - len(set(x)) # 6

Or you can use collections.Counter, subtract 1 from each value, then sum:

from collections import Counter

c = Counter("loooooveee")

res = sum(i-1 for i in c.values()) # 6

New question: `repetitions must be sequential`

You can use itertools.groupby to group sequential identical characters:

from itertools import groupby

g = groupby("aooooaooaoo")

res = sum(sum(1 for _ in j) - 1 for i, j in g) # 5

edited 1 hour ago

answered 2 hours ago

jpp

79.1k184593

add a commentÂ |Â

up vote
10
down vote

accepted

Original question: `order of repetition does not matter`

You can subtract the number of unique letters by the number of total letters. set applied to a string will return a unique collection of letters.

x = "loooooveee"
res = len(x) - len(set(x)) # 6

Or you can use collections.Counter, subtract 1 from each value, then sum:

from collections import Counter

c = Counter("loooooveee")

res = sum(i-1 for i in c.values()) # 6

New question: `repetitions must be sequential`

You can use itertools.groupby to group sequential identical characters:

from itertools import groupby

g = groupby("aooooaooaoo")

res = sum(sum(1 for _ in j) - 1 for i, j in g) # 5

edited 1 hour ago

answered 2 hours ago

jpp

79.1k184593

add a commentÂ |Â

up vote
10
down vote

accepted

Original question: `order of repetition does not matter`

You can subtract the number of unique letters by the number of total letters. set applied to a string will return a unique collection of letters.

x = "loooooveee"
res = len(x) - len(set(x)) # 6

Or you can use collections.Counter, subtract 1 from each value, then sum:

from collections import Counter

c = Counter("loooooveee")

res = sum(i-1 for i in c.values()) # 6

New question: `repetitions must be sequential`

You can use itertools.groupby to group sequential identical characters:

from itertools import groupby

g = groupby("aooooaooaoo")

res = sum(sum(1 for _ in j) - 1 for i, j in g) # 5

edited 1 hour ago

answered 2 hours ago

jpp

79.1k184593

Original question: `order of repetition does not matter`

You can subtract the number of unique letters by the number of total letters. set applied to a string will return a unique collection of letters.

x = "loooooveee"
res = len(x) - len(set(x)) # 6

Or you can use collections.Counter, subtract 1 from each value, then sum:

from collections import Counter

c = Counter("loooooveee")

res = sum(i-1 for i in c.values()) # 6

New question: `repetitions must be sequential`

You can use itertools.groupby to group sequential identical characters:

from itertools import groupby

g = groupby("aooooaooaoo")

res = sum(sum(1 for _ in j) - 1 for i, j in g) # 5

edited 1 hour ago

answered 2 hours ago

jpp

79.1k184593

edited 1 hour ago

answered 2 hours ago

jpp

79.1k184593

answered 2 hours ago

jpp

79.1k184593

answered 2 hours ago

jpp

79.1k184593

add a commentÂ |Â

up vote
2
down vote

You could use a regular expression if you want:

import re

rx = re.compile(r'(w)1+')

repeating = sum([m.span()[1] - m.span()[0] - 1 for m in rx.finditer("loooooveee")])
print(repeating)

This correctly yields 6 and makes use of the .span() function.

The expression is

(w)1+

which captures a word character (one of a-zA-Z0-9_) and tries to repeat it as often as possible.

See a demo on regex101.com for the repeating pattern.

If you want to match any character (that is, not only word characters), change your expression to:

(.)1+

See another demo on regex101.com.

edited 2 hours ago

answered 2 hours ago

Jan

23.7k52347

Very clean solution! Thank you so much.
â€“Â emremrah
1 hour ago

@emremrah: You're welcome.
â€“Â Jan
1 hour ago

add a commentÂ |Â

up vote
2
down vote

You could use a regular expression if you want:

import re

rx = re.compile(r'(w)1+')

repeating = sum([m.span()[1] - m.span()[0] - 1 for m in rx.finditer("loooooveee")])
print(repeating)

This correctly yields 6 and makes use of the .span() function.

The expression is

(w)1+

which captures a word character (one of a-zA-Z0-9_) and tries to repeat it as often as possible.

See a demo on regex101.com for the repeating pattern.

