I have a list of numbers in a file
cat to_delete.txt
2
3
6
9
11
and many txt files in one folder. Each file has tab delimited lines (can be more lines than this).
3 0.55667 0.66778 0.54321 0.12345
6 0.99999 0.44444 0.55555 0.66666
7 0.33333 0.34567 0.56789 0.34543
I want to remove the lines that the first number ($1 for awk) is in to_delete.txt and print only the lines that the first number is not in to_delete.txt. The change should be replacing the old file.
Expected output
7 0.33333 0.34567 0.56789 0.34543
This is what I got so far, which doesn't remove anything;
for file in *.txt; do awk '$1 != /2|3|6|9|11/' "$file" > "$tmp" && mv "$tmp" "$file"; done
I've looked through so many similar questions here but still cannot make it work. I also tried grep -v -f to_delete.txt and sed -n -i '/$to_delete/!p'
Any help is appreciated. Thanks!
In awk:
$ awk 'NR==FNR{a[$1];next}!($1 in a)' delete file
Output:
7 0.33333 0.34567 0.56789 0.34543
Explained:
$ awk '
NR==FNR { # hash records in delete file to a hash
a[$1]
next
}
!($1 in a) # if $1 not found in record in files after the first, output
' delete files* # mind the file order
My first idea was this:
printf "%s\n" *.txt | xargs -n1 sed -i "$(sed 's!.*!/& /d!' to_delete.txt)"
printf "%s\n" *.txt - outputs the *.txt files each on separate lines
| xargs -n1 execute the following command for each line passing the line content as the input
sed -i - edit file in place
$( ... ) - command substitution
sed 's!.*!/^& /d!' to_delete.txt - for each line in to_delete.txt, append the line with /^ and suffix with /d. That way from the list of numbers I get a list of regexes to delete, like:
/^2 /d
/^3 /d
/^6 /d
and so on. Which tells sed to delete lines matching the regex - line starting with the number followed by a space.
But I think awk would be simpler. You could do:
awk '$1 != 2 && $1 != 3 && $1 != 6 ... and so on ...`
but that would be longish, unreadable. It's easier to read the map from the file and then check if the number is in the array:
awk 'FNR==NR{ map[$1] } FNR!=NR && !($1 in map)' to_delete.txt "$file"
The FNR==NR is true only for the first file. So when we read it, we set the map[$1] (we "set" it, just so such element exists). Then FNR!=NR is true for the second file, for which we check if the first element is the key in the map. If it is not, the expression is true and the line gets printed out.
all together:
for file in *.txt; do awk 'FNR==NR{ map[$1] } FNR!=NR && !($1 in map)' to_delete.txt "$file" > "$tmp"; mv "$tmp" "$file"; done
Related
I am here wondering that if I can read each line of a.txt and compare it to each line in b.txt. If any line in a.txt matches the beginning part of the line in b.txt, we replace the matched line with the line we found in a.txt. So let's say there are two lines: alias cd /correct/path/ and alias cd /wrong/path/sth in a.txt b.txt respectively. Now after I execute my command I would like the lines to be all like: alias cd /correct/path/ on both files. My own solution is to do two while...read.. functions and use sed -i /// to replace the line, but I think it is very clumsy and not efficient. I am looking to be enlightened with a more clean & efficient solution. Here is my code if it helps by any chance:
awk 'NR==FNR { array[$0]; next } { delete array[$0] } END{for (key in array) { print key } }' a.txt b.txt > tmp
input="tmp"
while IFS= read -r line
do
echo "$line"
cat b.txt > n_tmp
n_input="$n_tmp"
while IFS= read -r n_line
do
if $n_line | awk '{print $1, $2}' == $line | awk '{print $1, $2}'; then
sed -i "s/$n_line/$line/" b.txt
fi
done < "$n_input"
rm -rf n_tmp
done < "$input"
rm -rf tmp```
There are a few mistakes in this script and most of them are within the line: if $n_line | awk '{print $1, $2}' == $line | awk '{print $1, $2}'; then. First of all the way to get result from $n_line | awk '{print $1, $2}' is wrong as there is no action for n_line variable. There needs to be added an echo so that we can get the output of the string and the awk command can follow up. Secondly there is no double quotes for strings or whatever I was trying to get from the $n_line | awk '{print $1, $2}' command. Lastly, there is a double bracket needed to wrap around the two sides of the comparator. So in the end it should look something like this:
b_string=`echo "$n_line" | awk '{print $1, $2}'`
if [[ "$a_string" == "$b_string" ]]; then
I figured to declare the echoing part into a variable as well, it may look a bit cleaner and easier to handle. There are still some other problems with this script, but as of now I think the primary issue is solved.
