Split file by number of words

Shell Programming and Scripting
1

Dear all
I am trying to divide a file using the number of words as a condition. Alternatively, I would at least like to be able to retrieve the first x words of a given file. Any tips?
Thanks in advance.

2

get a fixed number of words:

wcnt=200 # get first 200 words
awk -v w=$wcnt  '{for(i=1;i<=NF && cnt<w; i++) 
                                    {printf("%s ", $i); cnt++}   look here 
                           print
                          if(cnt>=w) {exit} } '  inputfile > outputfile
3

Thanks for the reply. I got an error:

awk: syntax error at source line 1
context is
{for(i=1;i<=NF && cnt<w; >>> i++, <<<
awk: illegal statement at source line 1
awk: illegal statement at source line 1

Thank you

4

Made a change - my awk liked what I did, I forgot other awks may not. My bad. See old post above

If you are on Solaris use nawk not awk.

1 Like
5

N=200
while read L
do
for W in $L
do
((i++))
echo $W
((i>=N)) && break 2
done
done <file.txt

1 Like
6

Thank you for your replies. jim mcnamara's suggestion worked best, as frans way of doing it works, but unfortunately messes up the original text format (writes one word in each line), which I'd like to retain. Thank you both, though.

7

just modified line 7. as follows

      echo -n "$W "

and added an echo between lines 9 and 10.

8 
xargs -n1 <infile |head -200