If you want to match any character (that is, not only word characters), change your expression to:

(.)1+

See another demo on regex101.com.

edited 2 hours ago

answered 2 hours ago

Jan

23.7k52347

Very clean solution! Thank you so much.
â€“Â emremrah
1 hour ago

@emremrah: You're welcome.
â€“Â Jan
1 hour ago

add a commentÂ |Â

up vote
2
down vote

You could use a regular expression if you want:

import re

rx = re.compile(r'(w)1+')

repeating = sum([m.span()[1] - m.span()[0] - 1 for m in rx.finditer("loooooveee")])
print(repeating)

This correctly yields 6 and makes use of the .span() function.

The expression is

(w)1+

which captures a word character (one of a-zA-Z0-9_) and tries to repeat it as often as possible.

See a demo on regex101.com for the repeating pattern.

If you want to match any character (that is, not only word characters), change your expression to:

(.)1+

See another demo on regex101.com.

edited 2 hours ago

answered 2 hours ago

Jan

23.7k52347

You could use a regular expression if you want:

import re

rx = re.compile(r'(w)1+')

repeating = sum([m.span()[1] - m.span()[0] - 1 for m in rx.finditer("loooooveee")])
print(repeating)

This correctly yields 6 and makes use of the .span() function.

The expression is

(w)1+

which captures a word character (one of a-zA-Z0-9_) and tries to repeat it as often as possible.

See a demo on regex101.com for the repeating pattern.

If you want to match any character (that is, not only word characters), change your expression to:

(.)1+

See another demo on regex101.com.

edited 2 hours ago

answered 2 hours ago

Jan

23.7k52347

edited 2 hours ago

answered 2 hours ago

Jan

23.7k52347

answered 2 hours ago

Jan

23.7k52347

answered 2 hours ago

Jan

23.7k52347

Very clean solution! Thank you so much.
â€“Â emremrah
1 hour ago

@emremrah: You're welcome.
â€“Â Jan
1 hour ago

add a commentÂ |Â

Very clean solution! Thank you so much.
â€“Â emremrah
1 hour ago

@emremrah: You're welcome.
â€“Â Jan
1 hour ago

Very clean solution! Thank you so much.
â€“Â emremrah
1 hour ago

@emremrah: You're welcome.
â€“Â Jan
1 hour ago

add a commentÂ |Â

up vote
0
down vote

try this:

word=input('something:')

sum = 0

chars=set(list(word)) #get the set of unique characters

for item in chars: #iterate over the set and output the count for each item
 if word.count(char)>1:
 sum+=word.count(char)
 print('|'.format(item,str(word.count(char)))

print('Total:'+str(sum))

EDIT:

added total count of repetitions

edited 2 hours ago

answered 2 hours ago

vencaslac

520114

add a commentÂ |Â

up vote
0
down vote

try this:

word=input('something:')

sum = 0

chars=set(list(word)) #get the set of unique characters

for item in chars: #iterate over the set and output the count for each item
 if word.count(char)>1:
 sum+=word.count(char)
 print('|'.format(item,str(word.count(char)))

print('Total:'+str(sum))

EDIT:

added total count of repetitions

edited 2 hours ago

answered 2 hours ago

vencaslac

520114

add a commentÂ |Â

up vote
0
down vote

try this:

word=input('something:')

sum = 0

chars=set(list(word)) #get the set of unique characters

for item in chars: #iterate over the set and output the count for each item
 if word.count(char)>1:
 sum+=word.count(char)
 print('|'.format(item,str(word.count(char)))

print('Total:'+str(sum))

EDIT:

added total count of repetitions

edited 2 hours ago

answered 2 hours ago

vencaslac

520114

try this:

word=input('something:')

sum = 0

chars=set(list(word)) #get the set of unique characters

for item in chars: #iterate over the set and output the count for each item
 if word.count(char)>1:
 sum+=word.count(char)
 print('|'.format(item,str(word.count(char)))

print('Total:'+str(sum))

EDIT:

added total count of repetitions

edited 2 hours ago

answered 2 hours ago

vencaslac

520114

edited 2 hours ago

answered 2 hours ago

vencaslac

520114

answered 2 hours ago

vencaslac

520114

answered 2 hours ago

vencaslac

520114

add a commentÂ |Â

up vote
0
down vote

i'm not giving you a better solution, many have done it.
I'll just correct the one you gave.

