I am having trouble getting this to work:
awk -F ";" '{for(i=1; i < NF;i++) $i ~ /^_.*/ {print $i}}'
I want to iterate over all the fields (records can have 7-9) and print only those that start with an _ except the line above gives me a syntax error at the print statement and if I omit the {print $i} I dont get any output.
How is the correct way to do this?
You're missing an if:
awk -F ";" '{for(i=1; i < NF;i++) if($i ~ /^_.*/) {print $i}}'
The structure of an awk program is condition { action } but what you have currently is all within an action block (the condition is true by default). Within the action block the if isn't implicit.
As an aside, the .* in the pattern is redundant; you may as well use /^_/ to match any string starting with _.
Note: since fields are 1-indexed, likely that the right loop condition is i <= NF. If you are sure that the last field is unnecessary, condition i < NF will do the job.
Related
Consider this source:
field1;field2;"data;data field3";field4;"data;data field5";field6
field1;"data;data field2";field3;field4;field5;"data;data field6"
As you can see, the field delimiter is being used inside certain fields, enclosed between ". I cannot directly parse with awk because there is no way of avoiding unwanted splitting, at least I haven't found a way. Moreover, those special fields have a variable position within a line and they can occur once, twice, 4 times etc.
I thought of a solution involving a pre-parsing step, where I replace the ; in those fields with a code of some sort. The problem is that sed / awk perform greedy REGEX match. So in the above example, I can only replace ; within the last field enclosed in quotes in each line.
How can I match each instance of quotes and replace the specific ; within them? I do not want to use perl or python etc.
Using gnu awk you can use special FPAT variable to have a regex for your fields.
You can use this command to replace all ; by | inside the double quotes:
awk -v OFS=';' -v FPAT='"[^"]*"|[^;]*' '{for (i=1; i<=NF; i++) gsub(/;/, "|", $i)} 1' file
field1;field2;"data|data field3";field4;"data|data field5";field6
field1;"data|data field2";field3;field4;field5;"data|data field6"
As an alternative to FPAT you can set the awk FS to be double quotes and then swap out your semicolon delimiter for every other field:
awk -F"\"" '{for(i=1;i<=NF;++i){ if(i%2==0) gsub(/;/, "|", $i)}} {print $0}' yourfile
Here awk is:
Splitting the record by double quote (-F"\"")
Looping through each field that it finds ({for(i=1;i<=NF;++i))
Testing the field ordinal's mod 2 if it's 0 (if(i%2==0))
If it's even then it swaps out the semicolons with pipes (gsub(/;/, "|", $i))
Prints out the transformed record ({print $0})
I am trying to replace a word in awk using an if else statement.
Below is what I have already attempted. Basically if the thrid word is "Normal", change it to 0, else change it to 1.
awk -F "," '{(if $3=='Normal') {$3=0}; else {$3=1} }' filename
You need a proper if - else condition, followed with a trigger to print the line:
awk '{if ($3=="Normal") {$3=0} else {$3=1}}1' file
Here, 1 acts as a True condition, which triggers awk's default behaviour, consisting in printing the current line.
You can use a ternary operator for this, though:
awk '{$3=($3=="Normal") ? 0 : 1}1' file
Note also that you are dealing with ,-delimited fields, so you need to preserve them afterwards. For that, use OFS (output field separator). All together:
awk 'BEGIN{FS=OFS=","} {$3=($3=="Normal") ? 0 : 1}1' file
Given a sample file like this:
$ cat a
hello,how,are,you
i,am,Normal,buuu
This would return:
$ awk 'BEGIN{FS=OFS=","} {$3=($3=="Normal") ? 0 : 1}1' a
hello,how,1,you
i,am,0,buuu
What was wrong with your approach?
awk -F "," '{(if $3=='Normal') {$3=0}; else {$3=1} }' filename
You are using {(if condition) {action}; else {action}}. This has several problems:
the condition in if is the one within parentheses: if (condition) instead of (if condition).
You say (if condition) {action} and then ;. With this semi-colon, the if-statement gets finished and the next action will be executed anyways. So just drop it and say if (condition) {...} else {...} without this ; before else.
You are using single quotes in $3=='Normal'. This is closing the awk-statement and hence not being interpreted properly, so you need to use double quotes: $3=="Normal".
Missing a print to have the line printed.
