sort Find how many distinct values are among the certain column

Question

For example, this is the list.txt

Joe 3
Jack 1
Ulysses 6
Fox 2
Cassidy 1
Jones 6
Kevin 7

Then the output should be 5 because there are 5 different values in the 2rd column.

How should I finish this by only using sort cut wc uniq?

I have an idea, first use sort -k2n to sort the second column in increasing order and then use uniq to eliminate the second-column duplicated rows, and then the result would be like

Cassidy 1
Fox 2
Joe 3
Jones 6
Kevin 7

and then I use cut -d ' ' -f2 to list all the numbers like 1 2 3 6 7 and then I use wc -d to count the number of distinct values and this will return 5.

What should I do in the uniq part to eliminate the duplicated rows with the same number?

Is there a simple way to accomplish this?

Answer 1

I would start with cut since you only care about uniqueness in column 2:

cut -d' ' list.txt

results in:

1
2
1
3
6
7
6

Now you want unique values; uniq will do that, but only if it's sorted. If you're going to sort, go ahead and use sort's -u flag:

cut -d' ' -f2 list.txt | sort -u

Results in:

1
2
3
6
7

and now you can use wc to count the number of lines of output:

cut -d' ' -f2 list.txt | sort -u  | wc -l

which gives you:

5

Note that we're relying a specific format for the list.txt file -- no spaces in people's names!

Answer 2

I would go with:

sort -k2,2 -u names | wc -l
5

Where names has this content:

cat names
Joe 3
Jack 1
Ulysses 6
Fox 2
Cassidy 1
Jones 6
Kevin 7

Answer 3

Since you are required to use -sort -cut -wc -uniq, then it seems that the required command line should be something like this:

$ cut -d' ' -f2 file.txt | sort -n | uniq | wc -l
5