Given input such as:
1
1a
1.1b
2.0c
How to extract the integer/decimal number at beginning of each input line, using only Linux/Unix command line utilities?
Using awk, you could say:
awk '{print $0+0}'
Awk is available in Linux, BSD, and many other Unix-like operating systems. It helps in this way:
echo "1" | awk '{a+=$0; print a}' # output 1
echo "1a" | awk '{a+=$0; print a}' # output 1
echo "1.1b" | awk '{a+=$0; print a}' # output 1.1
echo "2.0c" | awk '{a+=$0; print a}' # output 2
Some more awk
For extracting only digits
$ awk 'gsub(/[[:alpha:]].*/,x,$1) + 1' << EOF
1
1a
1.1b
2.0c
EOF
1
1
1.1
2.0
For integer
$ awk '{print int($0)}' << EOF
1
1a
1.1b
2.0c
EOF
1
1
1
2
---edit---
If there is any blank line in file, you can avoid printing zero from following
$ awk 'NF{$0+=0}1' << EOF
1
1a
1.1b
2foot4c
2
EOF
1
1
1.1
2
2
Here is a way to do this with sed:
echo "12.3abc" | sed -n 's/^\([0-9.][0-9.]*\).*/\1/p'
Output:
12.3
The block in parentheses matches all numbers or periods '.' that occur at the beginning of the line. Everything after that is match by the '.*'.
The \1 says to replace the entire line with just the portion that was matched in the parentheses.
Assuming your version of grep supports -o:
grep -o '^[0-9.]\+' data.in
NB: This will match any sequence of digits and decimal points at the start of the line.
Related
How do I print the number of characters on lines x - y of a text file?
I tried using wc -m filename.txt
but I couldn't figure out how to limit the search.
You could use
head -y filename | tail -(y-x+1) | wc -m
You can use the sed command to select the lines you want and then pipe the output into wc. Something like this would select lines 6-10 and print the number of characters:
sed -n '6,10p' filename.txt | wc -m
Try this:
awk '{ print NR, "-", length($0)}' filename.txt
It will print the line number NR and the characters per line length($0) of filename.txt so output will be something like:
1 - 3 # line 1 with 3 characters
2 - 0 # line 2 with no characters
...
In case you just want to print the number of characters for a specific range, let's say from line 1 to 3, this could be used:
awk 'NR>=1 && NR<=3 { print length($0)}' filename.txt
I want to sum all the numbers in a file (columns and lines) given by the first parameter, but my program shows sum=sum+$i instead of the numeric sum:
sum=0;
file=$1
for i in $file
do
sum=sum+$i;
done;
echo "The sum is: " $sum
Input file:
$cat file.txt
10 20 10
40
50
Expected output :
The sum is: 21
Maybe if there is an awk method to solve this?
Try this -
$cat file1.txt
10 20 10
40
50
$awk '{for(i=1;i<=NF;i++) {sum+=$i}} END {print sum}' file1.txt
130
OR
$xargs < file1.txt| tr ' ' + | bc
130
cat file.txt | xargs | sed -e 's/\ /+/g' | bc
You can also use a simple read and an array to sum the value relying on word splitting to separate the values into an array via the default IFS (Internal Field Separator), e.g.
#!/bin/bash
declare -i sum=0
fn="${1:-/dev/stdin}" ## read from file as 1st argument (default stdin)
while read -r line; do ## read each line
a=( $line ) ## separate values into array
for i in ${a[#]}; do ## for each value in array
((sum += i)) ## add to sum
done
done <"$fn"
echo "sum: $sum"
Example Input File
$ cat dat/numfile.txt
10 20 10
40
50
Example Use/Output
$ bash sumnumfile.sh dat/numfile.txt
sum: 130
Another for some awks (at least mawk and gawk):
$ awk -v RS="[^0-9]" '{s+=$1}END{print s}' file
130
I need to count all lines of an unix file. The file has 3 lines but wc -l gives only 2 count.
I understand that it is not counting last line because it does not have end of line character
Could any one please tell me how to count that line as well ?
grep -c returns the number of matching lines. Just use an empty string "" as your matching expression:
$ echo -n $'a\nb\nc' > 2or3.txt
$ cat 2or3.txt | wc -l
2
$ grep -c "" 2or3.txt
3
It is better to have all lines ending with EOL \n in Unix files. You can do:
{ cat file; echo ''; } | wc -l
Or this awk:
awk 'END{print NR}' file
This approach will give the correct line count regardless of whether the last line in the file ends with a newline or not.
awk will make sure that, in its output, each line it prints ends with a new line character. Thus, to be sure each line ends in a newline before sending the line to wc, use:
awk '1' file | wc -l
Here, we use the trivial awk program that consists solely of the number 1. awk interprets this cryptic statement to mean "print the line" which it does, being assured that a trailing newline is present.