def mne(text):
char = text[0]
emphasis_size = 0
for i in range(1,len(text)):
 print(i, text[i], char, emphasis_size)
 if text[i] == char:
 emphasis_size += 1
 else:
 char = text[i]

return emphasis_size

is giving me:

>>>1 o l 0
 2 o o 0
 3 o o 1
 4 o o 2
 5 o o 3
 6 v o 4
 7 e v 4
 8 e e 4
 9 e e 5
 6

Which is what you wanted. no need for going backward, no need of [i-1]. just go forward and use too indices in the list (i and i-1)

answered 2 hours ago

Alexis

789113

add a commentÂ |Â

up vote
0
down vote

i'm not giving you a better solution, many have done it.
I'll just correct the one you gave.

def mne(text):
char = text[0]
emphasis_size = 0
for i in range(1,len(text)):
 print(i, text[i], char, emphasis_size)
 if text[i] == char:
 emphasis_size += 1
 else:
 char = text[i]

return emphasis_size

is giving me:

>>>1 o l 0
 2 o o 0
 3 o o 1
 4 o o 2
 5 o o 3
 6 v o 4
 7 e v 4
 8 e e 4
 9 e e 5
 6

Which is what you wanted. no need for going backward, no need of [i-1]. just go forward and use too indices in the list (i and i-1)

answered 2 hours ago

Alexis

789113

add a commentÂ |Â

up vote
0
down vote

i'm not giving you a better solution, many have done it.
I'll just correct the one you gave.

def mne(text):
char = text[0]
emphasis_size = 0
for i in range(1,len(text)):
 print(i, text[i], char, emphasis_size)
 if text[i] == char:
 emphasis_size += 1
 else:
 char = text[i]

return emphasis_size

is giving me:

>>>1 o l 0
 2 o o 0
 3 o o 1
 4 o o 2
 5 o o 3
 6 v o 4
 7 e v 4
 8 e e 4
 9 e e 5
 6

Which is what you wanted. no need for going backward, no need of [i-1]. just go forward and use too indices in the list (i and i-1)

answered 2 hours ago

Alexis

789113

i'm not giving you a better solution, many have done it.
I'll just correct the one you gave.

def mne(text):
char = text[0]
emphasis_size = 0
for i in range(1,len(text)):
 print(i, text[i], char, emphasis_size)
 if text[i] == char:
 emphasis_size += 1
 else:
 char = text[i]

return emphasis_size

is giving me:

>>>1 o l 0
 2 o o 0
 3 o o 1
 4 o o 2
 5 o o 3
 6 v o 4
 7 e v 4
 8 e e 4
 9 e e 5
 6

Which is what you wanted. no need for going backward, no need of [i-1]. just go forward and use too indices in the list (i and i-1)

answered 2 hours ago

Alexis

789113

answered 2 hours ago

Alexis

789113

answered 2 hours ago

Alexis

789113

answered 2 hours ago

Alexis

789113

add a commentÂ |Â

up vote
0
down vote

Therefore, the solution would look something like this:

def measure_normalized_emphasis(text):
 return len(text) - len(set(text))

This will give you the exact result.

Also, make sure to look out for some edge cases, which you should as it is a good practice.

answered 2 hours ago

Dhruv Joshi

161

New contributor

I think you missed the "The repetitions must be sequential" part of the question.
â€“Â Muhammad Ahmad
2 hours ago

@MuhammadAhmad Ahh. The question got edited while I was answering. However, it seems that jpp has already given a clear explanation.
â€“Â Dhruv Joshi
1 hour ago

add a commentÂ |Â

up vote
0
down vote

Therefore, the solution would look something like this:

def measure_normalized_emphasis(text):
 return len(text) - len(set(text))

This will give you the exact result.

Also, make sure to look out for some edge cases, which you should as it is a good practice.

answered 2 hours ago

Dhruv Joshi

161

New contributor

I think you missed the "The repetitions must be sequential" part of the question.
â€“Â Muhammad Ahmad
2 hours ago

@MuhammadAhmad Ahh. The question got edited while I was answering. However, it seems that jpp has already given a clear explanation.
â€“Â Dhruv Joshi
1 hour ago

add a commentÂ |Â

up vote
0
down vote

Therefore, the solution would look something like this:

def measure_normalized_emphasis(text):
 return len(text) - len(set(text))

This will give you the exact result.

