I want to count the amount of lines in a CSV file.
But only those with "<" at the beginning.
Use grep with c:
grep -c '^<' <FILE>
Use the awk, Luke:
$ cat file
foo
< this foo
not this foo <
< another of those foos
$ awk '/^</{c++}END{print c}' file
2
You can get all the line containing '<' at the begining with grep '^<' and count remaining line with wc -l.
grep "^<" file.csv | wc -l
grep -E "^<" filename | wc -l
Use regular expressions with grep to check for < at the start of the line (^) and then pipe through wc -l to get number of lines.
Related
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"
I trying make oneline command with operation where I can do:
in folder "data" have 570 files - each file have some text line - file are called from 1 to 570.txt
I want cat each file, grep by word and count how manny that word occurs.
For the moment he is trying to get this using ' for '
for FILES in $(find /home/my/data/ -type f -print -exec cat {} \;) ; do echo $FILES; cat $FILES |grep word ; done |wc -l
but if I do that they correctly counts but does not display the counted file
I would like it to look :
----> 1.txt <----
210
---> 2.txt <----
15
etc, etc, etc..
How to get it
grep -o word * | uniq -c
is practically all you need.
grep -o word * gives a line for each hit, but only prints the match, in this case "word". Each line is prefixed with the filename it was found in.
uniq -c gives only one line per file so to say and prefixes it with the count.
You can further format it to your needs with awk or whatever, though, for example like this:
grep -o word * | uniq -c | cut -f1 -d':' | awk '{print "File: " $2 " Count: " $1}'
You can try this :
for file in /path/to/folder/data/* ; do echo "----> $file <----" ; grep -c "word_to_count" /path/to/folder/data/$file ; done
for loop will ierate over file inside folder "data".
For each of these file, print the name and search for number of occurrence of "word_to_count" (grep -c will directly output a count of matching lines).
Be carefull, if there is more than one iteration of your search word inside a line, this solution will count only one for these iteration.
Bit of awk should do it?
awk '{s+=$1} END {print s}' mydatafile
Note: some versions of awk have some odd behaviours if you are going to be adding anything exceeding 2^31 (2147483647). See comments for more background. One suggestion is to use printf rather than print:
awk '{s+=$1} END {printf "%.0f", s}' mydatafile
$ python -c "import sys; print(sum(int(l) for l in sys.stdin))"
If you only want the total number of lines, you could use
find /home/my/data/ -type f -exec cat {} + | wc -l
I have many files on a server which contains many lines:
201701010530.contentState.csv.gz
201701020530.contentState.csv.gz
201701030530.contentState.csv.gz
201701040530.contentState.csv.gz
I would like with one line command this result:
170033|20170101
169865|20170102
170010|20170103
170715|20170104
The goal is to have the number of lines of each file, just by keeping the date which is already in the filename of the file.
I tried this but the result is not in one line but two...
for f in $(ls -1 2017*gz);do zcat $f | wc -l;echo $f | awk '{print substr($0,1,8)}';done
Thanks in advance guys.
Just use zcat file | wc -l to get the number of lines.
For the name, I understand it is enough to extract the first 8 characters:
$ t="201701030530.contentState.csv.gz"
$ echo "${t:0:8}"
20170103
All together:
for file in 2017*gz;
do
lines=$(zcat "$file" | wc -l)
printf "%s|%s\n" "$lines" "${file:0:8}"
done > myresult.csv
Note the usage of for file in 2017*gz; to go through the files matching the 2017*gz pattern: this suffices, no need to parse ls!
Use zgrep -c ^ file to count the lines, here encapsulated in awk:
$ awk 'FNR==1{ "zgrep -c ^ " FILENAME | getline s; print s "|" substr(FILENAME,1,8) }' *.gz
12|20170101
The whole "zgrep -c ^ " FILENAME should probably be in a var (s) and then s | getline s.
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)'
This question already has answers here:
Get just the integer from wc in bash
(19 answers)
Closed 8 years ago.
I want to get only the number of lines in a file:
so I do:
$wc -l countlines.py
9 countlines.py
I do not want the filename, so I tried
$wc -l countlines.py | cut -d ' ' -f1
but this just echo empty line.
I just want number 9 to be printed
Use stdin and you won't have issue with wc printing filename
wc -l < countlines.py
You can also use awk to count lines. (reference)
awk 'END { print NR }' countlines.py
where countlines.py is the file you want to count
If your file doesn't ends with a \n (new line) the wc -l gives a wrong result. Try it with the next simulated example:
echo "line1" > testfile #correct line with a \n at the end
echo -n "line2" >> testfile #added another line - but without the \n
the
$ wc -l < testfile
1
returns 1. (The wc counts the number of newlines (\n) in a file.)
Therefore, for counting lines (and not the \n characters) in a file, you should to use
grep -c '' testfile
e.g. find empty character in a file (this is true for every line) and count the occurences -c. For the above testfile it returns the correct 2.
Additionally, if you want count the non-empty lines, you can do it with
grep -c '.' file
Don't trust wc :)
Ps: one of the strangest use of wc is
grep 'pattern' file | wc -l
instead of
grep -c 'pattern' file
cut is being confused by the leading whitespace.
I'd use awk to print the 1st field here:
% wc -l countlines.py | awk '{ print $1 }'
As an alternative, wc won't print the file name if it is being piped input from stdin
$ cat countlines.py | wc -l
9
yet another way :
cnt=$(wc -l < countlines.py )
echo "total is $cnt "
Piping the file name into wc removes it from the output, then translate away the whitespace:
wc -l <countlines.py |tr -d ' '
Use awk like this:
wc -l countlines.py | awk {'print $1'}