I have 18 csv files, all between 1mb and 14mb. The sum of all files is 64mb. I want to create a new csv file that contains a subset of those files-- only the lines featuring the pattern "Hello" (or "HELLO", or "hello" ...). Here's what I'm doing
cat *.csv | head -n 1 > new.csv # I want to create a header first
cat *.csv | grep -i "hello" >> new.csv
I'm running Debian on WSL. The output file is much, much larger than the original 64mb (I stopped the process after 1+ hour, and the file was 300+ GB).
How can a subset of a text file be larger than the original files? Does it have anything to do with WSL?
This is not an OS issue. When you redirect your output to new.csv, shell creates that file first, before the glob expression *.csv is evaluated. That means the expansion of *.csv would include new.csv as well. That seems like the root cause of the recursive grep issue you are facing.
You are reading all the files twice, which is not necessary. You can make your operation a lot simpler and efficient with a single awk command:
awk 'NR==1 {print} tolower($0) ~ /hello/ {print}' *.csv > csv.new
mv csv.new new.csv
since the output file is named csv.new it won't interfere with the glob *.csv
NR==1 picks up the first line (header) from the very first file
The awk command can be written more succinctly as:
awk 'NR==1 || tolower($0) ~ /hello/' *.csv > csv.new
You are using *.csv and redirecting the output to new.csv which falls under *.csv which is causing recursion in grep result. perhaps you can try,
grep -i hello *.csv --exclude="new.csv" >> new.csv
Related
I'm trying to grep multiple strings which look like this (there's a few hundred) against a file which contains data:string
Example strings: (no sensitive data is provided, they have been modified).
$H$9a...DcuCqC/rMVmfiFNm2rqhK5vFW1
$H$9n...AHZAV.sTefg8ap8qI8U4A5fY91
$H$9o...Bi6Z3E04x6ev1ZCz0hItSh2JJ/
$H$9w...CFva1ddp8IRBkgwww3COVLf/K1
I've been researching how to grep a file of patterns against another file, and came across the following commands
grep -f strings.txt datastring.txt > output.txt
grep -Ff strings.txt datastring.txt > output.txt
But unfortunately, these commands do NOT work successfully, and only print out a handful of results to my output file. I think it may be something to do with the symbols contained in strings.txt, but I'm unsure. Any help/advice would be great.
To further mention, I'm using Cygwin on Windows (if this is relevant).
Here's an updated example:
strings.txt contains the following:
$H$9a...DcuCqC/rMVmfiFNm2rqhK5vFW1
$H$9n...AHZAV.sTefg8ap8qI8U4A5fY91
$H$9o...Bi6Z3E04x6ev1ZCz0hItSh2JJ/
$H$9w...CFva1ddp8IRBkgwww3COVLf/K1
datastring.txt contains the following:
$H$9a...DcuCqC/rMVmfiFNm2rqhK5vFW1:53491
$H$9n...AHZAV.sTefg8ap8qI8U4A5fY91:03221
$H$9o...Bi6Z3E04x6ev1ZCz0hItSh2JJ/:20521
$H$9w...CFva1ddp8IRBkgwww3COVLf/K1:30142
So technically, all lines should be included in the OUTPUT file, but only this line is outputted:
$H$9w...CFva1ddp8IRBkgwww3COVLf/K1:30142
I just don't understand.
You have showed the output of cat -A strings.txt elsewhere, which includes ^M representing a CR (carriage return) character at the end of each line:
This indicates your file has Windows line endings (CR LF) instead of the Unix line endings (only LF) that grep would expect.
You can convert files with dos2unix strings.txt and back with unix2dos strings.txt.
Alternatively, if you don't have dos2unix installed in your Cygwin environment, you can also do that with sed.
sed -i 's/\r$//' strings.txt # dos2unix
sed -i 's/$/\r/' strings.txt # unix2dos
cat file.txt | grep -x "\d*"
grep: \Documents and Settings: Is a directory
I want to search file.txt for any lines that are numbers only but grep seems to be viewing \d* as a wildcard for files and not the pattern. How can I specify that it's the pattern and it should use stdin for what to grep over?
The file is full of lines of datetime stamps, some end with a letter, some don't.
