I want to replace the first value (in first column and line so here 1) and add one to this value, so I have a file like this
1
1 1
2 5
1 6
I use this sentence
read -r a < file
echo $aa
sed "s/$aa/$(($aa + 1))/" file
# or
sed 's/$aa/$(($aa + 1))/' file
But when I make that, he change all first column one into two. I have try to change the quote but it make nothing.
restrict the script to first line only, i.e.
sed '1s/old/new/'
awk might be a better tool for this.
awk 'NR==1{$1=$1+1}1'
for the first line add 1 to the first field and print. Can be rewritten as
awk 'NR==1{$1+=1}1'
or
awk 'NR==1{$1++}1'
perl -p0e 's/(\d+)/$1+1/e' file
Related
I am new to using the Mac terminal. I need to add a tab delimited column to a text file with 3 existing columns. The columns look pretty much like this:
org1 1-20 1-40
org2 3-35 6-68
org3 16-38 40-16
etc.
I need them to look like this:
org1 1-20 1-40 1
org2 3-35 6-68 2
org3 16-38 40-16 3
etc.
My apologies if this question has been covered. Answers to similar questions are sometimes exceedingly esoteric and are not easily translatable to this specific situation.
In awk. print the record and the required tab and row count after it:
$ awk '{print $0 "\t" NR }' foo
org1 1-20 1-40 1
org2 3-35 6-68 2
org3 16-38 40-16 3
If you want to add the line numbers to the last column:
perl -i -npe 's/$/"\t$."/e' file
where
-i replaces the file in-pace (remove, if you want to print the result to the standard output);
-n causes Perl to apply the substitution to each line from the file, just like sed;
-p prints the result of expression;
-e accepts Perl expression;
s/.../.../e substitutes the first part to the second (delimited with slash), and the e flag causes Perl to evaluate the replacement as Perl expression;
$ is the end-of-line anchor;
$. variable keeps the number of the current line
In other words, the command replaces the end of the line ($) with a tab followed by the line number $..
You can paste the file next to the same file with line numbers prepended (nl), and all the other columns removed (cut -f 1):
$ paste infile <(nl infile | cut -f 1)
org1 1-20 1-40 1
org2 3-35 6-68 2
org3 16-38 40-16 3
The <(...) construct is called process substitution and basically allows you to treat the output of a command like a file.
I have a consistent file with numbers like
0123456
0234566
.
.
.
etc
With bash tools, command line preferable, how can I remove each line if the third digit equals 2 .
eg, with cut -c3 I can get the correct digit but I cannot combine it effectively with sed or something similar. I am not looking for a pattern, only the 3rd digit.
(I have done it in a script in python but I was wondering how its done through a one-line bash command). Thank you!
EDIT: Additionally, if I want to delete the lines where the third digit NOT equals to 2 (opposite question)
You can just do this with sed
sed -i '/^..2/d' file
If you want to do the opposite you can do:
sed -i '/^..[^2]/d' file
since you are dealing with a specific character.
I would use awk:
$ awk -F "" '$3!=2' file
0234566
by setting the field separator to "" (empty, just valid on GNU-awk), every character is stored in a different field. Then, saying $3 != 2 checks if the 3rd character is not 2 and, if so, the line is printed.
Or with pure bash, using Using shell parameter expansion ${parameter:offset:length}:
while IFS= read -r line
do
[ "${line:2:1}" != "2" ] && echo "$line"
done < file
I have a file contains some lines. Now I want to read the lines and get the line numbers. As below:
while read line
do
string=$line
number=`awk '{print NR}'` # This way is not right, gets all the line numbers.
done
Here is my scenario: I have one file, contains some lines, such as below:
2015Y7M3D0H0Mi44S7941
2015Y7M3D22H24Mi3S7927
2015Y7M3D21H28Mi21S5001
I want to read each line of this file, print out the last characters starts with "S" and the line number of it. it shoud looks like:
1 S7941
2 S7927
3 S5001
So, what should I properly do to get this?
Thanks.
Can anyone help me out ???
The UNIX shell is simply an environment from which to call tools and a language to sequence those calls. The UNIX general purpose text processing tool is awk so just use it:
$ awk '{sub(/.*S/,NR" S")}1' file
1 S7941
2 S7927
3 S5001
If you're going to be doing any text manipulation, get the book Effective Awk Programming, 4th Edition, by Arnold Robbins.
I just asked one of my friend, Found a simple way:
cat -n $file |while read line
do
number=echo $line | cut -d " " -f 1
echo $number
done
That means if we can not get line number from the file itself, we pass it with a line number.
I have these two lines within a file:
<first-value system-property="unique.setting.limit">3</first-value>
<second-value-limit>50000</second-value-limit>
where I'd like to get the following as output using awk or sed:
3
50000
Using this sed command does not work as I had hoped, and I suspect this is due to the presence of the quotes and delimiters in my line entry.
sed -n '/WORD1/,/WORD2/p' /path/to/file
How can I extract the values I want from the file?
awk -F'[<>]' '{print $3}' input.txt
input.txt:
<first-value system-property="unique.setting.limit">3</first-value>
<second-value-limit>50000</second-value-limit>
Output:
3
50000
sed -e 's/[a-zA-Z.<\/>= \-]//g' file
Using sed:
sed -E 's/.*limit"*>([0-9]+)<.*/\1/' file
Explanation:
.* takes care of everything that comes before the string limit
limit"* takes care of both the lines, one with limit" and the other one with just limit
([0-9]+) takes care of matching numbers and only numbers as stated in your requirement.