Pretty new to AWK programming. I have a file1 with entries as:
15>000000513609200>000000513609200>B>I>0011>>238/PLMN/000100>File Ef141109.txt>0100-75607-16156-14 09-11-2014
15>000000513609200>000000513609200>B>I>0011>Danske Politi>238/PLMN/000200>>0100-75607-16156-14 09-11-2014
15>000050354428060>000050354428060>B>I>0011>Danske Politi>238/PLMN/000200>>4100-75607-01302-14 31-10-2014
I want to write a awk script, where if 2nd field subtracted from 3rd field is a 0, then it prints field 2. Else if the (difference > 0), then it prints all intermediate digits incremented by 1 starting from 2nd field ending at 3rd field. There will be no scenario where 3rd field is less than 2nd. So ignoring that condition.
I was doing something as:
awk 'NR > 2 { print p } { p = $0 }' file1 | awk -F">" '{if ($($3 - $2) == 0) print $2; else l = $($3 - $2); for(i=0;i<l;i++) print $2++; }'
(( Someone told me awk is close to C in terms of syntax ))
But from the output it looks to me that the String to numeric or numeric to string conversions are not taking place at right place at right time. Shouldn't it be taken care by AWK automatically ?
The OUTPUT that I get:
513609200
513609201
513609200
Which is not quiet as expected. One evident issue is its ignoring the preceding 0s.
Kindly help me modify the AWK script to get the desired result.
NOTE:
awk 'NR > 2 { print p } { p = $0 }' file1 is just to remove the 1st and last entry in my original file1. So the part that needs to be fixed is:
awk -F">" '{if ($($3 - $2) == 0) print $2; else l = $($3 - $2); for(i=0;i<l;i++) print $2++; }'
In awk, think of $ as an operator to retrieve the value of the named field number ($0 being a special case)
$1 is the value of field 1
$NF is the value of the field given in the NF variable
So, $($3 - $2) will try to get the value of the field number given by the expression ($3 - $2).
You need fewer $ signs
awk -F">" '{
if ($3 == $2)
print $2
else {
v=$2
while (v < $3)
print v++
}
}'
Normally, this will work, but your numbers are beyond awk integer bounds so you need another solution to handle them. I'm posting this to initiate other solutions and better illustrate your specifications.
$ awk -F'>' '{for(i=$2;i<=$3;i++) print i}' file
note that this will skip the rows that you say impossible to happen
A small scale example
$ cat file_0
x>1000>1000>etc
x>2000>2003>etc
x>3000>2999>etc
$ awk -F'>' '{for(i=$2;i<=$3;i++) print i}' file_0
1000
2000
2001
2002
2003
Apparently, newer versions of gawk has --bignum options for arbitrary precision integers, if you have a compatible version that may solve your problem but I don't have access to verify.
For anyone who does not have ready access to gawk with bigint support, it may be simpler to consider other options if some kind of "big integer" support is required. Since ruby has an awk-like mode of operation,
let's consider ruby here.
To get started, there are just four things to remember:
invoke ruby with the -n and -a options (-n for the awk-like loop; -a for automatic parsing of lines into fields ($F[i]));
awk's $n becomes $F[n-1];
explicit conversion of numeric strings to integers is required;
To specify the lines to be executed on the command line, use the '-e TEXT' option.
Thus a direct translation of:
awk -F'>' '{for(i=$2;i<=$3;i++) print i}' file
would be:
ruby -an -F'>' -e '($F[1].to_i .. $F[2].to_i).each {|i| puts i }' file
To guard against empty lines, the following script would be slightly better:
($F[1].to_i .. $F[2].to_i).each {|i| puts i } if $F.length > 2
This could be called as above, or if the script is in a file (say script.rb) using the incantation:
ruby -an -F'>' script.rb file
Given the OP input data, the output is:
513609200
513609200
50354428060
The left-padding can be accomplished in several ways -- see for example this SO page.
I am looking for a sed in which I can recognize all of the text in between two indicators and then replace it with a place holder.
For instance, the 1st indicator is a list of words
(no|noone|haven't)
and the 2nd indicator is a list of punctuation
Code:
(.|,|!)
From an input text such as
"Noone understands the plot. There is no storyline. I haven't
recommended this movie to my friends! Did you understand it?"
The desired result would be.
"Noone understands_AFFIX me_AFFIX. There is no storyline_AFFIX. I
haven't recommended_AFFIX this_AFFIX movie_AFFIX to_AFFIX my_AFFIX
friends_AFFIX! Did you understand it?"
