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I've got a situation where I've got several lines, and I need two different fields out of each of them. To be specific, I have a list of references in a bibliography and I want to get the last name and the year.

Sample input:

Aloise-Young, P.A. (1993). The development of self-presentation.  Self-promotion in 6- to 10-year-old children. Social Cognition, 11, 201-222.
Banerjee, R. (2002). Children's understanding of self-presentational behavior: Links with mental-state reasoning and the attribution of embarrassment. Merril-Palmer Quarterly, 48, 378-404.
Bennett, M., & Wellman, H. (1989). The role of second-order belief-understanding and social context in children's self-attribution of social emotions. Social Development, 9, 126-130.

Desired output:

Aloise-Young 1993
Banerjee 2002
Bennett 1989

I can get the last names with cat file | cut -d, -f1

I can get the years with cat file | grep -o "[[:digit:]]\{4\}"

My problem is that now I have two separate outputs and I don't know how to combine them in the way I want. Any ideas? I suspect maybe awk can do what I need, but I'm a bash noob and I don't know awk yet.

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6 Answers

up vote 3 down vote accepted

I put your snippet in a file called alois:

sed -r 's/^([^ ,]+)[^0-9]+([0-9]+).*$/\1 \2/' alois 
Aloise-Young 1993
Banerjee 2002
Bennett 1989

Quick explanation: we use sed's search & replace function s/pattern/replacement/

^([^ ,]+) means: from the beginning of line take anything that's not a space or a , and remember those. (that's that the parentheses are there for).

[^0-9]+ look for anything that's not a numeric value, but ignore it.

([0-9]+) take and remember any consecutive digits

.*$ match everything else to the end of line.

\1 \2 replace everything matched (the whole line) with the remembered values from above.

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I'm on a Mac, so I had to use -E instead of -r flag, but that worked great after that. Thanks! –  Nathan Wallace May 2 '13 at 1:28
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When the text processing gets too tough for basic tools, try Awk.

awk -F , '{last_name = $1; sub(/\).*/, ""); sub(/.*\(/, ""); print last_name, $0}'

Here sed is about on par — it's less readable, but awk suffers from the lack of backreferences.

sed -n 's/^\([^,]*\),[^(]*(\([^()]*\)).*/\1 \2/p'

For this particular task, Perl is a little easier overall. You can use the non-greedy repetition operator *? to ensure that you capture the first parenthesized part of the line.

perl -l -ne '/^([^,]*),.*?\(([^()]*)\)/ and print "$1 $2"'
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With your sed command I get this: sed -n 's/^([^,]*),[^(]*([^()])$/\1 \2/p' alois sed: -e expression #1, char 34: invalid reference \2 on `s' command's RHS –  tink May 2 '13 at 1:23
    
I also got an error with the sed command and with awk one as well. Could be because I'm running Mac, not Linux. Perl worked straight away, though –  Nathan Wallace May 2 '13 at 1:29
    
@tink Sorry, I shouldn't have typed the commands directly in my browser. There, I've tested them and corrected them. –  Gilles May 2 '13 at 1:29
    
They all work for me now. Thanks for all those! I'm definitely going to study these and figure out what they're doing. I'd upvote you if I were allowed yet –  Nathan Wallace May 2 '13 at 1:31
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In general, you can join the output of commands with paste and process substitution <(...), so in your case this works:

 paste -d ' ' <(cut -d, -f1 file) <(grep -o "[[:digit:]]\{4\}" file)

Output:

Aloise-Young 1993
Banerjee 2002
Bennett 1989

But this entails passing file twice which is unnecessary, so you should probably be using a tool that can grab both items in one go, e.g. sed, awk, etc.

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You may find out that you need to refine your requirements. For instance, none of the solutions given so far work for input like:

Smith, J., & 3Com(Inc.) research (1999), XYZ statistics (1960 - 1998)

To report everything up to the first coma along with the first occurrence of a sequence of 4 digits enclosed in parenthesis, you could do:

perl -ne 'print "$1 $2\n" if /^(.*?),.*?\((\d{4})\)/'
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cut + sed:

cut -d\) -f1 testfile | sed 's/\,.*(/ /g'
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Not sexy but you can trim the text from the first comma to the first open paren. Substitute all that with a space. Then just trim the character from the first closing paren to the end of the line.

1. convert this => , ..... ( to a space
2. convert this => )........ to nothing

The command

$ cat file | sed 's/,.*(/ /' | sed 's/).*//'
Aloise-Young 1993
Banerjee 2002
Bennett 1989
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This doesn't work if the part after the year contains (, but you could use 's/,[^(]*(/ /;s/).*//'. –  Lauri Ranta May 2 '13 at 15:39
    
@LauriRanta - Thanks. It works for the sample data that the OP provided. If he wanted it to be more robust then he should've included more examples 8-). –  slm May 2 '13 at 15:48
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