How to extract first sentence of a file in bash?

Question

I want to extract the first sentence of a file in bash. I used the following command:

sed 's/(\?|\.|!).*//' filename

However, it does not work. What is wrong with my regex?

If I have the following sentence in the file: Stack overflow is the best? I am also the best., the output needs to be Stack overflow is the best?

Note: the question mark needs to be there at the end. Also the sentence may end with full stop / question mark / exclamation mark.

Edit: The sentence might contain words like Mr. etc

It's a complicated issue, as it depends on the language used. Think of `Mr. Stack overflow is the best.` A naive solution will stop at `Mr.`. — pawamoy, Jan 10 '18 at 13:18

score 1 · Accepted Answer · edited Jan 10 '18 at 16:20

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sed -r 's/([.*\?]|[.*\.]|[.*!]).*/\1/' file
              ^       ^     ^   ^  ^
              |_______|_____|___|__|_> Any symbols before first `?` **OR**
                      |_____|___|__|_> Any symbols before first `.` **OR**
                            |___|__|_> Any symbols before first `!`
                                |__|_> Any symbols
                                   |_> Print all found symbols in first pair of brackets

My solution will find:

"(Any symbols before first ? found or Any symbols before first . found or Any symbols before first ! found ), any symbols after that. --> print found in brackets symbols".

edited Jan 10 '18 at 16:20

alseether

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answered Jan 10 '18 at 13:12

Viktor Khilin

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Can you please tell me what is wrong in my solution? – humble Jan 10 '18 at 13:16
@rjmessibarca in your solution sed use "base" regex, not extened (in my case, it's `-r` key). Your sed just can't find matches so it do nothing. Also if you add `-r` key in your solution, It will just delete all the symbols after first `?`, or after first `!`, or after first `.`, not printing all the symbols before it. – Viktor Khilin Jan 10 '18 at 13:20
@rjmessibarca nope – Viktor Khilin Jan 10 '18 at 13:30
An explanation would help a lot. – humble Jan 10 '18 at 13:31
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My solution will find: `(`"Any symbols before first `?` found or Any symbols before first `.` found or Any symbols before firts `!`found`)`, any symbols after that. --> print found in brackets symbols". Your solution will: "Find **all** symbols, with `?` at the **end** of string, or with `.` at the end of string, or `!` at the end of string, print found match in brackets. – Viktor Khilin Jan 10 '18 at 13:36
@ViktorKhilin is preferable to make an edit in your post with this explanation than explain in comments. It's also more readable. – alseether Jan 10 '18 at 13:43
@ViktorKhilin that's the hard way ;) – alseether Jan 10 '18 at 13:56
I don't think the bracket expressions do what you think they do. What is the idea of `[.*\?]`? This matches any one character of `.`, `*`, ``\`` or `?`. And alternation between character classes doesn't make sense either. The whole expression `([.*\?]|[.*\.]|[.*!])` is equivalent to `[.*\?!]`. – Benjamin W. Jan 10 '18 at 15:02

alseether · Answer 2 · 2018-01-10T13:36:41.197

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I think you're not matching the beginning of the line. My solution is:

^.*?[.?!]

Which means:

^ : The match must be at the beginning of the line
.*? : any number of characters (greedy, aka as less as possible)
[.?!]] : match one of the chars inside []

Working example here

Note that solution its working for python. I think there are no greedy searches with sed

edited Jan 10 '18 at 13:36

answered Jan 10 '18 at 13:13

alseether

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It's `sed`, `([.?!]).*` – revo Jan 10 '18 at 13:18
```$ cat filename Stack overflow is the best? I am also the best. [sahaquiel@sahaquiel-PC Stackoverflow]$ sed '^.*?[.?!]' filename sed: -e expression #1, char 1: unknown command: `^'``` You sure it works? (off.: how can I place line break in comments? :( ) – Viktor Khilin Jan 10 '18 at 13:25
@ViktorKhilin You can't. PS: Thanks for pointing. I've tried the solution and didn't work with sed. Greedy searches are not available. – alseether Jan 10 '18 at 13:37

score 1 · Answer 3 · answered Jan 10 '18 at 15:08

If your input file consists of just one line, you can use

$ grep -o '^[^.!?]*[.!?]' <<< 'Stack overflow is the best? I am also the best.'
Stack overflow is the best?

If there are multiple lines and your first sentence might be across multiple lines, you can use -z with GNU grep to treat the file as a single line:

$ grep -zo '^[^.!?]*[.!?]' <<< $'Stack overflow\nis the best? I am also the best.'
Stack overflow
is the best?

The regex consists of these components:

^ anchor to start of line
[^.!?]* zero or more characters other than ., ! or ?
[.!?] on of ., ! or ?

RavinderSingh13 · Answer 4 · 2018-01-10T13:29:01.783

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If your actual Input_file is same as shown example then following sed may help you in same.

sed 's/[\?\.\!].*/\?/'   Input_file

Output will be as follows.

Stack overflow is the best?

edited Jan 10 '18 at 13:29

answered Jan 10 '18 at 13:12

RavinderSingh13

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Sentences could be finished with comma or `!`, not only `?` as I understood. – Viktor Khilin Jan 10 '18 at 13:14
@ViktorKhilin, OP has not mentioned it and I had mentioned that as per shown sample only this code is. – RavinderSingh13 Jan 10 '18 at 13:14
I think it was just an example, but I like your solution for the single defined symbol in the end. – Viktor Khilin Jan 10 '18 at 13:16
@RavinderSingh13 Although, I have not downvoted. But, the sentence can end with ! or . or ? – humble Jan 10 '18 at 13:27

How to extract first sentence of a file in bash?

4 Answers4