Counting grep results - bash

I'm new to bash so I'm finding trouble doing something very basic.
Through playing with various scripts I found out that the following script prints the lines that contain the "word"
for file in*; do
cat $file | grep "word"
done
doing the following:
for file in*; do
cat $file | grep "word" | wc -l
done
had a result of printing in every iteration how many times did the "word" appeared on file.
How can I implement a counter for all those appearances and in the
end just echo the counter?
I used a counter that way but it appeared 0.
let x+=cat $filename | grep "word"

You can pipe the entire loop to wc -l.
for file in *; do
cat $file | grep "word"
done | wc -l
This is a useless use of cat. How about:
for file in *; do
grep "word" "$file"
done | wc -l
Actually, the entire loop is unnecessary if you pass all the file names to grep at once.
grep "word" * | wc -l
Note that if word shows up more than once on the same line these solutions will only count the entire line once. If you want to count same-line occurrences separately you can use -o to print each match on a separate line:
grep -o "word" * | wc -l

The oneliner in John's answer is the way to go. Just to satisfy your curiosity:
sum=0
for f in *; do
x="$(grep 'word' "$f" | wc -l)"
echo "x: $x"
(( sum += x ))
done
echo "sum: $sum"
If the line containing the grep and wc does not yield a number you are SOL. That is why you should stick to the other solution or do a pure bash implementation with things like read, 'case and *word*)' or if [[ "$line" =~ "$re_containing_word" ]]; then ...

Related

count all the lines in all folders in bash [duplicate]

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

Count mutiple occurences of a word on the same line using grep

Here I made a small script that take input from user searching some pattern from a file and displays required no of lines from that file where the pattern is found. Although this code is searching the pattern line wise due to standard grep practice. I mean if the pattern occurs twice on the same line, i want the output to print twice. Hope I make some sense.
#!/bin/sh
cat /dev/null>copy.txt
echo "Please enter the sentence you want to search:"
read "inputVar"
echo "Please enter the name of the file in which you want to search:"
read "inputFileName"
echo "Please enter the number of lines you want to copy:"
read "inputLineNumber"
[[-z "$inputLineNumber"]] || inputLineNumber=20
cat /dev/null > copy.txt
for N in `grep -n $inputVar $inputFileName | cut -d ":" -f1`
do
LIMIT=`expr $N + $inputLineNumber`
sed -n $N,${LIMIT}p $inputFileName >> copy.txt
echo "-----------------------" >> copy.txt
done
cat copy.txt
As I understood, the task is to count number of pattern occurrences in line. It can be done like so:
count=$((`echo "$line" | sed -e "s|$pattern|\n|g" | wc -l` - 1))
Suppose you have one file to read. Then, code will be following:
#!/bin/bash
file=$1
pattern="an."
#reading file line by line
cat -n $file | while read input
do
#storing line to $tmp
tmp=`echo $input | grep "$pattern"`
#counting occurrences count
count=$((`echo "$tmp" | sed -e "s|$pattern|\n|g" | wc -l` - 1))
#printing $tmp line $count times
for i in `seq 1 $count`
do
echo $tmp
done
done
I checked this for pattern "an." and input:
I pass here an example of many 'an' letters
an
ananas
an-an-as
Output is:
$ ./test.sh input
1 I pass here an example of many 'an' letters
1 I pass here an example of many 'an' letters
1 I pass here an example of many 'an' letters
3 ananas
4 an-an-as
4 an-an-as
Adapt this to your needs.
How about using awk?
Assume the pattern you are searching for is in variable $pattern and the file you are checking is $file
The
count=`awk 'BEGIN{n=0}{n+=split($0,a,"'$pattern'")-1}END {print n}' $file`
or for a line
count=`echo $line | awk '{n=split($0,a,"'$pattern'")-1;print n}`

How to get "wc -l" to print just the number of lines without file name?

