A minor point is that if you just need to compare distances you don't need to
compute the hypotenuse, its square will do so no subtractions etc etc.
--
Robin Becker
--
https://mail.python.org/mailman/listinfo/python-list
On Mon, May 25, 2015 at 11:11 PM, Steven D'Aprano st...@pearwood.info wrote:
Let's compare three methods.
def naive(a, b):
return math.sqrt(a**2 + b**2)
def alternate(a, b):
a, b = min(a, b), max(a, b)
if a == 0: return b
if b == 0: return a
return a * math.sqrt(1
On 27 May 2015 at 19:00, Brian Blais bbl...@gmail.com wrote:
On Mon, May 25, 2015 at 11:11 PM, Steven D'Aprano st...@pearwood.info wrote:
Let's compare three methods.
def naive(a, b):
return math.sqrt(a**2 + b**2)
def alternate(a, b):
a, b = min(a, b), max(a, b)
if a == 0:
On Mon, May 25, 2015, at 15:21, ravas wrote:
Is this valid? Does it apply to python?
Any other thoughts? :D
The math.hypot function uses the C library's function which should deal
with such concerns internally. There is a fallback version in case the C
library does not have this function, in
On Tue, May 26, 2015, at 09:40, random...@fastmail.us wrote:
On Mon, May 25, 2015, at 15:21, ravas wrote:
Is this valid? Does it apply to python?
Any other thoughts? :D
The math.hypot function uses the C library's function which should deal
with such concerns internally. There is a
Am 25.05.15 um 21:21 schrieb ravas:
I read an interesting comment:
The coolest thing I've ever discovered about Pythagorean's Theorem is an
alternate way to calculate it. If you write a program that uses the distance
form c = sqrt(a^2 + b^2) you will suffer from the lose of half of your
El 25/05/15 15:21, ravas escribió:
I read an interesting comment:
The coolest thing I've ever discovered about Pythagorean's Theorem is an
alternate way to calculate it. If you write a program that uses the distance
form c = sqrt(a^2 + b^2) you will suffer from the lose of half of your
On Monday, May 25, 2015 at 1:27:43 PM UTC-7, Gary Herron wrote:
This is a statement about floating point numeric calculations on a
computer,. As such, it does apply to Python which uses the underlying
hardware for floating point calculations.
Validity is another matter. Where did you
On Monday, May 25, 2015 at 1:27:24 PM UTC-7, Christian Gollwitzer wrote:
Wrong. Just use the built-in function Math.hypot() - it should handle
these cases and also overflow, infinity etc. in the best possible way.
Apfelkiste:~ chris$ python
Python 2.7.2 (default, Oct 11 2012, 20:14:37)
On 05/25/2015 12:21 PM, ravas wrote:
I read an interesting comment:
The coolest thing I've ever discovered about Pythagorean's Theorem is an
alternate way to calculate it. If you write a program that uses the distance
form c = sqrt(a^2 + b^2) you will suffer from the lose of half of your
I read an interesting comment:
The coolest thing I've ever discovered about Pythagorean's Theorem is an
alternate way to calculate it. If you write a program that uses the distance
form c = sqrt(a^2 + b^2) you will suffer from the lose of half of your
available precision because the square
On Tue, 26 May 2015 05:21 am, ravas wrote:
I read an interesting comment:
The coolest thing I've ever discovered about Pythagorean's Theorem is an
alternate way to calculate it. If you write a program that uses the
distance form c = sqrt(a^2 + b^2) you will suffer from the lose of half of
Am 26.05.15 um 05:11 schrieb Steven D'Aprano:
mismatch after 3 trials
naive: 767.3916150255787
alternate: 767.3916150255789
hypot: 767.3916150255787
which shows that:
(1) It's not hard to find mismatches;
(2) It's not obvious which of the three methods is more accurate.
The main problem is
On Monday, May 25, 2015 at 8:11:25 PM UTC-7, Steven D'Aprano wrote:
Let's compare three methods.
...
which shows that:
(1) It's not hard to find mismatches;
(2) It's not obvious which of the three methods is more accurate.
Thank you; that is very helpful!
I'm curious: what about the
On Mon, May 25, 2015 at 1:21 PM, ravas ra...@outlook.com wrote:
I read an interesting comment:
The coolest thing I've ever discovered about Pythagorean's Theorem is an
alternate way to calculate it. If you write a program that uses the distance
form c = sqrt(a^2 + b^2) you will suffer from
Oh ya... true _
Thanks :D
On Monday, May 25, 2015 at 9:43:47 PM UTC-7, Ian wrote:
def distance(A, B):
A B are objects with x and y attributes
:return: the distance between A and B
dx = B.x - A.x
dy = B.y - A.y
a = min(dx, dy)
b = max(dx, dy)
On 05/25/2015 09:13 PM, ravas wrote:
On Monday, May 25, 2015 at 8:11:25 PM UTC-7, Steven D'Aprano wrote:
Let's compare three methods.
...
which shows that:
(1) It's not hard to find mismatches;
(2) It's not obvious which of the three methods is more accurate.
Thank you; that is very helpful!
On Monday, May 25, 2015 at 10:16:02 PM UTC-7, Gary Herron wrote:
It's probably not the square root that's causing the inaccuracies. In
many other cases, and probably here also, it's the summing of two
numbers that have vastly different values that loses precision. A
demonstration:
big
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