I have file.txt exemplary here:
This line contains ABC
This line contains DEF
This line contains GHI
and here the following list.txt:
contains ABC<TAB>ABC
contains DEF<TAB>DEF
Now I am writing a script that executes the following commands for each line of this external file list.txt:
take the string from column 1 of list.txt and search in a third file file.txt
if the first command is positive, return the string from column 2 of list.txt
So my output.txt is:
ABC
DEF
This is my code for grep/echo with putting the query/return strings manually:
if grep -i -q 'contains abc' file.txt
then
echo ABC >output.txt
else
echo -n
fi
if grep -i -q 'contains def' file.txt
then
echo DEF >>output.txt
else
echo -n
fi
I have about 100 search terms, which makes the task laborious if done manually. So how do I include while read line; do [commands]; done<list.txt together with the commands about column1 and column2 inside that script?
I would like to use simple grep/echo/awkcommands if possible.
Something like this?
$ awk -F'\t' 'FNR==NR { a[$1] = $2; next } {for (x in a) if (index($0, x)) {print a[x]}} ' list.txt file.txt
ABC
DEF
For the lines of the first file (FNR==NR), read the key-value pairs to array a. Then for the lines of the second line, loop through the array, check if the key is found on the line, and if so, print the stored value. index($0, x) tries to find the contents of x from (the current line) $0. $0 ~ x would instead take x as a regex to match with.
If you want to do it in the shell, starting a separate grep for each and every line of list.txt, something like this:
while IFS=$'\t' read k v ; do
grep -qFe "$k" file.txt && echo "$v"
done < list.txt
read k v reads a line of input and splits it (based on IFS) into k and v.
grep -F takes the pattern as a fixed string, not a regex, and -q prevents it from outputting the matching line. grep returns true if any matching lines are found, so $v is printed if $k is found in file.txt.
Using awk and grep:
for text in `awk '{print $4}' file.txt `
do
grep "contains $text" list.txt |awk -F $'\t' '{print $2}'
done
I have shell script variable var="7,8,9"
These are the line number use to delete to file using sed.
Here I tried:
sed -i "$var"'d' test_file.txt
But i got error `sed: -e expression #1, char 4: unknown command: ,'
Is there any other way to remove the line?
sed command doesn't accept comma delimited line numbers.
You can use this awk command that uses a bit if BASH string manipulation to form a regex with the given comma separated line numbers:
awk -v var="^(${var//,/|})$" 'NR !~ var' test_file.txt
This will set awk variable var as this regex:
^(7|8|9)$
And then condition NR !~ var ensures that we print only those lines that don't match above regex.
For inline editing, if you gnu-awk with version > 4.0 then use:
awk -i inplace -v var="^(${var//,/|})$" 'NR !~ var' test_file.txt
Or for older awk use:
awk -v var="^(${var//,/|})$" 'NR !~ var' test_file.txt > $$.tmp && mv $$.tmp test_file.txt
I like sed, you were close to it. You just need to split each line number into a separate command. How about this:
sed -e "$(echo 1,3,4 | tr ',' '\n' | while read N; do printf '%dd;' $N; done)"
do like this:
sed -i "`echo $var|sed 's/,/d;/g'`d;" file
Another option to consider would be ed, with printf '%s\n' to put commands onto separate lines:
lines=( 9 8 7 )
printf '%s\n' "${lines[#]/%/d}" w | ed -s file
The array lines contains the line numbers to be deleted; it's important to put these in descending order! The expansion ${lines[#]/%/d} adds a d (delete) command to each line number and w writes to the file at the end. You can change this to ,p instead, to check the output before overwriting your file.
As an aside, for this example, you could also just use 7,9 as a single entry in the array.
I want a unix command to find the lines between first & last occurence of a word
For example:
let's imagine we have 1000 lines. Tenth line contains word "stackoverflow", thirty fifth line also contains word "stackoverflow".
I want to print lines between 10 and 35 and write it to a new file.
You can make it in two steps. The basic idea is to:
1) get the line number of the first and last match.
2) print the range of lines in between these range.
$ read first last <<< $(grep -n stackoverflow your_file | awk -F: 'NR==1 {printf "%d ", $1}; END{print $1}')
$ awk -v f=$first -v l=$last 'NR>=f && NR<=l' your_file
Explanation
read first last reads two values and stores them in $first and $last.
grep -n stackoverflow your_file greps and shows the output like this: number_of_line:output
awk -F: 'NR==1 {printf "%d ", $1}; END{print $1}') prints the number of line of the first and last match of stackoverflow in the file.
And
awk -v f=$first -v l=$last 'NR>=f && NR<=l' your_file prints all lines from $first line number till $last line number.