20140110122200
20131208041510M
...
I'm trying to only get the lines that don't end in a letter.
EDIT: I've also tried setting the filename instead of piping it with cat. Not much different.
C:\long\path>grep -ex "\d*" -f file.txt
grep: \Dell: Is a directory
grep: \Documents and Settings: Is a directory
Why are you using cat to pass the file to grep? Why not just give grep the filename directly?
grep -x '\d*' file.txt
I think the actual problem you're seeing is that the * wildcard is being expanded. That's why grep is giving you errors that mention actual directories (beginning with 'd') on your system.
I am in the process of creating a script that lists all files opened via lsof output. I would like to checksum specific files and ignore directories from that output but am at a loss to do so EFFECTIVELY. For example: (I'm using FreeBSD btw)
lsof | awk '/\//{print $9}' | sort -u | head -n 5
prints:
/
/bin/sleep
/dev/bpf
What I'd like to do is: FROM that output, ignore any directories and perform an md5 on FILES (not directories).
Any pointers?
Give a try to following perl command:
lsof | perl -MDigest::MD5=md5_hex -ane '
$f = $F[ $#F ];
-f $f and printf qq|%s %s\n|, $f, md5_hex( $f )
'
It filters lsof output to plain files (-f). Take a look into perlfunc to change it to add different kind of files.
It outputs each file and its md5 separated by a space character. An example in my system is like:
/usr/lib/libm-2.17.so a2d3b2de9a1f59fb99427714fefb49ca
/usr/lib/libdl-2.17.so d74d8ac16c2d13128964353d4be7061a
/usr/lib/libnsl-2.17.so 34b6909ec60c337c21b044642b9baa3d
/usr/lib/ld-2.17.so 3d0e7b5b5c4e59c5c4b6a858cc79fcf1
/usr/sbin/lsof b9b8fbc8f296e47969713f6369d97c0d
/usr/lib/locale/locale-archive 3ea56273193198a718b9a5de33d553db
/usr/lib/libc-2.17.so ba51eeb4025b7f5d7f400f1968f4b5f9
/usr/lib/ld-2.17.so 3d0e7b5b5c4e59c5c4b6a858cc79fcf1
...
I am trying to use a shell script (well a "one liner") to find any common lines between around 50 files.
Edit: Note I am looking for a line (lines) that appears in all the files
So far i've tried grep grep -v -x -f file1.sp * which just matches that files contents across ALL the other files.
I've also tried grep -v -x -f file1.sp file2.sp | grep -v -x -f - file3.sp | grep -v -x -f - file4.sp | grep -v -x -f - file5.sp etc... but I believe that searches using the files to be searched as STD in not the pattern to match on.
Does anyone know how to do this with grep or another tool?
I don't mind if it takes a while to run, I've got to add a few lines of code to around 500 files and wanted to find a common line in each of them for it to insert 'after' (they were originally just c&p from one file so hopefully there are some common lines!)
Thanks for your time,
When I first read this I thought you were trying to find 'any common lines'. I took this as meaning "find duplicate lines". If this is the case, the following should suffice:
sort *.sp | uniq -d
Upon re-reading your question, it seems that you are actually trying to find lines that 'appear in all the files'. If this is the case, you will need to know the number of files in your directory:
find . -type f -name "*.sp" | wc -l
If this returns the number 50, you can then use awk like this:
WHINY_USERS=1 awk '{ array[$0]++ } END { for (i in array) if (array[i] == 50) print i }' *.sp
You can consolidate this process and write a one-liner like this:
WHINY_USERS=1 awk -v find=$(find . -type f -name "*.sp" | wc -l) '{ array[$0]++ } END { for (i in array) if (array[i] == find) print i }' *.sp
old, bash answer (O(n); opens 2 * n files)
From #mjgpy3 answer, you just have to make a for loop and use comm, like this:
#!/bin/bash
tmp1="/tmp/tmp1$RANDOM"
tmp2="/tmp/tmp2$RANDOM"
cp "$1" "$tmp1"
shift
for file in "$#"
do
comm -1 -2 "$tmp1" "$file" > "$tmp2"
mv "$tmp2" "$tmp1"
done
cat "$tmp1"
rm "$tmp1"
Save in a comm.sh, make it executable, and call
./comm.sh *.sp
assuming all your filenames end with .sp.
