File test
musically us
challenged a goat that day
spartacus was his name
ba ba ba blacksheep
grep -oic "[\s]*" test
grep -oic "[ ]*" test
grep -oic "[\t]*" test
grep -oic "[\n]*" test
All give me 4, when I expect 11
grep --version -> grep (BSD grep) 2.5.1-FreeBSD
Running this on OSX Sierra 10.12
Repeating spaces should not be counted as one space.
If you are open to tricks and alternatives you might like this one:
$ awk '{print --NF}' <(tr -d '\n' <file)
11
Above solution will count "whitespace" between words. As a result for a string of 'fifteen--> <--spaces' awk will measure 1, like grep.
If you need to count actual single spaces you can use this :
$ awk -F"[ ]" '{print --NF}' <<<"fifteen--> <--spaces"
15
$ awk -F"[ ]" '{print --NF}' <<<" 2 4 6 8 10"
10
$ awk -F"[ ]" '{print --NF}' <(tr -d '\n' <file)
11
One step forward, to count single spaces and tabs:
$ awk -F"[ ]|\t" '{print --NF}' <(echo -e " 2 4 6 8 10\t12 14")
13
tr is generally better for this (in most cases):
tr -d -C ' ' <file | wc -c
The grep solution relies on the fact that the output of grep -o is newline-separated — it will fail miserably for example in the following type of circumstance where there might be multiple spaces:
v='fifteen--> <--spaces'
echo "$v" | grep -o -E ' +' | wc -l
echo "$v" | tr -d -C ' ' | wc -c
grep only returns 1, when it should be 15.
EDIT: If you wanted to count multiple characters (eg. TAB and SPACE) you could use:
tr -dC $'[ \t]' <<< $'one \t' | wc -c
Just use awk:
$ awk -v RS=' ' 'END{print NR-1}' file
11
or if you want to handle empty files gracefully:
$ awk -v RS=' ' 'END{print NR - (NR?1:0)}' /dev/null
0
The -c option counts the number of lines that match, not individual matches. Use grep -o and then pipe to wc -l, which will count the number of lines.
grep -o ' ' test | wc -l
Related
I am trying to print only specific output from sentence like below
Before and after dot text should be printed
InputVar="ABC SDFSG XYZ.AFGAJK JKK"
Expected output :
XYZ.AFGAJK
I am using cut command not working
echo "$InputVar" | cut -d'' -f2
Any other approach ?
Here are a few suggestions. awk with RS set to a space seems easiest. YMMV
$ echo "$InputVar" | cut -d ' ' -f 3
XYZ.AFGAJK
$ echo "$InputVar" | awk '/\./' RS=' '
XYZ.AFGAJK
$ echo "$InputVar" | awk '{for(i=1;i<=NF;i++) if(match($i,"\\.")) print $i}'
XYZ.AFGAJK
$ echo "$InputVar" | sed -n 's/.* \([^ .]*[.][^ .]*\) .*/\1/p'
XYZ.AFGAJK
Using cut:
If you really want to use cut, then you could try:
echo "$InputVar" | cut -d' ' -f3
Which uses a space character as a delimiter (you originally had an empty string, which is not allowed), and extracts field 3 rather than field 2.
Using grep:
You can use grep rather than cut, to match & extract specifically what you want:
echo "$InputVar" | grep -Eo '[^ ]+\.[^ ]+'
Explanation:
The -E option is for extended regex
The -o option is for extracting the matched component only
The regex matches a literal ., surrounded by a non-empty sequence of non-space characters
Comparing the two methods:
Either of these will work with your shown example. But, suppose the input string was instead:
InputVar="ABC SDFSG XYZ.AFGAJK JKK XYZ.ABC"
The version using grep would give all the matches (a literal . with non-space characters on either side).
Using cut however, you would need to specify the specific fields you want, i.e.
$ echo "$InputVar" | cut -d' ' -f3,5
XYZ.AFGAJK
XYZ.ABC
If you instead wanted just the n-th match, using the grep approach, you could use sed to select the n-th match, e.g.
$ echo "$InputVar" | grep -Eo '[^ ]+\.[^ ]+'
XYZ.AFGAJK
XYZ.ABC
$ echo "$InputVar" | grep -Eo '[^ ]+\.[^ ]+' | sed '1q;d'
XYZ.AFGAJK
$ echo "$InputVar" | grep -Eo '[^ ]+\.[^ ]+' | sed '2q;d'
XYZ.ABC
I'm new to Linux shell. I know there are tools to do this thing, such as awk. But I'm wondering if I could do it using grep or wc or other commands? awk seems intimidating to me. Thanks.
