calculate punctuation percentage in a string in Python

I have been working on calculating the percentage of punctuations in a sentence. For some reason, my function works when doing double spacing, but counts all the characters and the white space. For example, I have a text DEACTIVATE: OK so total full length is 14 when I subtract the punctuation then length is 13, so percentage should be 1/13 = 7.63%, however, my function gives me 7.14%, which is basically 1/14 = 7.14%.

On the other side, if have just one white space, my function throws me an error

"ZeroDivisionError: division by zero".

Here is my code for your reference and a simple text samples

text= "Centre to position, remaining shift is still larger than maximum (retry nbr=1, centring_stroke.r=2.7662e-05, max centring stroke.r=2.5e-05)"

text2= "DEACTIVATE: KU-1421"

import string

def count_punct(text):

    count = sum([1 for char in text if char in string.punctuation])

    return round(count/(len(text) - text.count("  ")), 3)*100

df_sub['punct%'] = df_sub['Err_Text2'].apply(lambda x: count_punct(x))

df_sub.head(20)

edited Nov 14 '18 at 11:15

Vineeth Sai

2,48851323

asked Nov 14 '18 at 5:35

Vishwas

177

The code above is with two spaces (text.count(" ")). The code I am getting error is with single space (text.count(" ")).

– Vishwas
Nov 14 '18 at 5:39

That's not how percentage works. You don't subtract the number of punctuation characters, it's just (number of punctuation characters) / (length of string), unless you mean subtract the whitespace, not the punctuation.

– Tomothy32
Nov 14 '18 at 5:41

you mean like this "return round((count)/(len(text) - text.count(" ")), 3)*100"? I still get same error.

– Vishwas
Nov 14 '18 at 5:43

Your error is division by zero? That means you passed in some string that is either fully spaces or empty.

– Tomothy32
Nov 14 '18 at 5:47

Thats what I am not able to figure out. Any suggestion or advice will be much appreciated.

– Vishwas
Nov 14 '18 at 5:50

|
show 4 more comments

On the other side, if have just one white space, my function throws me an error

"ZeroDivisionError: division by zero".

Here is my code for your reference and a simple text samples

text= "Centre to position, remaining shift is still larger than maximum (retry nbr=1, centring_stroke.r=2.7662e-05, max centring stroke.r=2.5e-05)"

text2= "DEACTIVATE: KU-1421"

import string

def count_punct(text):

    count = sum([1 for char in text if char in string.punctuation])

    return round(count/(len(text) - text.count("  ")), 3)*100

df_sub['punct%'] = df_sub['Err_Text2'].apply(lambda x: count_punct(x))

df_sub.head(20)

edited Nov 14 '18 at 11:15

Vineeth Sai

2,48851323

asked Nov 14 '18 at 5:35

Vishwas

177

The code above is with two spaces (text.count(" ")). The code I am getting error is with single space (text.count(" ")).

– Vishwas
Nov 14 '18 at 5:39

That's not how percentage works. You don't subtract the number of punctuation characters, it's just (number of punctuation characters) / (length of string), unless you mean subtract the whitespace, not the punctuation.

– Tomothy32
Nov 14 '18 at 5:41

you mean like this "return round((count)/(len(text) - text.count(" ")), 3)*100"? I still get same error.

– Vishwas
Nov 14 '18 at 5:43

Your error is division by zero? That means you passed in some string that is either fully spaces or empty.

– Tomothy32
Nov 14 '18 at 5:47

Thats what I am not able to figure out. Any suggestion or advice will be much appreciated.

– Vishwas
Nov 14 '18 at 5:50

|
show 4 more comments

On the other side, if have just one white space, my function throws me an error

"ZeroDivisionError: division by zero".

Here is my code for your reference and a simple text samples

text= "Centre to position, remaining shift is still larger than maximum (retry nbr=1, centring_stroke.r=2.7662e-05, max centring stroke.r=2.5e-05)"

text2= "DEACTIVATE: KU-1421"

import string

def count_punct(text):

    count = sum([1 for char in text if char in string.punctuation])

    return round(count/(len(text) - text.count("  ")), 3)*100

df_sub['punct%'] = df_sub['Err_Text2'].apply(lambda x: count_punct(x))

df_sub.head(20)

edited Nov 14 '18 at 11:15

Vineeth Sai

2,48851323

asked Nov 14 '18 at 5:35

Vishwas

177

On the other side, if have just one white space, my function throws me an error

"ZeroDivisionError: division by zero".