Updated answer, python, opens only each file once
Looking at the other answers, I wanted to give one that opens once each file without using any temporary file, and supports duplicated lines. Additionally, let's process the files in parallel.
Here you go (in python3):
#!/bin/env python
import argparse
import sys
import multiprocessing
import os
EOLS = {'native': os.linesep.encode('ascii'), 'unix': b'\n', 'windows': b'\r\n'}
def extract_set(filename):
with open(filename, 'rb') as f:
return set(line.rstrip(b'\r\n') for line in f)
def find_common_lines(filenames):
pool = multiprocessing.Pool()
line_sets = pool.map(extract_set, filenames)
return set.intersection(*line_sets)
if __name__ == '__main__':
# usage info and argument parsing
parser = argparse.ArgumentParser()
parser.add_argument("in_files", nargs='+',
help="find common lines in these files")
parser.add_argument('--out', type=argparse.FileType('wb'),
help="the output file (default stdout)")
parser.add_argument('--eol-style', choices=EOLS.keys(), default='native',
help="(default: native)")
args = parser.parse_args()
# actual stuff
common_lines = find_common_lines(args.in_files)
# write results to output
to_print = EOLS[args.eol_style].join(common_lines)
if args.out is None:
# find out stdout's encoding, utf-8 if absent
encoding = sys.stdout.encoding or 'utf-8'
sys.stdout.write(to_print.decode(encoding))
else:
args.out.write(to_print)
Save it into a find_common_lines.py, and call
python ./find_common_lines.py *.sp
More usage info with the --help option.
Combining this two answers (ans1 and ans2) I think you can get the result you are needing without sorting the files:
#!/bin/bash
ans="matching_lines"
for file1 in *
do
for file2 in *
do
if [ "$file1" != "$ans" ] && [ "$file2" != "$ans" ] && [ "$file1" != "$file2" ] ; then
echo "Comparing: $file1 $file2 ..." >> $ans
perl -ne 'print if ($seen{$_} .= #ARGV) =~ /10$/' $file1 $file2 >> $ans
fi
done
done
Simply save it, give it execution rights (chmod +x compareFiles.sh) and run it. It will take all the files present in the current working directory and will make an all-vs-all comparison leaving in the "matching_lines" file the result.
Things to be improved:
Skip directories
Avoid comparing all the files two times (file1 vs file2 and file2 vs file1).
Maybe add the line number next to the matching string
Hope this helps.
Best,
Alan Karpovsky
See this answer. I originally though a diff sounded like what you were asking for, but this answer seems much more appropriate.
I have a CSV file (foo.csv) with 200,000 rows. I need to break it into four files (foo1.csv, foo2.csv... etc.) with 50,000 rows each.
I already tried simple ctrl-v/-c using gui text editors, but the my computer slows to a halt.
What unix command(s) could I use to accomplish this task?
I don't have a terminal handy to try it out, but it should be just split -d -l 50000 foo.csv.
Hopefully the naming isn't terribly important because with the -d option, the output files will be named foo.csv00 .. foo.csv03. You can add the -a 1 option so that the suffixes are 0-3, but there's no simple way to get the suffix to be injected into the middle of the filename.
you should use head and tail.
head -n 50000 myfile > part1.csv
head -n 100000 myfile | tail -n 50000 > part2.csv
head -n 150000 myfile | tail -n 50000 > part3.csv
etc ...
Else, but with no control on file names, you can use unix command split.
sed -n 2000,4000p somefile.txt
will print from lines 2000 to 4000 to stdout.
split -l50000 foo.csv
You can use sed
I wrote this little shell script for this topic very similar at yours.
This shell script + awk works fine for me:
#!/bin/bash
awk -v initial_line=$1 -v end_line=$2 '{
if (NR >= initial_line && NR <= end_line)
print $0
}' $3
Used with this sample file (file.txt):
one
two
three
four
five
six
The command (it will extract from second to fourth line in the file):
edu#debian5:~$./script.sh 2 4 file.txt
Output of this command:
two
three
four
Of course, you can improve it, for example by testing that all argument values are the expected :-)