Re: "undefined behavior"?

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!