Printing a line in a certain percentile of a larg text file

Question

I am looking for a single line command to print a line in a certain percentile of a large text file. My preferred solution is something based on sed, wc -l, and/or head/tail as I already know how to do that with awk and wc -l. To make it more clear, if my file has 1K lines of text, I need to print for example the (95%*1K)th line of that file.

score 1 · Answer 1 · answered Mar 01 '17 at 22:04

1

head -`wc -l file | awk '{print(int(0.95*$1))}'` file | tail -n 1

answered Mar 01 '17 at 22:04

user31264

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score 1 · Accepted Answer · answered Mar 01 '17 at 22:22

1

In bash:

head -`echo scale=0\;$(cat file|wc -l)\*95/100 | bc -l` file | tail -n 1

answered Mar 01 '17 at 22:22

user31264

6,557
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Thanks for both of your answers. – user2517676 Mar 01 '17 at 22:27

Printing a line in a certain percentile of a larg text file

2 Answers2