This is addressed in this question (and as R is free you also have access to it): https://stackoverflow.com/questions/8920145/count-the-number-of-all-words-in-a-string
I think the reviewer doesn't really know what they are asking for, because if there is only one question, then the mean number of words written by each participant is simply 80000/456, or approximately 175 words per person.
If that's what the reviewer is asking for --- and you have access to the word count for each individual --- it may useful to report, for words per person, the mean, standard deviation, median, 25th percentile, 75th percentile, minimum, and maximum .... But I have no idea if that is useful in your context. (But it doesn't hurt to give more than the reviewer was asking for.)
As Sal Mangiafico says, I thought your original question was the number of words for each participants, and as he says you usually report more than the mean per participants (which as David L Morgan notes would be trivial). Sometimes reviewers aren't clear in what they want, but Nikita Hayden , can you say if they did just want a single number of more?
Sal Mangiafico Daniel Wright yes I need to get the number of words per participant so that I can have the range as well as the mean so Daniel's code suggestion is perfect thanks
For reasons I don't really understand, Nikita has deleted the original question. But IIRC, she mentioned that she has/uses SPSS. AFAIK, SPSS still has no regular expressions functions. But I think one can still get a decent word count with the following (not too complicated) code.
* The following code is based on an example found here:
* https://www.ibm.com/support/pages/counting-events-substring-within-string-variable-spss
.
* Read in some data.
NEW FILE.
DATASET CLOSE ALL.
DATA LIST / stringvar 1-35 (a).
BEGIN DATA
Hi there.
How are you?
What about double spaces?
And what about triple spaces?
Quadruple spaces?
END DATA.
* First, replace double spaces with single spaces.
* Repeat the replacement several times in case there
* are longer runs of blank spaces. You may have to
* tweak the number of loops.
LOOP # = 1 to 5.
COMPUTE stringvar = REPLACE(stringvar," ", " ").
END LOOP.
LIST.
* Let the word count = the number of blank spaces + 1.
COMPUTE blanks = 0.
* Loop from 1 to (at least) the maximum string length.
LOOP #i=1 to 50.
. COMPUTE blanks = blanks + CHAR.INDEX(SUBSTR(stringvar,#i,1),' ').
END LOOP.
COMPUTE words = blanks + 1.
FORMATS blanks words (F5.0).
LIST blanks words stringvar.
Output from that LIST command:
blanks words stringvar
1 2 Hi there.
2 3 How are you?
3 4 What about double spaces?
4 5 And what about triple spaces?
1 2 Quadruple spaces?