I tried grep and wc, like this:
grep tol test.txt | wc -w
But grep will give me the whole line.
If I tried the following:
grep '^tol$*' test.txt | wc -w
It only counts the line begins with mol.
How can I grep the words starting with tol?
Something like that:
grep -o '\<tol[[:alpha:]]*\>' test.txt | wc -w
< - for beginning of the word,
> - the end of the word.
[[:alpha:]] - to avoid match of combinations like tol123 (You said you need only words).
-o - to show only matches, not the entire line.
You can do the same fairly simply with awk, e.g.
awk '{for(i=1;i<=NF;i++) $i~/^tol/ && n++} END {print n}'
Example
$ echo -e "tolerance topaz tolstoy\nbats toluene toledo" |
> awk '{for(i=1;i<=NF;i++) $i~/^tol/ && n++} END {print n}'
4
Another option is to translate all whitespace characters into linefeeds so that each word starts on a new line, then grep can count them itself:
echo -e "tolerance topaz\ttolstoy\nbats toluene toledo" | tr '[:space:]' '\n' | grep -c "^tol"
4
Or, if using a file called words.txt:
tr '[:space:]' '\n' < words.txt | grep -c "^tol"
wc -l file.txt
outputs number of lines and file name.
I need just the number itself (not the file name).
I can do this
wc -l file.txt | awk '{print $1}'
But maybe there is a better way?
Try this way:
wc -l < file.txt
cat file.txt | wc -l
According to the man page (for the BSD version, I don't have a GNU version to check):
If no files are specified, the standard input is used and no file
name is
displayed. The prompt will accept input until receiving EOF, or [^D] in
most environments.
To do this without the leading space, why not:
wc -l < file.txt | bc
Comparison of Techniques
I had a similar issue attempting to get a character count without the leading whitespace provided by wc, which led me to this page. After trying out the answers here, the following are the results from my personal testing on Mac (BSD Bash). Again, this is for character count; for line count you'd do wc -l. echo -n omits the trailing line break.
FOO="bar"
echo -n "$FOO" | wc -c # " 3" (x)
echo -n "$FOO" | wc -c | bc # "3" (√)
echo -n "$FOO" | wc -c | tr -d ' ' # "3" (√)
echo -n "$FOO" | wc -c | awk '{print $1}' # "3" (√)
echo -n "$FOO" | wc -c | cut -d ' ' -f1 # "" for -f < 8 (x)
echo -n "$FOO" | wc -c | cut -d ' ' -f8 # "3" (√)
echo -n "$FOO" | wc -c | perl -pe 's/^\s+//' # "3" (√)
echo -n "$FOO" | wc -c | grep -ch '^' # "1" (x)
echo $( printf '%s' "$FOO" | wc -c ) # "3" (√)
I wouldn't rely on the cut -f* method in general since it requires that you know the exact number of leading spaces that any given output may have. And the grep one works for counting lines, but not characters.
bc is the most concise, and awk and perl seem a bit overkill, but they should all be relatively fast and portable enough.
Also note that some of these can be adapted to trim surrounding whitespace from general strings, as well (along with echo `echo $FOO`, another neat trick).
How about
wc -l file.txt | cut -d' ' -f1
i.e. pipe the output of wc into cut (where delimiters are spaces and pick just the first field)
How about
grep -ch "^" file.txt
Obviously, there are a lot of solutions to this.
Here is another one though:
wc -l somefile | tr -d "[:alpha:][:blank:][:punct:]"
This only outputs the number of lines, but the trailing newline character (\n) is present, if you don't want that either, replace [:blank:] with [:space:].
Another way to strip the leading zeros without invoking an external command is to use Arithmetic expansion $((exp))
echo $(($(wc -l < file.txt)))
Best way would be first of all find all files in directory then use AWK NR (Number of Records Variable)
below is the command :
find <directory path> -type f | awk 'END{print NR}'
example : - find /tmp/ -type f | awk 'END{print NR}'
This works for me using the normal wc -l and sed to strip any char what is not a number.
wc -l big_file.log | sed -E "s/([a-z\-\_\.]|[[:space:]]*)//g"
# 9249133
I'm trying to get the count of a matching pattern from a variable to check the count of it, but it's only returning 1 as the results, here is what I'm trying to do:
x="HELLO|THIS|IS|TEST"
echo $x | grep -c "|"
Expected result: 3
Actual Result: 1
Do you know why is returning 1 instead of 3?