Here is my code for your reference and a simple text samples

text= "Centre to position, remaining shift is still larger than maximum (retry nbr=1, centring_stroke.r=2.7662e-05, max centring stroke.r=2.5e-05)"

text2= "DEACTIVATE: KU-1421"

import string

def count_punct(text):

    count = sum([1 for char in text if char in string.punctuation])

    return round(count/(len(text) - text.count("  ")), 3)*100

df_sub['punct%'] = df_sub['Err_Text2'].apply(lambda x: count_punct(x))

df_sub.head(20)

python percentage punctuation

edited Nov 14 '18 at 11:15

Vineeth Sai

2,48851323

asked Nov 14 '18 at 5:35

Vishwas

177

edited Nov 14 '18 at 11:15

Vineeth Sai

2,48851323

asked Nov 14 '18 at 5:35

Vishwas

177

edited Nov 14 '18 at 11:15

Vineeth Sai

2,48851323

edited Nov 14 '18 at 11:15

Vineeth Sai

2,48851323

edited Nov 14 '18 at 11:15

Vineeth Sai

2,48851323

asked Nov 14 '18 at 5:35

Vishwas

177

asked Nov 14 '18 at 5:35

Vishwas

177

asked Nov 14 '18 at 5:35

Vishwas

177

The code above is with two spaces (text.count(" ")). The code I am getting error is with single space (text.count(" ")).

– Vishwas
Nov 14 '18 at 5:39

That's not how percentage works. You don't subtract the number of punctuation characters, it's just (number of punctuation characters) / (length of string), unless you mean subtract the whitespace, not the punctuation.

– Tomothy32
Nov 14 '18 at 5:41

you mean like this "return round((count)/(len(text) - text.count(" ")), 3)*100"? I still get same error.

– Vishwas
Nov 14 '18 at 5:43

Your error is division by zero? That means you passed in some string that is either fully spaces or empty.

– Tomothy32
Nov 14 '18 at 5:47

Thats what I am not able to figure out. Any suggestion or advice will be much appreciated.

– Vishwas
Nov 14 '18 at 5:50

|
show 4 more comments

The code above is with two spaces (text.count(" ")). The code I am getting error is with single space (text.count(" ")).

– Vishwas
Nov 14 '18 at 5:39

That's not how percentage works. You don't subtract the number of punctuation characters, it's just (number of punctuation characters) / (length of string), unless you mean subtract the whitespace, not the punctuation.

– Tomothy32
Nov 14 '18 at 5:41

you mean like this "return round((count)/(len(text) - text.count(" ")), 3)*100"? I still get same error.

– Vishwas
Nov 14 '18 at 5:43

Your error is division by zero? That means you passed in some string that is either fully spaces or empty.

– Tomothy32
Nov 14 '18 at 5:47

Thats what I am not able to figure out. Any suggestion or advice will be much appreciated.

– Vishwas
Nov 14 '18 at 5:50

The code above is with two spaces (text.count(" ")). The code I am getting error is with single space (text.count(" ")).

– Vishwas
Nov 14 '18 at 5:39

That's not how percentage works. You don't subtract the number of punctuation characters, it's just (number of punctuation characters) / (length of string), unless you mean subtract the whitespace, not the punctuation.

– Tomothy32
Nov 14 '18 at 5:41

you mean like this "return round((count)/(len(text) - text.count(" ")), 3)*100"? I still get same error.

– Vishwas
Nov 14 '18 at 5:43

Your error is division by zero? That means you passed in some string that is either fully spaces or empty.

– Tomothy32
Nov 14 '18 at 5:47

Thats what I am not able to figure out. Any suggestion or advice will be much appreciated.

– Vishwas
Nov 14 '18 at 5:50

|
show 4 more comments

1 Answer
1

active

oldest

votes

Here, Make these small changes and your count_punct function should be up and running.. The reason your code was breaking is because you were checking for ___ instead of _. i.e 3 consecutive spaces instead of one space. That is why the difference always resulted in the same value.

import string

def count_punct(text):

    if text.strip() == "": # To take of care of all space input

        return 0

    count = sum([1 if char in string.punctuation else 0 for char in text ])

    spaces = text.count(" ") # Your error is here, Only check for 1 space instead of 3 spaces

    total_chars = len(text) - spaces



    return round(count / total_chars, 3)*100



text= "DEACTIVATE: OK"



print(count_punct(text))

Outputs:

7.7

And for the zero divide by error. It's a logic error when the total_chars is 0, because the length of string and number of spaces both are equal. Hence the difference is 0.