Examples
Let us create a file with three lines, each ending with a newline, and count the lines:
$ echo -n $'a\nb\nc\n' >file
$ awk '1' f | wc -l
3
The correct number is found.
Now, let's try again with the last new line missing:
$ echo -n $'a\nb\nc' >file
$ awk '1' f | wc -l
3
This still provides the right number. awk automatically corrects for a missing newline but leaves the file alone if the last newline is present.
Respect
I respect the answer from John1024 and would like to expand upon it.
Line Count function
I find myself comparing line counts A LOT especially from the clipboard, so I have defined a bash function. I'd like to modify it to show the filenames and when passed more than 1 file a total. However, it hasn't been important enough for me to do so far.
# semicolons used because this is a condensed to 1 line in my ~/.bash_profile
function wcl(){
if [[ -z "${1:-}" ]]; then
set -- /dev/stdin "$#";
fi;
for f in "$#"; do
awk 1 "$f" | wc -l;
done;
}
Counting lines without the function
# Line count of the file
$ cat file_with_newline | wc -l
3
# Line count of the file
$ cat file_without_newline | wc -l
2
# Line count of the file unchanged by cat
$ cat file_without_newline | cat | wc -l
2
# Line count of the file changed by awk
$ cat file_without_newline | awk 1 | wc -l
3
# Line count of the file changed by only the first call to awk
$ cat file_without_newline | awk 1 | awk 1 | awk 1 | wc -l
3
# Line count of the file unchanged by awk because it ends with a newline character
$ cat file_with_newline | awk 1 | awk 1 | awk 1 | wc -l
3
Counting characters (why you don't want to put a wrapper around wc)
# Character count of the file
$ cat file_with_newline | wc -c
6
# Character count of the file unchanged by awk because it ends with a newline character
$ cat file_with_newline | awk 1 | awk 1 | awk 1 | wc -c
6
# Character count of the file
$ cat file_without_newline | wc -c
5
# Character count of the file changed by awk
$ cat file_without_newline | awk 1 | wc -c
6
Counting lines with the function
# Line count function used on stdin
$ cat file_with_newline | wcl
3
# Line count function used on stdin
$ cat file_without_newline | wcl
3
# Line count function used on filenames passed as arguments
$ wcl file_without_newline file_with_newline
3
3
I want to find the last appearance of a string in a text file with linux commands. For example
1 a 1
2 a 2
3 a 3
1 b 1
2 b 2
3 b 3
1 c 1
2 c 2
3 c 3
In such a text file, i want to find the line number of the last appearance of b which is 6.
I can find the first appearance with
awk '/ b / {print NR;exit}' textFile.txt
but I have no idea how to do it for the last occurrence.
cat -n textfile.txt | grep " b " | tail -1 | cut -f 1
cat -n prints the file to STDOUT prepending line numbers.
grep greps out all lines containing "b" (you can use egrep for more advanced patterns or fgrep for faster grep of fixed strings)
tail -1 prints last line of those lines containing "b"
cut -f 1 prints first column, which is line # from cat -n
Or you can use Perl if you wish (It's very similar to what you'd do in awk, but frankly, I personally don't ever use awk if I have Perl handy - Perl supports 100% of what awk can do, by design, as 1-liners - YMMV):
perl -ne '{$n=$. if / b /} END {print "$n\n"}' textfile.txt
This can work:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
We check every second file being "b" and we record the number of line. It is appended, so by the time we finish reading the file, it will be the last one.
Test:
$ awk '{if ($2~"b") a=NR} END{print a}' your_file
6
Update based on sudo_O advise:
$ awk '{if ($2=="b") a=NR} END{print a}' your_file
to avoid having some abc in 2nd field.