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

Get just the integer from wc in bash

Is there a way to get the integer that wc returns in bash?
Basically I want to write the line numbers and word counts to the screen after the file name.
output: filename linecount wordcount
Here is what I have so far:
files=\`ls`
for f in $files;
do
if [ ! -d $f ] #only print out information about files !directories
then
# some way of getting the wc integers into shell variables and then printing them
echo "$f $lines $words"
fi
done
Most simple answer ever:
wc < filename
Just:
wc -l < file_name
will do the job. But this output includes prefixed whitespace as wc right-aligns the number.
You can use the cut command to get just the first word of wc's output (which is the line or word count):
lines=`wc -l $f | cut -f1 -d' '`
words=`wc -w $f | cut -f1 -d' '`
wc $file | awk {'print "$4" "$2" "$1"'}
Adjust as necessary for your layout.
It's also nicer to use positive logic ("is a file") over negative ("not a directory")
[ -f $file ] && wc $file | awk {'print "$4" "$2" "$1"'}
Sometimes wc outputs in different formats in different platforms. For example:
In OS X:
$ echo aa | wc -l
1
In Centos:
$ echo aa | wc -l
1
So using only cut may not retrieve the number. Instead try tr to delete space characters:
$ echo aa | wc -l | tr -d ' '
The accepted/popular answers do not work on OSX.
Any of the following should be portable on bsd and linux.
wc -l < "$f" | tr -d ' '
OR
wc -l "$f" | tr -s ' ' | cut -d ' ' -f 2
OR
wc -l "$f" | awk '{print $1}'
If you redirect the filename into wc it omits the filename on output.
Bash:
read lines words characters <<< $(wc < filename)
or
read lines words characters <<EOF
$(wc < filename)
EOF
Instead of using for to iterate over the output of ls, do this:
for f in *
which will work if there are filenames that include spaces.
If you can't use globbing, you should pipe into a while read loop:
find ... | while read -r f
or use process substitution
while read -r f
do
something
done < <(find ...)
If the file is small you can afford calling wc twice, and use something like the following, which avoids piping into an extra process:
lines=$((`wc -l "$f"`))
words=$((`wc -w "$f"`))
The $((...)) is the Arithmetic Expansion of bash. It removes any whitespace from the output of wc in this case.
This solution makes more sense if you need either the linecount or the wordcount.
How about with sed?
wc -l /path/to/file.ext | sed 's/ *\([0-9]* \).*/\1/'
typeset -i a=$(wc -l fileName.dat | xargs echo | cut -d' ' -f1)
Try this for numeric result:
nlines=$( wc -l < $myfile )
Something like this may help:
#!/bin/bash
printf '%-10s %-10s %-10s\n' 'File' 'Lines' 'Words'
for fname in file_name_pattern*; {
[[ -d $fname ]] && continue
lines=0
words=()
while read -r line; do
((lines++))
words+=($line)
done < "$fname"
printf '%-10s %-10s %-10s\n' "$fname" "$lines" "${#words[#]}"
}
To (1) run wc once, and (2) not assign any superfluous variables, use
read lines words <<< $(wc < $f | awk '{ print $1, $2 }')
Full code:
for f in *
do
if [ ! -d $f ]
then
read lines words <<< $(wc < $f | awk '{ print $1, $2 }')
echo "$f $lines $words"
fi
done
Example output:
$ find . -maxdepth 1 -type f -exec wc {} \; # without formatting
1 2 27 ./CNAME
21 169 1065 ./LICENSE
33 130 961 ./README.md
86 215 2997 ./404.html
71 168 2579 ./index.html
21 21 478 ./sitemap.xml
$ # the above code
404.html 86 215
CNAME 1 2
index.html 71 168
LICENSE 21 169
README.md 33 130
sitemap.xml 21 21
Solutions proposed in the answered question doesn't work for Darwin kernels.
Please, consider following solutions that work for all UNIX systems:
print exactly the number of lines of a file:
wc -l < file.txt | xargs
print exactly the number of characters of a file:
wc -m < file.txt | xargs
print exactly the number of bytes of a file:
wc -c < file.txt | xargs
print exactly the number of words of a file:
wc -w < file.txt | xargs
There is a great solution with examples on stackoverflow here
I will copy the simplest solution here:
FOO="bar"
echo -n "$FOO" | wc -l | bc # "3"
Maybe these pages should be merged?
Try this:
wc `ls` | awk '{ LINE += $1; WC += $2 } END { print "lines: " LINE " words: " WC }'
It creates a line count, and word count (LINE and WC), and increase them with the values extracted from wc (using $1 for the first column's value and $2 for the second) and finally prints the results.
"Basically I want to write the line numbers and word counts to the screen after the file name."
answer=(`wc $f`)
echo -e"${answer[3]}
lines: ${answer[0]}
words: ${answer[1]}
bytes: ${answer[2]}"
Outputs :
myfile.txt
lines: 10
words: 20
bytes: 120
files=`ls`
echo "$files" | wc -l | perl -pe "s#^\s+##"
You have to use input redirection for wc:
number_of_lines=$(wc -l <myfile.txt)
respectively in your context
echo "$f $(wc -l <"$f") $(wc -w <"$f")"

