using sed to turn paragraph to lines

This awk solution might work for you:

awk '/^[0-9]:[0-9]\.[0-9]/{ if (pass_num) printf "%s, word count: %i\n", pass_num, word_count pass_num=$1 word_count=-1 } { word_count+=NF } END { printf "%s, word count: %i\n", pass_num, word_count } ' file 

Test input:

# cat file 0:1.1 I am le passage one. There are many words in me. 0:1.2 I am le passage two. One two three four five six Seven 0:1.3 I am "Hello world" 

Test output:

0:1.1, word count: 11 0:1.2, word count: 12 0:1.3, word count: 4 

Each word is separated by empty space, so each word can be represented by each field in awk, i.e. word count in a line is equal to NF. The word count is summed up every line until the next passage.

When it encounters a new passage (indicated by the presence of a passage number), it

• prints out the previous passage's number and word count.
• set passage number to this new passage number
• reset passage word count (-1 because we don't want the passage number be counted)

The END{..} block is needed because the final passage doesn't have a trigger that causes it to print out the passage number and word count.

The if (pass_num) is to suppress printf when awk encounters the first passage.

This might work for you (GNU sed):

sed -r ':a;$bb;N;/\n[0-9]+:[0-9]+\.[0-9]+/!s/\n/ /g;ta;:b;h;s/\n.*//;s/([0-9]+:[0-9]+\.[0-9]+)(.*)/echo "\1 = $(wc -w <<<"\2")"/ep;g;D' file 

It forms each section into a single line then counts the words in the section less the section number (newlines are replaced by spaces).

$ cat file 0:1.1 This is the first passage... welcome to the SO, you leart a lot of things here. 0:1.2 This is the second passage... wer qwerqrq ewqr e 0:1.3 This is the second passage... 

Using sed and GNU grep:

$ sed -n '/0:1.1/,/[0-9]:[0-9]\.[0-9]/{//!p}' file | grep -Eo '[[:alpha:]]*' | wc -l 11 

0:1.1 -> Give the passage number here in which you want to count.

Here's one way with GNU awk:

awk -v RS='[0-9]+:[0-9]+\\.[0-9]+' -v FS='[ \t\n]+' 'NF > 0 { print R ": " NF - 2 } { R = RT }' 

If it is run on the file listed by doubledown, the output is:

0:1.1: 11 0:1.2: 12 0:1.3: 4 

Explanation

This works by splitting the input into records according to [0-9]+:[0-9]+\\.[0-9]+ and splitting into fields at whitespace. The record separator is off by one, hence the {R = RT }, the field counter is off by two because each record starts and ends with an FS, hence the NF - 2.

Edit - only count fields containing [:alnum:]

The above also counts e.g. ellipsis (...) as words, to avoid this do something like this:

awk -v RS='[0-9]+:[0-9]+\\.[0-9]+' -v FS='[ \t\n]+' ' NF > 0 { wc = NF-2 for(i=2; i<NF; i++) if($i !~ /[[:alnum:]]+/) wc-- print R ": " wc } { R = RT }'