It is also valid this one (shorter, I keep the one above because it is the one I thought :D):
$ awk '$2=="b" {a=NR} END{print a}' your_file
Another approach if $2 is always grouped (may be more efficient then waiting until the end):
awk 'NR==1||$2=="b",$2=="b"{next} {print NR-1; exit}' file
or
awk '$2=="b"{f=1} f==1 && $2!="b" {print NR-1; exit}' file
Acttualy this is my assignment.I have three-four file,related by student record.Every file have two-three student record.like this
Course Name:Opreating System
Credit: 4
123456 1 1 0 1 1 0 1 0 0 0 1 5 8 0 12 10 25
243567 0 1 1 0 1 1 0 1 0 0 0 7 9 12 15 17 15
Every file have different coursename.I did every coursename and studentid move
in one file but now i don't know how to add all marks and move to another file on same place where is id? Can you please tell me how to do it?
It looks like this:
Student# Operating Systems JAVA C++ Web Programming GPA
123456 76 63 50 82 67.75
243567 80 - 34 63 59
I did like this:
#!/bin/sh
find ~/2011/Fall/StudentsRecord -name "*.rec" | xargs grep -l 'CREDITS' | xargs cat > rsh1
echo "STUDENT ID" > rsh2
sed -n /COURSE/p rsh1 | sed 's/COURSE NAME: //g' >> rsh2
echo "GPA" >> rsh2
sed -e :a -e '{N; s/\n/ /g; ta}' rsh2 > rshf
sed '/COURSE/d;/CREDIT/d' rsh1 | sort -uk 1,1 | cut -d' ' -f1 | paste -d' ' >> rshf
Some comments and a few pointers :
It would help to add 'comments' for each line of code that is not self evident ; i.e. code like mv f f.bak doesn't need to be commented, but I'm not sure what the intent of your many lines of code are.
You insert a comment with the '#' char, like
# concatenate all files that contain the word CREDITS into a file called rsh1
find ~/2011/Fall/StudentsRecord -name "*.rec" | xargs grep -l 'CREDITS' | xargs cat > rsh1
Also note that you consistently use all uppercase for your search targets, i.e. CREDITS, when your sample files shows mixed case. Either used correct case for your search targets, i.e.
`grep -l 'Credits'`
OR tell grep to -i(gnore case), i.e.
`grep -il 'Credits'
Your line
sed -n /COURSE/p rsh1 | sed 's/COURSE NAME: //g' >> rsh2
can be reduced to 1 call to sed (and you have the same case confusion thing going on), try
sed -n '/COURSE/i{;s/COURSE NAME: //gip;}' rsh1 >> rsh2
This means (-n don't print every line by default),
`gip` = global substitute,
= ignore case in matching
print only lines where substituion was made
So you're editing out the string COURSE NAME for any line that has COURSE in it, and only printing those lines' (you probably don't need the 'g' (global) specifier given that you expect only 1 instance per line)
Your line
sed -e :a -e '{N; s/\n/ /g; ta}' rsh2 > rshf
Actually looks pretty good, very advanced, you're trying to 'fold' each 2 lines together into 1 line, right?
But,
sed '/COURSE/d;/CREDIT/d' rsh1 | sort -uk 1,1 | cut -d' ' -f1 | paste -d' ' >> rshf
I'm really confused by this, is this where you're trying to total a students score? (with a sort embedded I guess not). Why do you think you need a sort,
While it is possible to perform arithmetic in sed, it is super-crazy hard, so you can either use bash variables to calculate the values OR use a unix tool that is designed to process text AND perform logical and mathematical operations of the data presented, awk or perl come to mind here
Anyway, one solution to total each score is to use awk
echo "123456 1 1 0 1 1 0 1 0 0 0 1 5 8 0 12 10 25" |\
awk '{for (i=2;i<=NF;i++) { tot+=$i }; print $1 "\t" tot }'
Will give you a clue on how to proceed for that.
Awk has predefined variables that it populates for each file, and each line of text that it reads, i.e.
$0 = complete line of text (as defined by the internal variables RS (RecordSeparator)
which defaults to '\n' new-line char, the unix end-of-line char
$1 = first field in text (as defined by the internal variables FS (FieldSeparator)
which defaults to (possibly multiple) space chars OR tab char
a line with 2 connected spaces chars and 1 tab char has 3 fields)
NF = Number(of)Fields in current line of data (again fields defined by value of FS as
described above)
(there are many others, besides, $0, $n, $NF, $FS, $RS).
you can programatically increment for values like $1, $2, $3, by using a variable as in the example code, like $i (i is a variable that has a number between 2 and NF. The leading '$'
says give me the value of field i (i.e. $2, $3, $4 ...)
Incidentally, your problem could be easily solved with a single awk script, but apparently, you're supposed to learn about cat, cut, grep, etc, which is a very worthwhile goal.
I hope this helps.