To fix this you can simply add an if statement (already added above)

if text.strip() == "":

    print(0)

edited Nov 14 '18 at 6:16

answered Nov 14 '18 at 6:10

Vineeth Sai

2,48851323

Note that OP commented quote "The code above is with two spaces (text.count(" ")). The code I am getting error is with single space (text.count(" "))." But you're right about division by zero.

– Tomothy32
Nov 14 '18 at 6:25

@Tomothy32 I'm not getting any error with single space. And the results are always consistent. And why would he want to check for double spaces anyway?

– Vineeth Sai
Nov 14 '18 at 6:32

1

What I meant to say was that OP knew about the fact that they accidentally used multiple spaces, they just didn't edit the post. (Nothing wrong with your answer, sorry for any confusion.)

– Tomothy32
Nov 14 '18 at 6:34

Oh :) No confusion. Thanks for pointing it out.

– Vineeth Sai
Nov 14 '18 at 6:41

Thank you very much, it works now and percentage is correct.

– Vishwas
Nov 14 '18 at 16:47

|
show 1 more comment

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1 Answer
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1 Answer
1

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oldest

votes

import string

def count_punct(text):

    if text.strip() == "": # To take of care of all space input

        return 0

    count = sum([1 if char in string.punctuation else 0 for char in text ])

    spaces = text.count(" ") # Your error is here, Only check for 1 space instead of 3 spaces

    total_chars = len(text) - spaces



    return round(count / total_chars, 3)*100



text= "DEACTIVATE: OK"



print(count_punct(text))

Outputs:

7.7

And for the zero divide by error. It's a logic error when the total_chars is 0, because the length of string and number of spaces both are equal. Hence the difference is 0.

To fix this you can simply add an if statement (already added above)

if text.strip() == "":

    print(0)

edited Nov 14 '18 at 6:16

answered Nov 14 '18 at 6:10

Vineeth Sai

2,48851323

Note that OP commented quote "The code above is with two spaces (text.count(" ")). The code I am getting error is with single space (text.count(" "))." But you're right about division by zero.

– Tomothy32
Nov 14 '18 at 6:25

@Tomothy32 I'm not getting any error with single space. And the results are always consistent. And why would he want to check for double spaces anyway?

– Vineeth Sai
Nov 14 '18 at 6:32

1

What I meant to say was that OP knew about the fact that they accidentally used multiple spaces, they just didn't edit the post. (Nothing wrong with your answer, sorry for any confusion.)

– Tomothy32
Nov 14 '18 at 6:34

Oh :) No confusion. Thanks for pointing it out.

– Vineeth Sai
Nov 14 '18 at 6:41

Thank you very much, it works now and percentage is correct.

– Vishwas
Nov 14 '18 at 16:47

|
show 1 more comment

import string

def count_punct(text):

    if text.strip() == "": # To take of care of all space input

        return 0

    count = sum([1 if char in string.punctuation else 0 for char in text ])

    spaces = text.count(" ") # Your error is here, Only check for 1 space instead of 3 spaces

    total_chars = len(text) - spaces



    return round(count / total_chars, 3)*100



text= "DEACTIVATE: OK"



print(count_punct(text))

Outputs:

7.7

And for the zero divide by error. It's a logic error when the total_chars is 0, because the length of string and number of spaces both are equal. Hence the difference is 0.

To fix this you can simply add an if statement (already added above)

if text.strip() == "":

    print(0)

edited Nov 14 '18 at 6:16

answered Nov 14 '18 at 6:10

Vineeth Sai

2,48851323

Note that OP commented quote "The code above is with two spaces (text.count(" ")). The code I am getting error is with single space (text.count(" "))." But you're right about division by zero.

– Tomothy32
Nov 14 '18 at 6:25

@Tomothy32 I'm not getting any error with single space. And the results are always consistent. And why would he want to check for double spaces anyway?

– Vineeth Sai
Nov 14 '18 at 6:32

1

What I meant to say was that OP knew about the fact that they accidentally used multiple spaces, they just didn't edit the post. (Nothing wrong with your answer, sorry for any confusion.)

– Tomothy32
Nov 14 '18 at 6:34

Oh :) No confusion. Thanks for pointing it out.

– Vineeth Sai
Nov 14 '18 at 6:41

Thank you very much, it works now and percentage is correct.

