How many hours for a book

Sarmat · 12-10-2018, 10:07 AM

I have a plugin which shows how many words are in a book. Taking in consideration average reading speed is it possible to have a column with predicted hours per book value?

The math is like:
(number of words)/(average words per hour)

I know my reading speed, calibre doesn't have to calculate it! I just need it to make one simple calculation

I tried making a column based on another column:
{#words}/11160

but it didn't work

gbm · 12-10-2018, 10:35 AM

Quote:

Originally Posted by Sarmat

I have a plugin which shows how many words are in a book. Taking in consideration average reading speed is it possible to have a column with predicted hours per book value?

The math is like:
(number of words)/(average words per hour)

I know my reading speed, calibre doesn't have to calculate it! I just need it to make one simple calculation

I tried making a column based on another column:
{#words}/11160

but it didn't work

You can use the Template Functions utility.

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bernie

EDIT

https://manual.calibre-ebook.com/template_lang.html

Sarmat · 12-10-2018, 11:10 AM

I have experimented dozen times with divide(x, y) in different formats and styles
but i failed

examples:
{divide(10, 2)}
divide(10, 2)

NSILMike · 12-10-2018, 11:14 AM

Just do it yourself since you know how many words per book.

Sarmat · 12-10-2018, 11:17 AM

I am too lazy for this
and besides I have a lot of books

gbm · 12-10-2018, 11:22 AM

Quote:

Originally Posted by Sarmat

I have experimented dozen times with divide(x, y) in different formats and styles
but i failed

examples:
{divide(10, 2)}
divide(10, 2)

You will need to use Using functions in templates - template program mode or Using general program mode.

bernie

davidfor · 12-10-2018, 10:24 PM

The following template should work:

Code:

program:
format_number(divide(raw_field('#word_count'),11160),'{0:5.2f}')

You will need to use the correct column name and output template to suit.

Sarmat · 12-11-2018, 07:34 AM

Quote:

Originally Posted by davidfor

The following template should work:

Code:

program:
format_number(divide(raw_field('#word_count'),11160),'{0:5.2f}')

You will need to use the correct column name and output template to suit.

It has worked, thank you.

DNSB · 12-11-2018, 06:51 PM

Quote:

Originally Posted by NSILMike

Just do it yourself since you know how many words per book.

Quote:

Originally Posted by Sarmat

I am too lazy for this
and besides I have a lot of books

All us computer users are lazy and dedicated to the principal that simple, repetitive tasks should be done by computers freeing us for the important decisions such as which book to read next.

Sarmat · 12-12-2018, 05:39 AM

Quote:

Originally Posted by davidfor

The following template should work:

Code:

program:
format_number(divide(raw_field('#word_count'),11160),'{0:5.2f}')

You will need to use the correct column name and output template to suit.

Is it possible to make Calibre think that
12 is bigger than 3
or that
23 is bigger than 4?

for now Calibre considers only first character in the number when sorting
can I have it sort everything right?

davidfor · 12-12-2018, 08:55 AM

Quote:

Originally Posted by Sarmat

Is it possible to make Calibre think that
12 is bigger than 3
or that
23 is bigger than 4?

for now Calibre considers only first character in the number when sorting
can I have it sort everything right?

In the definition of the column, there is an option to set the type for searching and sorting. Setting that to "Number" works here.

12-10-2018, 10:07 AM	#1
Sarmat Enthusiast Posts: 38 Karma: 10 Join Date: Nov 2018 Device: Kindle Paperwhite 2	How many hours for a book I have a plugin which shows how many words are in a book. Taking in consideration average reading speed is it possible to have a column with predicted hours per book value? The math is like: (number of words)/(average words per hour) I know my reading speed, calibre doesn't have to calculate it! I just need it to make one simple calculation I tried making a column based on another column: {#words}/11160 but it didn't work Last edited by Sarmat; 12-10-2018 at 10:11 AM.

12-10-2018, 10:24 PM	#7
davidfor Grand Sorcerer Posts: 24,907 Karma: 47303748 Join Date: Jul 2011 Location: Sydney, Australia Device: Kobo:Touch,Glo, AuraH2O, GloHD,AuraONE, ClaraHD, Libra H2O; tolinoepos	The following template should work: Code: program: format_number(divide(raw_field('#word_count'),11160),'{0:5.2f}') You will need to use the correct column name and output template to suit.

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12-10-2018, 11:10 AM	#3
Sarmat Enthusiast Posts: 38 Karma: 10 Join Date: Nov 2018 Device: Kindle Paperwhite 2	I have experimented dozen times with divide(x, y) in different formats and styles but i failed examples: {divide(10, 2)} divide(10, 2)

12-10-2018, 11:14 AM	#4
NSILMike Guru Posts: 735 Karma: 35936 Join Date: Apr 2011 Location: Shrewsury, MA Device: Lenovo Android Tablet	Just do it yourself since you know how many words per book.

12-10-2018, 11:17 AM	#5
Sarmat Enthusiast Posts: 38 Karma: 10 Join Date: Nov 2018 Device: Kindle Paperwhite 2	I am too lazy for this and besides I have a lot of books

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