\1 is actually a shortcut for capturing pattern. When a pattern groups all or part of its content into a pair of parentheses, it captures that content and stores it temporarily in memory. For more details, please refer https://www.inkling.com/read/introducing-regular-expressions-michael-fitzgerald-1st/chapter-4/capturing-groups-and
The script solution with parameter expansion:
#!/bin/bash
while read line || test -n "$line" ; do
value="${line%<*}"
printf "%s\n" "${value##*\>}"
done <"$1"
output:
$ ./ltags.sh dat/ltags.txt
3
50000
Looks like XML to me, so assuming it forms part of some valid XML, e.g.
<root>
<first-value system-property="unique.setting.limit">3</first-value>
<second-value-limit>50000</second-value-limit>
</root>
You can use Perl's XML::Simple and do something like this:
perl -MXML::Simple -E '$xml = XMLin("file"); say $xml->{"first-value"}->{"content"}; say $xml->{"second-value-limit"}'
Output:
3
50000
If the XML structure is more complicated, then you may have to drill down a bit deeper to get to the values you want. If that's the case, you should edit the question to show the bigger picture.
Ashkan's awk solution is straightforward, but let me suggest a sed solution that accepts non-integer numbers:
sed -n 's/[^>]*>\([.[:digit:]]*\)<.*/\1/p' input.txt
This extracts the number between the first > character of the line and the following <. In my RE this "number" can be the empty string, if you don't want to accept an empty string please add the -r option to sed and replace \([.[:digit:]]*\) by ([.[:digit:]]+).
This question already has answers here:
How to get the part of a file after the first line that matches a regular expression
(12 answers)
Closed 7 years ago.
I have a file that contains a list of URLs. It looks like below:
file1:
http://www.google.com
http://www.bing.com
http://www.yahoo.com
http://www.baidu.com
http://www.yandex.com
....
I want to get all the records after: http://www.yahoo.com, results looks like below:
file2:
http://www.baidu.com
http://www.yandex.com
....
I know that I could use grep to find the line number of where yahoo.com lies using
grep -n 'http://www.yahoo.com' file1
3 http://www.yahoo.com
But I don't know how to get the file after line number 3. Also, I know there is a flag in grep -A print the lines after your match. However, you need to specify how many lines you want after the match. I am wondering is there something to get around that issue. Like:
Pseudocode:
grep -n 'http://www.yahoo.com' -A all file1 > file2
I know we could use the line number I got and wc -l to get the number of lines after yahoo.com, however... it feels pretty lame.
AWK
If you don't mind using AWK:
awk '/yahoo/{y=1;next}y' data.txt
This script has two parts:
/yahoo/ { y = 1; next }
y
The first part states that if we encounter a line with yahoo, we set the variable y=1, and then skip that line (the next command will jump to the next line, thus skip any further processing on the current line). Without the next command, the line yahoo will be printed.
The second part is a short hand for:
y != 0 { print }
Which means, for each line, if variable y is non-zero, we print that line. In AWK, if you refer to a variable, that variable will be created and is either zero or empty string, depending on context. Before encounter yahoo, variable y is 0, so the script does not print anything. After encounter yahoo, y is 1, so every line after that will be printed.
Sed
Or, using sed, the following will delete everything up to and including the line with yahoo:
sed '1,/yahoo/d' data.txt
This is much easier done with sed than grep. sed can apply any of its one-letter commands to an inclusive range of lines; the general syntax for this is
START , STOP COMMAND
except without any spaces. START and STOP can each be a number (meaning "line number N", starting from 1); a dollar sign (meaning "the end of the file"), or a regexp enclosed in slashes, meaning "the first line that matches this regexp". (The exact rules are slightly more complicated; the GNU sed manual has more detail.)
So, you can do what you want like so:
sed -n -e '/http:\/\/www\.yahoo\.com/,$p' file1 > file2
The -n means "don't print anything unless specifically told to", and the -e directive means "from the first appearance of a line that matches the regexp /http:\/\/www\.yahoo\.com/ to the end of the file, print."
This will include the line with http://www.yahoo.com/ on it in the output. If you want everything after that point but not that line itself, the easiest way to do that is to invert the operation:
sed -e '1,/http:\/\/www\.yahoo\.com/d' file1 > file2
which means "for line 1 through the first line matching the regexp /http:\/\/www\.yahoo\.com/, delete the line" (and then, implicitly, print everything else; note that -n is not used this time).
awk '/yahoo/ ? c++ : c' file1
Or golfed
awk '/yahoo/?c++:c' file1
Result
http://www.baidu.com
http://www.yandex.com
This is most easily done in Perl:
perl -ne 'print unless 1 .. m(http://www\.yahoo\.com)' file
In other words, print all lines that aren’t between line 1 and the first occurrence of that pattern.
Using this script:
# Get index of the "yahoo" word
index=`grep -n "yahoo" filepath | cut -d':' -f1`
# Get the total number of lines in the file
totallines=`wc -l filepath | cut -d' ' -f1`
# Subtract totallines with index
result=`expr $total - $index`
# Gives the desired output
grep -A $result "yahoo" filepath