Randomizing arg order for a bash for statement

I have a bash script that processes all of the files in a directory using a loop like
for i in *.txt
do
ops.....
done
There are thousands of files and they are always processed in alphanumerical order because of '*.txt' expansion.
Is there a simple way to random the order and still insure that I process all of the files only once?
Assuming the filenames do not have spaces, just substitute the output of List::Util::shuffle.
for i in `perl -MList::Util=shuffle -e'$,=$";print shuffle<*.txt>'`; do
....
done
If filenames do have spaces but don't have embedded newlines or backslashes, read a line at a time.
perl -MList::Util=shuffle -le'$,=$\;print shuffle<*.txt>' | while read i; do
....
done
To be completely safe in Bash, use NUL-terminated strings.
perl -MList::Util=shuffle -0 -le'$,=$\;print shuffle<*.txt>' |
while read -r -d '' i; do
....
done
Not very efficient, but it is possible to do this in pure Bash if desired. sort -R does something like this, internally.
declare -a a # create an integer-indexed associative array
for i in *.txt; do
j=$RANDOM # find an unused slot
while [[ -n ${a[$j]} ]]; do
j=$RANDOM
done
a[$j]=$i # fill that slot
done
for i in "${a[#]}"; do # iterate in index order (which is random)
....
done
Or use a traditional Fisher-Yates shuffle.
a=(*.txt)
for ((i=${#a[*]}; i>1; i--)); do
j=$[RANDOM%i]
tmp=${a[$j]}
a[$j]=${a[$[i-1]]}
a[$[i-1]]=$tmp
done
for i in "${a[#]}"; do
....
done
You could pipe your filenames through the sort command:
ls | sort --random-sort | xargs ....
Here's an answer that relies on very basic functions within awk so it should be portable between unices.
ls -1 | awk '{print rand()*100, $0}' | sort -n | awk '{print $2}'
EDIT:
ephemient makes a good point that the above is not space-safe. Here's a version that is:
ls -1 | awk '{print rand()*100, $0}' | sort -n | sed 's/[0-9\.]* //'
If you have GNU coreutils, you can use shuf:
while read -d '' f
do
# some stuff with $f
done < <(shuf -ze *)
This will work with files with spaces or newlines in their names.
Off-topic Edit:
To illustrate SiegeX's point in the comment:
$ a=42; echo "Don't Panic" | while read line; do echo $line; echo $a; a=0; echo $a; done; echo $a
Don't Panic
42
0
42
$ a=42; while read line; do echo $line; echo $a; a=0; echo $a; done < <(echo "Don't Panic"); echo $a
Don't Panic
42
0
0
The pipe causes the while to be executed in a subshell and so changes to variables in the child don't flow back to the parent.
Here's a solution with standard unix commands:
for i in $(ls); do echo $RANDOM-$i; done | sort | cut -d- -f 2-
Here's a Python solution, if its available on your system
import glob
import random
files = glob.glob("*.txt")
if files:
for file in random.shuffle(files):
print file

Resources