How do I count lines which specific column has two patterns? - linux

Question

year start year end location topic data type data value
2016 2017 AL Alcohol Crude Prevalence 16.9
2016 2017 CA Alcohol Other 15
2016 2017 AZ Neuropathy Other 13.1
2016 2017 HI Smoke Crude Prevalence 20
2016 2017 IL Cancer Other 20
2016 2017 KS Cancer Other 14
2016 2017 AZ Smoke Crude Prevalence 16.9
2016 2017 KY Cancer Other 13.8
2016 2017 LA Alcohol Crude Prevalence 18
The answer is required to count lines which are associated with the disease “topic”s "Alcohol" and "Cancer".
I already got the index of column named as "topic" , but the contents I am going to extract from "topic" is not correct, then I am not able to count the lines which is containing "Alcohol" and "Cancer", how to solve it?
Here is my code:
awk '{print $4}' AAA.csv > topic.txt
head -n5 topic.txt | less

Answer 1

You could try the following:
the call to awk gets the column in question, the grep filters the keywords, and the word count counts the lines
$ awk '{ print $4 }' data.txt | grep -e Alcohol -e Cancer | wc -l
6

Answer 2

Using a regexp with grep:
cat data.txt|tr -s " "|cut -d " " -f 4|grep -E '(Alcohol|Cancer)'|wc -l
If you are sure that words "Alcohol" and "Cancer" only appear in the 4th column you can just do
grep -E '(Alcohol|Cancer)' data.txt|wc -l
Addition
The OP asks in the comment:
If there are many columns, and I don't know the index of them. How can I extract the columns just based on their name ("topic")?
This code will store in the variable i the column containing "topic". Essentially, the code stores the first line of data.txt as an array variable s, and then parse the array elements until it finds the desired word. (You have to increase i by one at the end because the array index starts from 0).
Note: the code works only if actually a column "topic" is found.
head -n 1 data.txt|read -a s
for (( i=0; i<${#s[#]}; i++ ))
do
if [ "${s[$i]}" == "topic" ]
then
break
fi
done
i=$(( $i + 1 ))