– Vishwas
Nov 14 '18 at 16:47

|
show 1 more comment

import string

def count_punct(text):

    if text.strip() == "": # To take of care of all space input

        return 0

    count = sum([1 if char in string.punctuation else 0 for char in text ])

    spaces = text.count(" ") # Your error is here, Only check for 1 space instead of 3 spaces

    total_chars = len(text) - spaces



    return round(count / total_chars, 3)*100



text= "DEACTIVATE: OK"



print(count_punct(text))

Outputs:

7.7

And for the zero divide by error. It's a logic error when the total_chars is 0, because the length of string and number of spaces both are equal. Hence the difference is 0.

To fix this you can simply add an if statement (already added above)

if text.strip() == "":

    print(0)

edited Nov 14 '18 at 6:16

answered Nov 14 '18 at 6:10

Vineeth Sai

2,48851323

import string

def count_punct(text):

    if text.strip() == "": # To take of care of all space input

        return 0

    count = sum([1 if char in string.punctuation else 0 for char in text ])

    spaces = text.count(" ") # Your error is here, Only check for 1 space instead of 3 spaces

    total_chars = len(text) - spaces



    return round(count / total_chars, 3)*100



text= "DEACTIVATE: OK"



print(count_punct(text))

Outputs:

7.7

And for the zero divide by error. It's a logic error when the total_chars is 0, because the length of string and number of spaces both are equal. Hence the difference is 0.

To fix this you can simply add an if statement (already added above)

if text.strip() == "":

    print(0)

edited Nov 14 '18 at 6:16

answered Nov 14 '18 at 6:10

Vineeth Sai

2,48851323

edited Nov 14 '18 at 6:16

answered Nov 14 '18 at 6:10

Vineeth Sai

2,48851323

answered Nov 14 '18 at 6:10

Vineeth Sai

2,48851323

answered Nov 14 '18 at 6:10

Vineeth Sai

2,48851323

Note that OP commented quote "The code above is with two spaces (text.count(" ")). The code I am getting error is with single space (text.count(" "))." But you're right about division by zero.

– Tomothy32
Nov 14 '18 at 6:25

@Tomothy32 I'm not getting any error with single space. And the results are always consistent. And why would he want to check for double spaces anyway?

– Vineeth Sai
Nov 14 '18 at 6:32

1

What I meant to say was that OP knew about the fact that they accidentally used multiple spaces, they just didn't edit the post. (Nothing wrong with your answer, sorry for any confusion.)

– Tomothy32
Nov 14 '18 at 6:34

Oh :) No confusion. Thanks for pointing it out.

– Vineeth Sai
Nov 14 '18 at 6:41

Thank you very much, it works now and percentage is correct.

– Vishwas
Nov 14 '18 at 16:47

|
show 1 more comment

Note that OP commented quote "The code above is with two spaces (text.count(" ")). The code I am getting error is with single space (text.count(" "))." But you're right about division by zero.

– Tomothy32
Nov 14 '18 at 6:25

@Tomothy32 I'm not getting any error with single space. And the results are always consistent. And why would he want to check for double spaces anyway?

– Vineeth Sai
Nov 14 '18 at 6:32

1

What I meant to say was that OP knew about the fact that they accidentally used multiple spaces, they just didn't edit the post. (Nothing wrong with your answer, sorry for any confusion.)

– Tomothy32
Nov 14 '18 at 6:34

Oh :) No confusion. Thanks for pointing it out.

– Vineeth Sai
Nov 14 '18 at 6:41

Thank you very much, it works now and percentage is correct.

– Vishwas
Nov 14 '18 at 16:47

Note that OP commented quote "The code above is with two spaces (text.count(" ")). The code I am getting error is with single space (text.count(" "))." But you're right about division by zero.

– Tomothy32
Nov 14 '18 at 6:25

@Tomothy32 I'm not getting any error with single space. And the results are always consistent. And why would he want to check for double spaces anyway?

– Vineeth Sai
Nov 14 '18 at 6:32

What I meant to say was that OP knew about the fact that they accidentally used multiple spaces, they just didn't edit the post. (Nothing wrong with your answer, sorry for any confusion.)

– Tomothy32
Nov 14 '18 at 6:34

Oh :) No confusion. Thanks for pointing it out.

– Vineeth Sai
Nov 14 '18 at 6:41

Thank you very much, it works now and percentage is correct.

– Vishwas
Nov 14 '18 at 16:47

|
show 1 more comment

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