Thanks.
grep -c counts lines not matches within a line.
You can use awk to get a count:
x="HELLO|THIS|IS|TEST"
echo "$x" | awk -F '|' '{print NF-1}'
3
Alternatively you can use tr and wc:
echo "$x" | tr -dc '|' | wc -c
3
$ echo "$x" | grep -o '|' | grep -c .
3
grep -c does not count the number of matches. It counts the number of lines that match. By using grep -o, we put the matches on separate lines.
This approach works just as well with multiple lines:
$ cat file
hello|this|is
a|test
$ grep -o '|' file | grep -c .
3
The grep manual says:
grep, egrep, fgrep - print lines matching a pattern
and for the -c flag:
instead print a count of matching lines for each input file
and there is just one line that match
You don't need grep for this.
pipe_only=${x//[^|]} # remove everything except | from the value of x
echo "${#pipe_only}" # output the length of pipe_only
Try this :
$ x="HELLO|THIS|IS|TEST"; echo -n "$x" | sed 's/[^|]//g' | wc -c
3
With only one pipe with perl:
echo "$x" |
perl -lne 'print scalar(() = /\|/g)'
wc -l file.txt
outputs number of lines and file name.
I need just the number itself (not the file name).
I can do this
wc -l file.txt | awk '{print $1}'
But maybe there is a better way?
Try this way:
wc -l < file.txt
cat file.txt | wc -l
According to the man page (for the BSD version, I don't have a GNU version to check):
If no files are specified, the standard input is used and no file
name is
displayed. The prompt will accept input until receiving EOF, or [^D] in
most environments.
To do this without the leading space, why not:
wc -l < file.txt | bc
Comparison of Techniques
I had a similar issue attempting to get a character count without the leading whitespace provided by wc, which led me to this page. After trying out the answers here, the following are the results from my personal testing on Mac (BSD Bash). Again, this is for character count; for line count you'd do wc -l. echo -n omits the trailing line break.
FOO="bar"
echo -n "$FOO" | wc -c # " 3" (x)
echo -n "$FOO" | wc -c | bc # "3" (√)
echo -n "$FOO" | wc -c | tr -d ' ' # "3" (√)
echo -n "$FOO" | wc -c | awk '{print $1}' # "3" (√)
echo -n "$FOO" | wc -c | cut -d ' ' -f1 # "" for -f < 8 (x)
echo -n "$FOO" | wc -c | cut -d ' ' -f8 # "3" (√)
echo -n "$FOO" | wc -c | perl -pe 's/^\s+//' # "3" (√)
echo -n "$FOO" | wc -c | grep -ch '^' # "1" (x)
echo $( printf '%s' "$FOO" | wc -c ) # "3" (√)
I wouldn't rely on the cut -f* method in general since it requires that you know the exact number of leading spaces that any given output may have. And the grep one works for counting lines, but not characters.
bc is the most concise, and awk and perl seem a bit overkill, but they should all be relatively fast and portable enough.
Also note that some of these can be adapted to trim surrounding whitespace from general strings, as well (along with echo `echo $FOO`, another neat trick).
How about
wc -l file.txt | cut -d' ' -f1
i.e. pipe the output of wc into cut (where delimiters are spaces and pick just the first field)
How about
grep -ch "^" file.txt
Obviously, there are a lot of solutions to this.
Here is another one though:
wc -l somefile | tr -d "[:alpha:][:blank:][:punct:]"
This only outputs the number of lines, but the trailing newline character (\n) is present, if you don't want that either, replace [:blank:] with [:space:].
Another way to strip the leading zeros without invoking an external command is to use Arithmetic expansion $((exp))
echo $(($(wc -l < file.txt)))
Best way would be first of all find all files in directory then use AWK NR (Number of Records Variable)
below is the command :
find <directory path> -type f | awk 'END{print NR}'
example : - find /tmp/ -type f | awk 'END{print NR}'
This works for me using the normal wc -l and sed to strip any char what is not a number.
wc -l big_file.log | sed -E "s/([a-z\-\_\.]|[[:space:]]*)//g"
# 9249133