extract pattern using grep/sed
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21 months ago
Nelo ▴ 20

Hi

Pm4.1LM10m04850 0.24924Pm4.1LM01m05240 0.02328Pm4.1LM01m11200 -0.02328Pm4.1LM01m11050 0.02899Pm4.1LM03m10920 0.04638Pm4.1LM00m08740 -0.04638Pm4.1LM09m10890 0.18085Pm4.1LM05m02500 0.23509Pm4.1LM03m01390

This is my query data above.

I want to exclude those digits that are in bold and the rest in rows like this:

 Pm4.1LM10m04850 
 Pm4.1LM01m05240
 Pm4.1LM01m11200

and so on..

I tried using grep command: grep "Pm4.1LM[0-9]" FILENAME of course it didn't work because there is character in between [0-9] pattern I used in command.

Can anyone help please

Appreciated if help is given with bit of detail explanation!!!

Thanks in advance

grep linux sed • 1.4k views
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21 months ago
 tr " " "\n" < input.txt | sed 's/^[0-9\.\-]*//'
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0
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It works. Thank you.

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21 months ago
lacb ▴ 120
sed -r 's/\s[0-9\.\-]+Pm/ Pm/g' input.txt

outputs: Pm4.1LM10m04850 Pm4.1LM01m05240 Pm4.1LM01m11200 Pm4.1LM01m11050 Pm4.1LM03m10920 Pm4.1LM00m08740 Pm4.1LM09m10890 Pm4.1LM05m02500 Pm4.1LM03m01390

This script uses sed and a regular expression (regex): \s[0-9\.\-]+Pm

This regex matches the number with [0-9\.\-]+ (a series of numerical characters, decimal point and sign) and additional context (space before, "Pm" after)

The sed command replaces this match by only the " Pm" to remove the number.

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