# Help with Median
**URL:** https://community.unix.com/t/help-with-median/244778
**Category:** UNIX for Dummies Questions & Answers
**Created:** [July 16, 2009, 8:05am UTC](https://community.unix.com/t/help-with-median/244778 "2009-07-16T08:05:39Z")
**Posts on this page:** 12
**Page:** 1
### Author:  [@zajtat](https://community.unix.com/u/zajtat)
#### Post date: [July 16, 2009, 8:05am UTC](https://community.unix.com/t/help-with-median/244778/1 "2009-07-16T08:05:39Z")
Hi,
I know this question has been asked before, but I'd like to ask for some explanation rather than solution regarding finding median in unix.
I've got a simple one column file with over 2 million numbers and I need to find its median. The file looks like this:
0.123
0.235
0.890
0.000 etc (just plain numbers).
The solution that I've found in forum answers was:
1) find the number of lines with this code:
wc -l filename
In my case it is: 2543887
2) if the number is not even (like in my case) than the median can be found with this code:
cat filename| head -$2543887 | tail -1
My question is how head -2543887 is different to head -$2543887? What does the dollar sign do?
Many thanks in advance!
---
### Author:  [@rakeshawasthi](https://community.unix.com/u/rakeshawasthi)
#### Post date: [July 16, 2009, 8:46am UTC](https://community.unix.com/t/help-with-median/244778/2 "2009-07-16T08:46:39Z")
```nohighlight
cat filename| head -$2543887 | tail -1
```
Is it a correct command? Did you try this?? and cant see any maths here to get the median. am confused.
---
### Author:  [@zajtat](https://community.unix.com/u/zajtat)
#### Post date: [July 16, 2009, 8:59am UTC](https://community.unix.com/t/help-with-median/244778/3 "2009-07-16T08:59:40Z")
Yes, I've tried it and the answer was 0.0751, while typing just
cat filename | head -$254887 | tail -1 gives 9.9997. So, these two command lines do something different?
I'm new to Unix. Do you think what I found won't give me the median? What do you think I should use?
Many thanks!
---------- Post updated at 07:59 AM ---------- Previous update was at 07:56 AM ----------
sorry, I meant that typing
cat filename | head -2543887 |tail -1 gives 9.9997
So, typing it with or without the dollar sign gives different outputs...
---
### Author:  [@rakeshawasthi](https://community.unix.com/u/rakeshawasthi)
#### Post date: [July 16, 2009, 9:05am UTC](https://community.unix.com/t/help-with-median/244778/4 "2009-07-16T09:05:01Z")
i did not knew this $ thing with head...
and for median, i think you need to do the statistcs part. I dont know the formula 🙂
---
### Author:  [@vgersh99](https://community.unix.com/u/vgersh99)
#### Post date: [July 16, 2009, 9:07am UTC](https://community.unix.com/t/help-with-median/244778/5 "2009-07-16T09:07:21Z")
I don't follow, how the head|tail can calculate the [median](http://en.wikipedia.org/wiki/Median).
---
### Author:  [@zajtat](https://community.unix.com/u/zajtat)
#### Post date: [July 16, 2009, 9:14am UTC](https://community.unix.com/t/help-with-median/244778/6 "2009-07-16T09:14:33Z")
Unofrtunately, I'm new to Unix and don't know how the head and tail can calculate the median, but I thought that the dollar sign is a trick. This is why I asked the question in the first place.
Does anyone know how to calculate the mean, then?
---------- Post updated at 08:14 AM ---------- Previous update was at 08:10 AM ----------
sorry, I meant how to calculate the median
---
### Author:  [@vgersh99](https://community.unix.com/u/vgersh99)
#### Post date: [July 16, 2009, 9:45am UTC](https://community.unix.com/t/help-with-median/244778/7 "2009-07-16T09:45:32Z")
Google is your [friend](http://dada.perl.it/shootout/moments_allsrc.html)
---
### Author:  [@zajtat](https://community.unix.com/u/zajtat)
#### Post date: [July 16, 2009, 10:03am UTC](https://community.unix.com/t/help-with-median/244778/8 "2009-07-16T10:03:44Z")
Thank you Vgersh99, but I'm afraid it's a bit too complicated for me. Being a beginer, I don't get half of what is written there... Sorry!