I know that there is the following sed:
sed -n '/WORD1/,/WORD2/p' /path/to/file
which recognizes the content between two indicators. I have also found a lot of great information and resources here. However, I still cannot find a way to append the affix to each token of text that occurs between the two indicators.
I have also considered to use awk, such as
awk '{sub(/.*indic1 /,"");sub(/ indic2.*/,"");print;}' < infile
yet still, it does not allow me to append the affix.
Does anyone have a suggestion to do so, either with awk or sed?
Little more compact awk
$ awk 'BEGIN{RS=ORS=" ";s="_AFFIX"}
/[.,!]$/{f=0; $0=gensub(/(.)$/,"s\\1","g")}
f{$0=$0s}
/Noone|no|haven'\''t/{f=1}1' story
Noone understands_AFFIX the_AFFIX plot_AFFIX. There is no storyline_AFFIX. I haven't recommended_AFFIX this_AFFIX movie_AFFIX to_AFFIX my_AFFIX friends_AFFIX! Did you understand it?
Perl to the rescue!
perl -pe 's/(?:no(?:one)?|haven'\''t)\s*\K([^.,!]+)/
join " ", map "${_}_AFFIX", split " ", $1/egi
' infile > outfile
\K matches what's on its left, but excludes it from the replacement. In this case, it verifies the 1st indicator. (\K needs Perl 5.10+.)
/e evaluates the replacement part as code. In this case, the code splits $1 on whitespace, map adds _AFFIX to each of the members, and join joins them back into a string.
Here is one verbose awk command for the same:
s="Noone understands the plot. There is no storyline. I haven't recommended this movie to my friends! Did you understand it?"
awk -v IGNORECASE=1 -v kw="no|noone|haven't" -v pct='\\.|,|!' '{
a=0
for (i=2; i<=NF; i++) {
if ($(i-1) ~ "\\y" kw "\\y")
a=1
if (a && $i ~ pct "$") {
p = substr($i, length($i), 1)
$i = substr($i, 1, length($i)-1)
}
if (a)
$i=$i "_AFFIX" p
if(p) {
p=""
a=0
}
}
} 1'
Output:
Noone understands_AFFIX the_AFFIX plot_AFFIX. There is no storyline_AFFIX. I haven't recommended_AFFIX this_AFFIX movie_AFFIX to_AFFIX my_AFFIX friends_AFFIX! Did you understand it?
I want to print all the lines of a file where the first element of each line begins with a number using awk. Below are the details on the data contained in the file and command used:
filename contents:
12.44.4444goad ABCDEF/END
LMNOP/START joker
98.0 kites
command used:
awk '{ $1 ~ /^\d[a-zA-Z0-9]*/ }' filename
After running the above command, no results are displayed on the prompt.
Please let me know if there is any correction that needs to be made to the above command.
To print the lines starting with a digit, you can try the following:
awk '/^[[:digit:]]+/' file
as pointed out by #HenkLangeveld your syntax is incorrect. Also the regex \d is not available in awk.
If you only need to match at least one digit at the start of the line, all you need is ^ to match the start of a line and [0-9] to match a digit.
You can use curly brackets with an if statement:
awk '{if($1 ~ /^[0-9]/) print $0}' filename
But that would just be longhand for this:
awk '$1 ~ /^[0-9]/' filename
From your attempted solution, it looks like you want:
awk 'NF>1 && $1 ~ /^[0-9.]*$/' filename
You need to explicitly match the . if you want to include the decimal point, and you need the $ anchor to make the * meaningful. This will miss lines in which the first column looks like 5e39 or -2.3. You can try to catch those cases with:
awk 'NF>1 && $1 ~ /^-?[0-9.]*(e[0-9*])?$/' filename
but at this point I would tell you to use perl and stop trying to be more robust with awk.
Perhaps (this will print blank lines...not sure which behavior you want):
perl -lane 'use POSIX qw(strtod); my ($num, $end) = strtod($F[0]);
print unless $end;' filename
This uses strtod to parse the number and tells you the number of characters at the end of the string that are not part of it.
Drop the braces and the \d, like this:
awk ' $1 ~ /^[0-9]/ ' filename
Awk programs come in chunks. A chunk is a pattern block pair, where the block
defaults to { print }. (An empty pattern defaults to true.)
The /\d/ is a perl-ism and might work in some versions awk - not in those that I tried*. You need either the traditional /^[0-9]/ or the POSIX /^[[:digit:]]/ notation.
*
gnu and ast