I am trying to print only the matched pattern in a CSV file. Example: all the columns value starting with 35=its value. Thanks.
CSV file:
35=A,D=35,C=129,ff=136
D=35,35=BCD,C=129,ff=136
900035=G,D=35,C=129,ff=136
35=EF,D=35,C=129,ff=136,35=G
36=o,D=35,k=1
Output:
35=A
35=BCD
35=EF
35=G
The command I used did not work:
sed -n '/35=[A-Z]*?/ s/.*\(35=[A-Z]*?\).*/\1/p' filename
Using tr to replace all commas with newlines, and then grep to get all lines that start with the string 35=:
$ tr ',' '\n' <data.in | grep '^35='
35=A
35=BCD
35=EF
35=G
With GNU grep which supports -o option to print only matched string, each on its own line
$ grep -oE '\b35=[^,]+' ip.csv
35=A
35=BCD
35=EF
35=G
\b is word boundary, so that 900035 won't match[^,]+ to match one or more non, characters,
With awk
$ awk -F, '{ for(i=1;i<=NF;i++){if($i~/^35=/) print $i} }' ip.csv
35=A
35=BCD
35=EF
35=G
-F, set , as input field separatorfor(i=1;i<=NF;i++) iterate over all fieldsif($i~/^35=/) if field starts with 35=
print $i print that fieldSimilar with perl
perl -F, -lane 'foreach (@F){print if /^35=/}' ip.csv
\b isn't the best choice of boundary ; it would for instance match 35=value from somethingsomething, 42.35=value, somethingelse. If the grep versions supports -Perl regex and the input contains no spaces, I'd use a lookbehind to assert that we match from the start of a field. If there can be extra spaces this would require variable-length lookbehinds, which aren't implemented in any regex flavour available to grep AFAIK.
\K can be used stackoverflow.com/documentation/regex/639/…
\K trick, I knew about the meta-character but had never thought about its ability to emulate basic variable-length lookbehinds :)
With perl:
$ perl -lne 'print for /(\b35=[^,]+)/g' filename
35=A
35=BCD
35=EF
35=G
or perhaps more generally/robustly using the Text::CSV module
$ perl -MText::CSV -lne '
BEGIN{$p = Text::CSV->new()}
print for grep { /^35=/ } $p->fields(), $p->parse($_)
' filename
35=A
35=BCD
35=EF
35=G
Perl lookarounds with grep work really well.
grep -oP '(?<=35\=).*?(?=,)'
This returns the exact information minus the 35= bit
grep -oP '(?<=35\=).*?(?=,)' file.csv will return this
A
BCD
G
EF
Pure Bash solution:
( # Use parentheses as scope for IFS
IFS=$',\n' # Split on both , or \n
for c in $(</tmp/file.csv) # For every column or row
do
[[ "$c" =~ ^35= ]] && echo ${line##35=} # Find ^35= and print while removing ^35=
done
) # Optionally >/tmp/filtered-output.txt
Note, only use this for its readability and flexibility - if you can read it, otherwise you can use the following approach:
# Read | Replace | Find | Remove
cat /tmp/file.csv | tr ',' '\n' | grep '^35=' | sed 's/^35=//'
which is more intuitive and efficient.
Input (/tmp/file.csv):
35=A,D=35,C=129,ff=136
D=35,35=BCD,C=129,ff=136
900035=G,D=35,C=129,ff=136
35=EF,D=35,C=129,ff=136,35=G
36=o,D=35,k=1
Output:
A
BCD
EF
G
sed -e 'y/,/\n/; /^35=/P; D' filename