Grep time command output

Question

Using time ls, I have the following output:

$ time ls -l 
total 2
-rwx------+ 1 FRIENDS None 97 Jun 23 08:59 location.txt
-rw-r--r--+ 1 FRIENDS None 10 Jun 23 09:06 welcome
real    0m0.040s
user    0m0.000s    
sys     0m0.031s

Now, when I try to grep only the real value line, the actual result is:

$ time ls -l | grep real
real    0m0.040s
user    0m0.000s
sys     0m0.031s

My question is, how to get only the real value as output? In this case, 0m0.040s.

`time` is sending its output to stderr, not stdout, so `grep` isn't processing *any* of its output. — Gabe, Jun 23 '13 at 04:39

rici · Accepted Answer · 2016-07-14T18:26:25.177

time writes its output to stderr, so you need to pipe stderr instead of stdout. But it's also important to remember that time is part of the syntax of bash, and it times an entire pipeline. Consequently, you need to wrap the pipeline in braces, or run it in a subshell:

 $ { time ls -l >/dev/null; } 2>&1 | grep real
 real   0m0.005s

With Bash v4.0 (probably universal on Linux distributions but still not standard on Mac OS X), you can use |& to pipe both stdout and stderr:

{ time ls -l >/dev/null; } |& grep real

Alternatively, you can use the time utility, which allows control of the output format. On my system, that utility is found in /usr/bin/time:

/usr/bin/time -f%e ls -l >/dev/null

man time for more details on the time utility.

`|&` does the trick for me. – Saurav Sahu Nov 21 '22 at 05:30 — Saurav Sahu, Nov 21 '22 at 05:30

score 19 · Answer 2 · answered Jun 23 '13 at 04:34

19

(time ls -l)  2>&1 > /dev/null |grep real

This redirects stderr (which is where time sends its output) to the same stream as stdout, then redirects stdout to dev/null so the output of ls is not captured, then pipes what is now the output of time into the stdin of grep.

answered Jun 23 '13 at 04:34

Clyde

7,389
5
31
57

6

To be complete, you should add `| awk '{print $2}'` to the end, as the OP requires only the time portion. – Burhan Khalid Jun 23 '13 at 04:47
1

@BurhanKhalid Indeed so - but we should leave _something_ for the OP to do :-) – Clyde Jun 23 '13 at 06:31

score 14 · Answer 3 · edited May 23 '17 at 12:25

14

If you just want to specify the output format of time builtin, you can modify the value of TIMEFORMAT environment variable instead of filtering it with grep.

In you case,

TIMEFORMAT=%R
time ls -l

would give you the "real" time only.

Here's the link to relevant information in Bash manual (under "TIMEFORMAT").

This is a similar question on SO about parsing the output of time.

edited May 23 '17 at 12:25

Community

1
1

answered Jun 23 '13 at 04:46

doubleDown

8,048
1
32
48

2

Even better output this way. serafina@box:~$ TIMEFORMAT=%R && time ls > /dev/null 0.001 – Erin Ishimoticha Feb 12 '14 at 00:56

score 3 · Answer 4 · answered Jun 23 '16 at 18:26

Look out.. bash has a built-in "time" command. Here are some of the differences..

# GNU time command (can also use $TIMEFORMAT variable instead of -f)
bash> /usr/bin/time -f%e ls >/dev/null
0.00


# BASH built-in time command (can also use $TIME variable instead of -f)
bash> time -f%e ls >/dev/null
-f%e: command not found

real    0m0.005s
user    0m0.004s
sys     0m0.000s

score -1 · Answer 5 · answered Apr 19 '15 at 22:25

-1

I think, it can be made a little easier:

time ls &> /dev/null | grep real

answered Apr 19 '15 at 22:25

VeLKerr

2,995
3
24
47

Grep time command output

5 Answers5

Linked