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On 6/14/2024 1:18 PM, David Brown wrote:
> On 13/06/2024 16:38, DFS wrote:
>>
>> I write a little code every few days.  Mostly python.
> 
> Certainly if I wanted to calculate some statistics from small data sets, 
> I'd go for Python - it would not consider C unless it was for an 
> embedded system.
> 
>>
>> I like C for it's blazing speed.  Very addicting.  And it's much more 
>> challenging/frustrating than python.
> 
> With small data sets, Python has blazing speed - /every/ language has 
> blazing speed.  And for large data sets, use numpy on Python and you 
> /still/ have blazing speeds - a lot faster than anything you would write 
> in C (because numpy's underlying code is written in C by people who are 
> much better at writing fast numeric code than you or I).
> 
> The only reason to use C for something like is is for the challenge and 
> fun, which is fair enough.


It was fun, especially when I got every stat to match the website exactly.


I just now ported that C stats program to python.  The original C took 
me ~2.5 days to write and test.

The port to python then took about 2 hours.

It mainly consisted of replacing printf with print, removing brackets 
{}, changing vars max and min to dmax and dmin, dropping the \n from 
printf's, replacing fabs() with abs(), etc.

Line count dropped about 20%.


During conversion, I got a Python error I don't remember seeing in the past:

"TypeError: list indices must be integers or slices, not float"

because division returns a float, and some of the array addressing was 
like this: nums[i/2].

My initial fix was this clunk (convert to int()):

	# median and quartiles
	# quartiles divide sorted dataset into four sections
	# Q1 = median of values less than Q2
	# Q2 = median of the data set
	# Q3 = median of values greater than Q2
	if N % 2 == 0:
		Q2 = median = (nums[int((N/2)-1)] + nums[int(N/2)]) / 2.0
		i = int(N/2)
		if i % 2 == 0:
			Q1 = (nums[int((i/2)-1)] + nums[int(i/2)]) / 2.0
			Q3 = (nums[int(i + ((i-1)/2))] + nums[int(i+(i/2))]) / 2.0 	
		else:
			Q1 = nums[int((i-1)/2)]
			Q3 = nums[int(i + ((i-1)/2))]
			
	if N % 2 != 0:
		Q2 = median = nums[int((N-1)/2)]
		i = int((N-1)/2)
		if i % 2 == 0:
			Q1 = (nums[int((i/2)-1)] + nums[int(i/2)]) / 2.0
			Q3 = (nums[int(i + (i/2))] + nums[int(i + (i/2) + 1)]) / 2.0
		else:
			Q1 = nums[int((i-1)/2)]
			Q3 = nums[int(i + ((i+1)/2))]


And then with some substitution:

	if N % 2 == 0:
		i = int(N/2)
		Q2 = median = (nums[i - 1] + nums[i]) / 2.0
		x = int(i/2)
		y = int((i-1)/2)
		if i % 2 == 0:
			Q1 = (nums[x - 1] + nums[x])     / 2.0
			Q3 = (nums[i + y] + nums[i + x]) / 2.0 	
		else:
			Q1 = nums[y]
			Q3 = nums[i + y]
			
	if N % 2 != 0:
		i = int((N-1)/2)
		Q2 = median = nums[i]
		x = int(i/2)
		y = int((i-1)/2)
		z = int((i+1)/2)
		if i % 2 == 0:
			Q1 = (nums[x - 1] + nums[x])         / 2.0
			Q3 = (nums[i + x] + nums[i + x + 1]) / 2.0
		else:
			Q1 = nums[y]
			Q3 = nums[i + z]


How would you do it?


If you have an easy to apply formula for computing the quartiles, let's 
hear it!