$(grep -w "xyz" prog.R | wc -l) -ge 3

Question

I am a bit new to unix and trying to figure out what the following command is doing:

$(grep -w "xyz" prog.R | wc -l) -ge 3

I assume that the outcome would be boolean. The first part (i.e., grep -w "xyz" prog.R) is looking for the text xyz in the prog.R file. But I cannot make out what this part as a whole is doing: $(grep -w "xyz" prog.R | wc -l)

Answer 1

The grep command extracts all lines in the file called prog.R that contains the word xyz. The -w option makes it only match complete words, so it won't match e.g. xyzz or vxyz.

The wc -l command reads the output from grep via the | pipe, and counts the number of lines that the grep command produces. This will be the output of the command substitution as a whole, i.e. the $(...) bit.

Assuming that the command substitution ($(...)) sits inside [ ... ] or as arguments to the test utility, the -ge 3 then is a test to see whether the number of lines counted by wc -l is greater than or equal to three.

test "$(grep -w "xyz" prog.R | wc -l)" -ge 3

or

[ "$(grep -w "xyz" prog.R | wc -l)" -ge 3 ]

This returns an exit status that could be use by an if statement (as a "boolean", like you say), for example, but it's hard to say much more without seeing more of the code.

Without [ ... ] or test, the command is nonsense.

You may want to check the manuals for the grep, wc and test utilities, and possibly to read up on pipes and command substitutions.

---

Note my quoting of "$( grep ... )" above. Without the double quotes, the shell would split the result of the command substitution on the characters in $IFS (space, tab, newline, by default), and would then apply filename globbing to the resulting strings. This is something that we don't really want here, especially since we don't know the value of $IFS (if it contains digits, for whatever reason, this may affect the result of the test).

See also:

• When is double-quoting necessary?

---

The grep and wc test could be shortened into just

[ "$(grep -c -w 'xyz' prog.R)" -ge 3 ]

The -c option to grep makes it report the number of lines matched, making the use of wc -l unnecessary.

With GNU grep (and some others), the whole test could made further effective (faster):

[ "$(grep -c -m 3 -w 'xyz' prog.R)" -eq 3 ]

where -m 3 makes grep stop after three matches. With -c, grep then outputs 3 if at least three lines matches the expression, so we test with -eq 3 to see if this was the case.

The -m and -w options to grep are not standard, but often (-w) or sometimes (-m) implemented.

Answer 2

Your command:

$(grep -w "xyz" prog.R | wc -l) -ge 3

when in the context of a shell test command:

if [ $(grep -w "xyz" prog.R | wc -l) -ge 3 ]; then
    do_something
fi

would be testing whether or not xyz exists as a complete "word" on at least 3 lines in prog.R and setting a zero exit status if so, non-zero otherwise. See @Kuslananadas answer for a breakdown of that command line.

Alternatively you could just use a single awk command, e.g. this in GNU awk:

if awk '/\<wyz\>/ && (++c == 3){ f=1; exit } END{exit !f}' prog.R; then
    do_something
fi

or this in any awk:

if awk '/(^|[^[:alnum:]_])wyz([^[:alnum:]_]|$)/ && (++c == 3){ f=1; exit } END{exit !f}' prog.R; then
    do_something
fi