Print first N words of a file

Question

Is there any way to print the first N words of a file? I've tried cut but it reads a document line-by-line. The only solution I came up with is:

sed ':a;N;$!ba;s/\n/δ/g' file | cut -d " " -f -20 | sed 's/δ/\n/g'

Essentially, replacing newlines with a character that doesn't not exist in the file, applying "cut" with space as delimiter and then restoring the newlines.

Is there any better solution?

Answer 1

You could use awk to print the first n words:

$ awk 'NR<=8{print;next}{exit}' RS='[[:blank:]]+|\n' file

This would print the first 8 words. Each word is output on a separate line, are you looking to keep the original format of the file?

Edit:

The following will preserve the original format of the file:

awk -v n=8 'n==c{exit}n-c>=NF{print;c+=NF;next}{for(i=1;i<=n-c;i++)printf "%s ",$i;print x;exit}' file

Demo:

$ cat file
one two
thre four five six
seven 8 9 
10

$ awk -v n=8 'n==c{exit}n-c>=NF{print;c+=NF;next}{for(i=1;i<=n-c;i++)printf "%s ",$i;print x;exit}' file
one two
thre four five six
seven 8 

A small caveat: if the last line printed doesn't use a single space as a separator this line will lose it's formatting.

$ cat file 
one     two
thre     four five six
seven        8 9 
10

# the 8th word fell on 3rd line: this line will be formatted with single spaces
$ awk -v n=8 'n==c{exit}n-c>=NF{print;c+=NF;next}{for(i=1;i<=n-c;i++)printf "%s ",$i;print x;exit}' file
one     two
thre     four five six
seven 8

Answer 2

Assuming words are non-white space separated by white space, you can use tr to convert the document to one-word-per-line format and then count the first N lines:

tr -s ' \011' '\012' < file | head -n $N

where N=20 or whatever value you want for the number of words. Note that tr is a pure filter; it only reads from standard input and only writes to standard output. The -s option 'squeezes' out duplicate replacements, so you get one newline per sequence of blanks or tabs in the input. (If there is leading white space in the file, you get an initial blank line. There are various ways to deal with that, such as grab the first N+1 lines out output after all, or filter out all blank lines.)

Answer 3

Using GNU awk so we can set the RS to a regexp and access the matching string using RT:

$ cat file
the quick
brown     fox     jumped over
the
lazy
dog's back

$ gawk -v c=3 -v RS='[[:space:]]+' 'NR<=c{ORS=(NR<c?RT:"\n");print}' file
the quick
brown

$ gawk -v c=6 -v RS='[[:space:]]+' 'NR<=c{ORS=(NR<c?RT:"\n");print}' file
the quick
brown     fox     jumped over

$ gawk -v c=9 -v RS='[[:space:]]+' 'NR<=c{ORS=(NR<c?RT:"\n");print}' file
the quick
brown     fox     jumped over
the
lazy
dog's