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I have a file whose content is similar to the following one.
0
0
0.2
0
0
0
0
I need to remove all the lines with a single zero.
I was thinking to use grep -v "0", but this removes also the line containing 0.2. I saw I could use the -w option, but this doesn't seem to work either.
How can I remove all the lines containing just a single 0 and keep all those lines starting with a 0?
grep -vx 0
From man grep:
-x, --line-regexp
Select only those matches that exactly match the whole line.
For a regular expression pattern, this is like parenthesizing
the pattern and then surrounding it with ^ and $.
-w fails because the first 0 in 0.02 is considered a "word", and hence this line is matched. This is because it is followed by a "non-word" character. You can see this if you run the original command without -v, i.e. grep -w "0".
-F option since we're not using regex patterns, just plain string matching
Feb 14, 2019 at 13:54
-F (surprisingly to me) appears to take a similar amount of time or even slightly slower (~5–10%). Hence, I'm not sure what the advantage would be.
grep presumably has a special case for regexes with no metacharacters, because that's a common use-case. It's surprising that fgrep would be slower, but it's not surprising that the overhead of noticing this special case while compiling a short pattern is negligible vs. the time to scan a large file. (If it requires a special case at all to go that fast, vs. a pattern with a character class or x.*y.)
Feb 15, 2019 at 14:54
grep recognizes any character other than \n newline as a line separator. If not, the implicit ^ and $ can still turn into a fixed-string search like strstr(big_buf, "\n0\n"). (Or 0\n at the start of a buffer.) But we're not just looking for the first match potentially far into a big buffer, we want to efficiently filter. But anyway, in theory yes it's just a 2-byte memcmp at the start of each line, and you'd hope that both fgrep and grep would see that.
Feb 15, 2019 at 15:04
With grep:
grep -v '^0$' file
^ means beginning of the line, $ means end of the line.
[a-Z0-9]
Feb 17, 2019 at 6:16
While grep can be used for this (as other answers clearly show), let’s take a step back and think about what you actually want:
Regex interpret character sequence data. They don’t know about numbers, only about individual digits (and regular combinations thereof). Although in your particular case there’s a simple hack around this limitation, it’s ultimately a requirement mismatch.
Unless there’s a very good reason to use grep here (e.g. because you’ve measured it, and it’s vastly more efficient, and efficiency is crucial in your case), I recommend using a different tool.
awk, for instance, can filter based on numeric comparisions, e.g.:
awk '$1 == 0' your_file
But also, to get all lines containing numbers greater than zero:
awk '$1 > 0' your_file
I love regex, it’s a great tool. But it’s not the only tool. As the saying goes, if all you have is grep, everything looks like a regular language.
printf '0\n1\n-1\na\nb\n0\n0 also\n0.0\n-0.0\n0*0\n' | awk '($1 == 0)' will match: 0, 0.0 and -0.0... and also 0 also ! Not just "0". (which is sometimes what's needed, sometimes not). If the user want only "0" : awk '/^0$/' (or grep '^0$'). Also you should edit: the user needs to add ! to negate the test, so it hides 0 (and other zeroes) and displays the rest. ie: awk '!( $0 == 0)'
Feb 14, 2019 at 15:20
> rather than != (or, equivalently, ! (… == …)) to highlight that this is an arbitrary numerical comparison, not just equality. As for your other comment, this is entirely true but then we’re essentially back in string comparison territory and the existing solution using grep works (though awk of course also works).
Feb 15, 2019 at 11:39
$0=="0"
Feb 15, 2019 at 18:59
grep's -w is a bit convoluted in a way that it splits up the original string into word and non-word constituents (anything except letters, digits or underscore) . Since it has already encountered a a valid word constituent 0 in 0.02 it had asserted the negation logic to remove the line.
Using sed is a bit easy in this context to just remove the whole words that match
sed '/^0$/d' file
When the lines you want to delete only contain a 0 followed by the next line you can select those lines by issuing the following command:
grep -v "^0$"
This will only print the occurrences of 0 that are at the end of a line and at the beginning of a line at the same time. The -v option then inverts our selection.
-v, so it doesn't work.
-v option, thanks!
Feb 14, 2019 at 8:10
Another answer for the sake of variety, assuming you have a PCRE-enabled grep
grep -Pv "^0(?!\.)"
this performs a negative lookahead to match the lines that start with 0 and are not followed by a dot. Then -v discards non-matching lines. You can see in action here
0123, which is not what the OP wants
Assuming any line which is not just a single 0 has a period
grep '\.' file
\b - word border
grep -v "\b0\b"
match beginning of line, your pattern and end of line
grep -v "^0$"
or as @Sparhawk suggested -vx lineregexp
Note that -w works, but in your case 0.2 are two words because dot character is a word separator.
grep -v "\b0\b" doesn't really work here. What version of grep do you use?
Feb 14, 2019 at 7:26
grep (BSD grep) 2.5.1-FreeBSD on macOS and grep (GNU grep) 2.16 on ubuntu
Feb 14, 2019 at 8:02
\< and\> as word boundaries, but that will have the same effect as -w
Feb 14, 2019 at 13:53
-w, which fails here.grepfor this task? And what exactly do you mean by a single zero? This sounds very much like an XY problem.