How to grep a data within a square brackets?

Question

I have following type of data in my .log file:

Cell[BoxData["0.8693473380926441`"]
Cell[BoxData["100.40266203596555`"]
Cell[BoxData["23.338724857049048`"]
Cell[BoxData["0.25720321805387686`"]
Cell[BoxData["66.85236878385265`"]

I want to grep only the numeric numbers, like:

0.8693473380926441
100.40266203596555
23.338724857049048
0.25720321805387686
66.85236878385265

I have tried, grep -Po '\[.*?]', but it only eliminates one brackets. Any help will be appreciated.

Answer 1

awk is better for this:

awk -F '["`]' '{print $2}' file

That sets the field separator as either " or a tilde and then prints the second column/field.

Output:

0.8693473380926441
100.40266203596555
23.338724857049048
0.25720321805387686
66.85236878385265

Answer 2

I tried with below mentioned command and worked fine

 sed 's/.*\["//g' filename| sed 's/`"\]//g'

output

0.8693473380926441
100.40266203596555
23.338724857049048
0.25720321805387686
66.85236878385265

Answer 3

Any chance anyone can answer the question in the subject instead of the example. I am trying to use awk to grab the number in “[nnn]” and while awk -F'\[' '{print $2}' | awk -F'\]' '{print $1}’ works, it sure seem ugly. I have tried -F’[[]]’ and -F’[\[\]]’ but neither work

Answer 4

grep for what it is you actually want. What you want from that is just the numbers. You want some digits, and a period, and some digits, so:

egrep  '[0-9]+\.[0-9]+' -o foo.txt

The -o says "Only print the part that matches, not the entire line".