---
### Author:  [@vgersh99](https://community.unix.com/u/vgersh99)
#### Post date: [July 16, 2009, 10:20am UTC](https://community.unix.com/t/help-with-median/244778/9 "2009-07-16T10:20:00Z")
[Courtesy of [Awk](http://dada.perl.it/shootout/gawk_allsrc.html)]
nawk -f stats.awk myFile
stats.awk:
```plaintext
# $Id: moments.gawk,v 1.1 2001/05/13 07:09:06 doug Exp $
# http://www.bagley.org/~doug/shootout/
BEGIN {
#delete ARGV;
sum = 0;
n = 0;
}
{
nums[n++] = $1;
sum += $1;
}
END {
mean = sum/n;
for (num in nums) {
dev = nums[num] - mean;
if (dev > 0) { avg_dev += dev; } else { avg_dev -= dev; }
vari += dev^2;
skew += dev^3;
kurt += dev^4;
}
avg_dev /= n;
vari /= (n - 1);
std_dev = sqrt(vari);
if (vari > 0) {
skew /= (n * vari * std_dev);
kurt = kurt/(n * vari * vari) - 3.0;
}
nums[n] = nums[0];
heapsort(n, nums);
mid = int(n/2)+1;
median = (n % 2) ? nums[mid] : (nums[mid] + nums[mid-1])/2;
printf("n: %d\n", n);
printf("median: %f\n", median);
printf("mean: %f\n", mean);
printf("average_deviation: %f\n", avg_dev);
printf("standard_deviation: %f\n", std_dev);
printf("variance: %f\n", vari);
printf("skew: %f\n", skew);
printf("kurtosis: %f\n", kurt);
}
function heapsort (n, ra) {
l = int(0.5+n/2) + 1
ir = n;
for (;;) {
if (l > 1) {
rra = ra[--l];
} else {
rra = ra[ir];
ra[ir] = ra[1];
if (--ir == 1) {
ra[1] = rra;
return;
}
}
i = l;
j = l * 2;
while (j <= ir) {
if (j < ir && ra[j] < ra[j+1]) { ++j; }
if (rra < ra[j]) {
ra = ra[j];
j += (i = j);
} else {
j = ir + 1;
}
}
ra = rra;
}
}
```
---
### Author:  [@zajtat](https://community.unix.com/u/zajtat)
#### Post date: [July 16, 2009, 10:28am UTC](https://community.unix.com/t/help-with-median/244778/10 "2009-07-16T10:28:28Z")
Yes, thank you! I have seen it through the link you've attached, but I don't understand it. I will look thnigs up, but was hoping there's a simpler way of doing it. Somebody's hinted that I should use awk, but then I saw something on this forum and decided to ask.
Does anybody know anything simpler than this huge script, please?
Many thanks in advance!
---------- Post updated at 09:28 AM ---------- Previous update was at 09:24 AM ----------
I have just tried ot run
nawk -f stats.awk myfile
but it doesn't work. Nawk is not recognized as a command...
---
### Author:  [@vgersh99](https://community.unix.com/u/vgersh99)
#### Post date: [July 16, 2009, 10:38am UTC](https://community.unix.com/t/help-with-median/244778/11 "2009-07-16T10:38:16Z")
sort myDataFile | awk -f median.awk
median.awk:
```plaintext
{
nums[++n] = $1;
}
END {
mid = int(n/2)+1;
median = (n % 2) ? nums[mid] : (nums[mid] + nums[mid-1])/2;
printf("median: %f\n", median);
}
```
---------- Post updated at 10:38 AM ---------- Previous update was at 10:37 AM ----------
> [@zajtat](#):
>
> Yes, thank you! I have seen it through the link you've attached, but I don't understand it. I will look thnigs up, but was hoping there's a simpler way of doing it. Somebody's hinted that I should use awk, but then I saw something on this forum and decided to ask.
> Does anybody know anything simpler than this huge script, please?
>
> Many thanks in advance!
>
> ---------- Post updated at 09:28 AM ---------- Previous update was at 09:24 AM ----------
>
> I have just tried ot run
> nawk -f stats.awk myfile
> but it doesn't work. Nawk is not recognized as a command...
then use either 'awk' or 'gawk'
---
### Author:  [@zajtat](https://community.unix.com/u/zajtat)
#### Post date: [July 16, 2009, 10:46am UTC](https://community.unix.com/t/help-with-median/244778/12 "2009-07-16T10:46:04Z")
It does work now! I've got loads to learn and understand now!!!!
Many thanks for your help!!!!