Find multiple occurences of same string

user_11404337 0
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In a range of files, I want to see which line has atleast 4 times the same occurence of the same word. This word can be any word.

So input:

a a a b b e e e
o o o o p p p y y y
w r r r u u i i o o r
x x o o i i p p z z y y

Output:

o o o o p p p y y y
w r r r u u i i o o r

What I have tried at the moment is to make sure that sentences are put separate, ready to be processed basically.

cat * |
    tr '\n' ' '|
    sed 's/[.!?;"]/ & /g' |
    sed 's/[.!?]/&\n/g'|
    grep -E -w '\b([[:alnum:]]*)\{4*\}\b'

But my grep doesn't get anything, so how do I get that Grep only prints out all sentences which contain a word which occurs atleast 4 times in it?

sed ubuntu grep
2 Answers

With GNU grep, you can use a PCRE regex like

grep -P '\b(\w+)\b(.*\b\1\b){3}'

See the regex demo.

Test in Ubuntu 18.04.4 LTS:

enter image description here

Details

  • \b(\w+)\b - a whole word (captured in Group 1) (\b is a word boundary and \w matches letters, digits or underscores)
  • (.*\b\1\b){3} - three occurrences ({3}) of any text followed with the same value as in Group 1 (as \1 is an inline backreference to Group 1 value) as a whole word (again, \b word boundaries are used.)
user_38329700

Awk is another possibility that will do the same thing with a simple array using the elements as counters without needing a REGEX (though the answer by @WiktorStribizew is the efficient way to go). For example, using the a[] array with the field as the index, the value at that index is incremented on each occurrence and when any one field (word) reaches 4 occurrences the line is printed and you skip to the next record.

awk '{delete a; for(i=1;i<=NF;i++) if (++a[$i]==4) {print $0; next}}' file

(delete a simply deletes the array before each records is processed)

Example Use/Output

$ awk '{delete a; for(i=1;i<=NF;i++) if (++a[$i]==4) {print $0; next}}' file
o o o o p p p y y y
w r r r u u i i o o r
user_34221020
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