I have created a simple script that prints the contents of a text file using cat command. Now I want to print a line along with its number, but at the same time I need to ignore blank lines. The following format is desired:
1 George Jones Berlin 2564536877
2 Mike Dixon Paris 2794321976
I tried using
cat -n catalog.txt | grep -v '^$' catalog.txt
But I get the following results:
George Jones Berlin 2564536877
Mike Dixon Paris 2794321976
I have managed to get rid of the blank lines, but line's number is not printed. What am I doing wrong?
Here are the contents of catalog.txt:
George Jones Berlin 2564536877
Mike Dixon Paris 2794321976
Your solution doesn't work because cat -n catalog.txt is already giving you non-blank lines.
You can pipe grep's output to cat -n:
grep -v '^$' yourFile | cat -n
Example:
test.txt:
Hello
how
are
you
?
$ grep -v '^$' test | cat -n
1 Hello
2 how
3 are
4 you
5 ?
At first glance, you should drop the file name in the command line to grep to make grep read from stdin:
cat -n catalog.txt | grep -v '^$'
^^^
In your code, you supplied catalog.txt to grep, which made it read from the file and ignore its standard input. So you're basically grepping from the file instead of the output of cat piped to its stdin.
To correctly ignore blank lines the prepend line numbers, switch the order of grep and cat:
grep -v '^$' catalog.txt | cat -n
Another awk
$ awk 'NF{$0=FNR " " $0}NF' 48488182
1 George Jones Berlin 2564536877
3 Mike Dixon Paris 2794321976
The second line was blank in this case.
single, simple, basic awk solution could help you here.
Solution 1st:
awk 'NF{print FNR,$0}' Input_file
Solution 2nd: Above will print line number including the line number of NULL lines, in case you want to leave empty lines line number then following may help you in same.
awk '!NF{FNR--;next} NF{print FNR,$0}' Input_file
Solution 3rd: Using only grep, though output will have a colon in between line number and the line.
grep -v '^$' Input_file | grep -n '.*'
Explanation of Solution 1st:
NF: Checking condition here if NF(Number of fields in current line, it is awk's out of the box variable which has the value of number of fields in a line) is NOT NULL, if this condition is TRUE then following the actions mentioned next to it.
{print FNR,$0}: Using print function of awk here to print FNR(Line number, which will have the line's number in it, it is awk's out of box variable) then print $0 which means current line.
By this we satisfy OP's both the conditions of leaving empty lines and print the line numbers along with lines too. I hope this helps you.
Related
I am trying to number the lines of a txt file and hide the empty ones . I use this code :
cat -n file.txt | grep . file.txt
But it doesnt work . It ignores the cat command . I want to display all the non-empty lines and number them ( the txt file is not a static one , like a list that a user can type in ).
edit : Given the great solutions below , i would also add that grep . file.txt | cat -n also worked .
I assume you want to number the lines that remain after the empty lines are removed.
Solution #1
Use sed '/^$/d' to delete the empty lines then pipe its output to cat -n to number them:
sed '/^$/d' file.txt | cat -n
The sed program contains only one command: d (delete the line). The sed commands can be prefixed by zero, one or two addresses that tell what lines the command applies to.
In this case there is only one address /^$/. It is a regex (enclosed in /) that selects the empty lines; the lines where start of the line (^) is followed by the end of the line ($).
Solution #2
You can also use grep -v '^$' to filter out the empty lines:
grep -v '^$' file.txt | cat -n
Again, ^$ is a regular expression that matches the empty lines. -v reverses the condition and tells grep to display the lines that do not match the regex.
The commands above do not modify the file. They read the content of file.txt, process it and display the result on screen.
Update
As #robc suggests in a comment, nl is even better than cat -n to number the lines. Thank you #robc, I didn't know about nl until now (I didn't know about cat -n either). It is never too late to learn new things.
This could be easily done with awk. This will print line with line numbers and ignore empty lines.
awk 'NF{print FNR,$0}' file.txt
Explanation: Adding detailed explanation for above code.
awk ' ##Starting awk program from here.
NF{ ##Checking condition if NF(number of fields) is NOT NULL in current line then do following.
print FNR,$0 ##Printing current line number by mentioning FNR and then current line value.
}
' file.txt ##Mentioning Input_file name which we are passing to awk program here.
I have a file with 10,000 lines. Using the following command, I am deleting all lines after line 10,000.
sed -i '10000,$ d' file.txt
However, now I would like to delete the first X lines so that the file has no more than 10,000 lines.
I think it would be something like this:
sed -i '1,$x d' file.txt
Where $x would be the number of lines over 10,000. I'm a little stuck on how to write the if, then part of it. Or, I was thinking I could use the original command and just cat the file in reverse?
For example, if we wanted just 3 lines from the bottom (seems simpler after a few helpful answers):
Input:
Example Line 1
Example Line 2
Example Line 3
Example Line 4
Example Line 5
Expected Output:
Example Line 3
Example Line 4
Example Line 5
Of course, if you know a more efficient way to write the command, I would be open to that too. Your positive input is highly appreciated.
tail can do exactly what you want.
tail -n 10000 file.txt
For simplicity, I would reverse the file, keep the first 10000 lines, then re-reverse the file.