Test
$ cat a
here we
have some text
stackoverflow
and other things
bla
bla
bla bla
stackoverflow
and whatever else
stackoverflow
to make more fun
blablabla
$ read first last <<< $(grep -n stackoverflow a | awk -F: 'NR==1 {printf "%d ", $1}; END{print $1}')
$ awk -v f=$first -v l=$last 'NR>=f && NR<=l' a
stackoverflow
and other things
bla
bla
bla bla
stackoverflow
and whatever else
stackoverflow
By steps:
$ grep -n stackoverflow a
3:stackoverflow
9:stackoverflow
11:stackoverflow
$ grep -n stackoverflow a | awk -F: 'NR==1 {printf "%d ", $1}; END{print $1}'
3 11
$ read first last <<< $(grep -n stackoverflow a | awk -F: 'NR==1 {printf "%d ", $1}; END{print $1}')
$ echo "first=$first, last=$last"
first=3, last=11
If you know an upper bound of how many lines there can be (say, a million), then you can use this simple abusive script:
(grep -A 100000 stackoverflow | grep -B 1000000 stackoverflow) < file
You can append | tail -n +2 | head -n -1 to strip the border lines as well:
(grep -A 100000 stackoverflow | grep -B 1000000 stackoverflow
| tail -n +2 | head -n -1) < file
I'm not 100% sure from the question whether the output should be inclusive of the first and last matching lines, so I'm assuming it is. But this can be easily changed if we want exclusive instead.
This pure-bash solution does it all in one step - i.e. the file (or pipe) is only read once:
#!/bin/bash
function midgrep {
while read ln; do
[ "$saveline" ] && linea[$((i++))]=$ln
if [[ $ln =~ $1 ]]; then
if [ "$saveline" ]; then
for ((j=0; j<i; j++)); do echo ${linea[$j]}; done
i=0
else
saveline=1
linea[$((i++))]=$ln
fi
fi
done
}
midgrep "$1"
Save this as a script (e.g. midgrep.sh) and pipe whatever output you like to it as follows:
$ cat input.txt | ./midgrep.sh stackoverflow
This works as follows:
find the first matching line and buffer in the first element of an array
continue reading lines until the next match, buffering to the array as we go
on each subsequent matches, flush the buffer array to output
continue reading file to the end. If there are no more matches, then the last buffer is simply discarded.
The advantage of this approach is that we only read through the input one time only. The disadvantage is that we buffer everything between each match - if there are many lines between each match, then these are all buffered to memory, until we hit the next match.
Also this uses the bash =~ regular expression operator to keep this pure bash. But you could replace this with a grep instead, if you are more comfortable with that.
Using perl :
perl -00 -lne '
chomp(my #arr = split /stackoverflow/);
print join "\nstackoverflow", #arr[1 .. $#arr -1 ]
' file.txt | tee newfile.txt
The idea behind this is to feed an array of the whole input file in to chunks using "stackoverflow" string to split. Next, we print the 2nd occurrences to the last -1 with join "stackoverflow".
How can I delete the first (!) line of a text file if it's empty, using e.g. sed or other standard UNIX tools. I tried this command:
sed '/^$/d' < somefile
But this will delete the first empty line, not the first line of the file, if it's empty. Can I give sed some condition, concerning the line number?
With Levon's answer I built this small script based on awk:
#!/bin/bash
for FILE in $(find some_directory -name "*.csv")
do
echo Processing ${FILE}
awk '{if (NR==1 && NF==0) next};1' < ${FILE} > ${FILE}.killfirstline
mv ${FILE}.killfirstline ${FILE}
done
The simplest thing in sed is:
sed '1{/^$/d}'
Note that this does not delete a line that contains all blanks, but only a line that contains nothing but a single newline. To get rid of blanks:
sed '1{/^ *$/d}'
and to eliminate all whitespace:
sed '1{/^[[:space:]]*$/d}'
Note that some versions of sed require a terminator inside the block, so you might need to add a semi-colon. eg sed '1{/^$/d;}'
Using sed, try this:
sed -e '2,$b' -e '/^$/d' < somefile
or to make the change in place:
sed -i~ -e '2,$b' -e '/^$/d' somefile
If you don't have to do this in-place, you can use awk and redirect the output into a different file.
awk '{if (NR==1 && NF==0) next};1' somefile
This will print the contents of the file except if it's the first line (NR == 1) and it doesn't contain any data (NF == 0).
NR the current line number,NF the number of fields on a given line separated by blanks/tabs
E.g.,
$ cat -n data.txt
1
2 this is some text
3 and here
4 too
5
6 blank above
7 the end
$ awk '{if (NR==1 && NF==0) next};1' data.txt | cat -n
1 this is some text
2 and here
3 too
4
5 blank above
6 the end
and
cat -n data2.txt
1 this is some text
2 and here
3 too
4
5 blank above
6 the end
$ awk '{if (NR==1 && NF==0) next};1' data2.txt | cat -n
1 this is some text
2 and here
3 too
4
5 blank above
6 the end
Update:
This sed solution should also work for in-place replacement:
sed -i.bak '1{/^$/d}' somefile
The original file will be saved with a .bak extension
Delete the first line of all files under the actual directory if the first line is empty :
find -type f | xargs sed -i -e '2,$b' -e '/^$/d'
This might work for you:
sed '1!b;/^$/d' file