Bash - remove all lines beginning with 'P'

Question

I have a text file that\'s about 300KB in size. I want to remove all lines from this file that begin with the letter \"P\". This is what I\'ve been using:

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Answer 1

Explanation

1. use ^ to anchor your pattern to the beginning of the line ;
2. delete lines matching the pattern using sed and the d flag.

Solution #1

cat file.txt | sed '/^P/d'

Better solution

Use sed-only:

sed '/^P/d' file.txt > new.txt

Answer 2

Use start of line mark and quotes:

 cat file.txt | egrep -v '^P.*'

P* means P zero or more times so together with -v gives you no lines

^P.* means start of line, then P, and any char zero or more times

Quoting is needed to prevent shell expansion.

This can be shortened to

egrep -v ^P file.txt

because .* is not needed, therefore quoting is not needed and egrep can read data from file.

As we don't use extended regular expressions grep will also work fine

grep -v ^P file.txt

Finally

grep -v ^P file.txt > new.txt

Answer 3

This works:

cat file.txt | egrep -v -e '^P'

-e indicates expression.

Answer 4

Use sed with inplace substitution (for GNU sed, will also for your cygwin)

sed -i '/^P/d' file.txt

BSD (Mac) sed

sed -i '' '/^P/d' file.txt

Answer 5

With awk:

awk '!/^P/' file.txt

Explanation

1. The condition starts with an ! (negation), that negates the following pattern ;
   • /^P/ means "match all lines starting with a capital P",
2. So, the pattern is negated to "ignore lines starting with a capital P".
3. Finally, it leverage awk's behavior when { … } (action block) is missing, that is to print the record validating the condition.

So, to rephrase, it ignores lines starting with a capital P and print everything else.

Note

sed is line oriented and awk column oriented. For your case you should use the first one, see Edouard Lopez's reponse.