It makes saving the file in-place a touch more complicated
source=file.txt
temp=$(mktemp)
tac "$source" | sed '10000 q' | tac > "$temp" && mv "$temp" "$source"
Without reversing the file, you'd count the number of lines and do some arithmetic:
sed -i "1,$(( $(wc -l < file.txt) - 10000 )) d" file.txt
$ awk -v n=3 '{a[NR%n]=$0} END{for (i=NR+1;i<=(NR+n);i++) print a[i%n]}' file
Example Line 3
Example Line 4
Example Line 5
Add -i inplace if you have GNU awk and want to do "inplace" editing.
To keep the first 10000 lines :
head -n 10000 file.txt
To keep the last 10000 lines :
tail -n 10000 file.txt
Test with your file Example
tail -n 3 file.txt
Example Line 3
Example Line 4
Example Line 5
tac file.txt | sed "$x q" | tac | sponge file.txt
The sponge command is useful here in avoiding an additional temporary file.
tail -10000 <<<"$(cat file.txt)" > file.txt
Okay, not «just» tail, but this way it`s capable of inplace truncation.
I understand that grep can extract the specific content from a file line by line.
Just wondering how can add another column before or after each line as an index.
For example:
grep "aaa" text.txt > tmp.txt
In the tmp.txt file, we can see the content as follows,
aaawekjre
qejrteraaa
wrgeaaaere
However, I would like to add a specific index as an extra column.
Therefore, the tmp.txt might look like this:
John aaawekjre
John qejrteraaa
John wrgeaaaere
You can use awk:
awk '/aaa/{print "John", $0}' text.txt > tmp.txt
$ sed -n '/aaa/ s/^/John /p' text.txt
John aaawekjre
John qejrteraaa
John wrgeaaaere
How it works
-n
This tells sed not to print anything unless we explicitly ask it to.
/aaa/ s/^/John /p
This selects lines that contain aaa. For those lines, we do a substitution (s/^/John /) to put John at the beginning of the line and we print the line (p).
In this way, lines that do not contain aaa are never printed. Thus, there is no need for a separate grep process.
try this
grep "aaa" text.txt | awk '{print "John " $0}' > tmp.txt
Hello I have a problem in bash.
i have a file and i am trying insert a point in the final line of each line:
cat file | sed s/"\n"/\\n./g > salida.csv
but not works =(.
Because i need count the lines with a word
I need count the lines with the same country
and if i do a grep the grep take colombia and colombias.
And other question how i can count lines with the same country?
for example
1 colombia
2 brazil
3 ecuador
4 colombias
5 colombia
colombia 2
colombias 1
ecuador 1
brazil 1
how about
cut -f2 -d' ' salida.csv | sort | uniq -c
since a sed solution was posted (probably the best tool for this task), I'll contribute an awk
awk '$NF=$NF"."' file > salida.csv
Update:
$ cat input
1 colombia
2 brazil
3 ecuador
4 colombias
5 colombia
$ awk '{a[$2]++}END{for (i in a) print i, a[i]}' input
brazil 1
colombias 1
ecuador 1
colombia 2
...and, please stop updating your question with different questions...
Your command line has a few problems. Some that matter, some that are style choices, but here's my take:
Unnecessary cat. sed can take a filename as an argument.
Your sed command doesn't need the g. Since each line only has one end, there's no reason to tell it to look for more.
Don't look for the newline character, just match the end of line with $.
That leaves you with:
sed s/$/./ file > salida.csv
Edit:
If your real question is "How do I grep for colombia, but not match colombias?", you just need to use the -w flag to match whole words:
grep -w colombia file
If you want to count them, just add -c:
grep -c -w colombia file
Read the grep(1) man page for more information.
In a very large file I need to find the position (line number) of a string, then extract the 2 lines above and below that string.
To do this right now - I launch vi, find the string, note it's line number, exit vi, then use sed to extract the lines surrounding that string.
Is there a way to streamline this process... ideally without having to run vi at all.
Maybe using grep like this:
grep -n -2 your_searched_for_string your_large_text_file
Will give you almost what you expect
-n : tells grep to print the line number
-2 : print 2 additional lines (and the wanted string, of course)
You can do
grep -C 2 yourSearch yourFile
To send it in a file, do
grep -C 2 yourSearch yourFile > result.txt
Use grep -n string file to find the line number without opening the file.
you can use cat -n to display the line numbers and then use awk to get the line number after a grep in order to extract line number:
cat -n FILE | grep WORD | awk '{print $1;}'
although grep already does what you mention if you give -C 2 (above/below 2 lines):
grep -C 2 WORD FILE
You can do it with grep -A and -B options, like this:
grep -B 2 -A 2 "searchstring" | sed 3d
grep will find the line and show two lines of context before and after, later remove the third one with sed.
If you want to automate this, simple you can do a Shell Script. You may try the following:
#!/bin/bash
VAL="your_search_keyword"
NUM1=`grep -n "$VAL" file.txt | cut -f1 -d ':'`
echo $NUM1 #show the line number of the matched keyword
MYNUMUP=$["NUM1"-1] #get above keyword
MYNUMDOWN=$["NUM1"+1] #get below keyword
sed -n "$MYNUMUP"p file.txt #display above keyword
sed -n "$MYNUMDOWN"p file.txt #display below keyword
The plus point of the script is you can change the keyword in VAL variable as you like and execute to get the needed output.