Also, make sure to look out for some edge cases, which you should as it is a good practice.

answered 2 hours ago

Dhruv Joshi

161

New contributor

Therefore, the solution would look something like this:

def measure_normalized_emphasis(text):
 return len(text) - len(set(text))

This will give you the exact result.

Also, make sure to look out for some edge cases, which you should as it is a good practice.

answered 2 hours ago

Dhruv Joshi

161

New contributor

answered 2 hours ago

Dhruv Joshi

161

New contributor

answered 2 hours ago

Dhruv Joshi

161

answered 2 hours ago

Dhruv Joshi

161

New contributor

Dhruv Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

I think you missed the "The repetitions must be sequential" part of the question.
â€“Â Muhammad Ahmad
2 hours ago

@MuhammadAhmad Ahh. The question got edited while I was answering. However, it seems that jpp has already given a clear explanation.
â€“Â Dhruv Joshi
1 hour ago

add a commentÂ |Â

I think you missed the "The repetitions must be sequential" part of the question.
â€“Â Muhammad Ahmad
2 hours ago

@MuhammadAhmad Ahh. The question got edited while I was answering. However, it seems that jpp has already given a clear explanation.
â€“Â Dhruv Joshi
1 hour ago

I think you missed the "The repetitions must be sequential" part of the question.
â€“Â Muhammad Ahmad
2 hours ago

@MuhammadAhmad Ahh. The question got edited while I was answering. However, it seems that jpp has already given a clear explanation.
â€“Â Dhruv Joshi
1 hour ago

add a commentÂ |Â

up vote
0
down vote

I think your code is comparing the wrong things

You start by finding the last character:

char = text[-1]

Then you compare this to itself:

for i in range(1, len(text)):
 if text[-i] == char: #<-- surely this is test[-1] to begin with?

Why not just run through the characters:

def measure_normalized_emphasis(text):
 char = text[0]
 emphasis_size = 0
 for i in range(1, len(text)):
 if text[i] == char:
 emphasis_size += 1
 else:
 char = text[i]

 return emphasis_size

This seems to work.

answered 2 hours ago

doctorlove

14.3k22850

add a commentÂ |Â

up vote
0
down vote

I think your code is comparing the wrong things

You start by finding the last character:

char = text[-1]

Then you compare this to itself:

for i in range(1, len(text)):
 if text[-i] == char: #<-- surely this is test[-1] to begin with?

Why not just run through the characters:

def measure_normalized_emphasis(text):
 char = text[0]
 emphasis_size = 0
 for i in range(1, len(text)):
 if text[i] == char:
 emphasis_size += 1
 else:
 char = text[i]

 return emphasis_size

This seems to work.

answered 2 hours ago

doctorlove

14.3k22850

add a commentÂ |Â

up vote
0
down vote

I think your code is comparing the wrong things

You start by finding the last character:

char = text[-1]

Then you compare this to itself:

for i in range(1, len(text)):
 if text[-i] == char: #<-- surely this is test[-1] to begin with?

Why not just run through the characters:

def measure_normalized_emphasis(text):
 char = text[0]
 emphasis_size = 0
 for i in range(1, len(text)):
 if text[i] == char:
 emphasis_size += 1
 else:
 char = text[i]

 return emphasis_size

This seems to work.

answered 2 hours ago

doctorlove

14.3k22850

I think your code is comparing the wrong things

You start by finding the last character:

char = text[-1]

Then you compare this to itself:

for i in range(1, len(text)):
 if text[-i] == char: #<-- surely this is test[-1] to begin with?

Why not just run through the characters:

def measure_normalized_emphasis(text):
 char = text[0]
 emphasis_size = 0
 for i in range(1, len(text)):
 if text[i] == char:
 emphasis_size += 1
 else:
 char = text[i]

 return emphasis_size

This seems to work.

answered 2 hours ago

doctorlove

14.3k22850

answered 2 hours ago

doctorlove

14.3k22850

answered 2 hours ago

doctorlove

14.3k22850

answered 2 hours ago

doctorlove

14.3k22850

add a commentÂ |Â

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