The following script puzzles me. It creates two nested lists that
compare identically. After identical element assignments, the lists
are different. In one case, a single element is replaced. In the
other, an entire column is replaced.
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Hi Alex,
With all due respect to your well-deserved standing in the Python
community, I'm not convinced that equality shouldn't imply invariance
under identical operations.
Perhaps the most fundamental notion is mathematics is that the left and
right sides of an equation remain identical after any
oops! last sentence of 2nd paragraph in previous message should read
"If my expectation is NOT met ..."
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Hi Tim,
In your example, a & b are references to the same object. I agree they
should compare equally. But please note that a==b is True at every
point in your example, even after the ValueError raised by b.remove(1).
That's good consistent behavior.
My original example is a little different.
Yes. You stated it quite precisely. I believe l1==l2 should always
return True and l1==l3 should always be False. (unless l3 is reassigned
as l3=l1). Your idea of a separate operator for 'all elements have
numerically equal values at the moment of comparision' is a good one.
For want of a better
Considering the number of new programmers who get bit by automatic
coercion, I wish Dennis Ritchie had made some different choices when he
designed C. But then I doubt he ever dreamed it would become so wildly
successful.
Being a curmudgeon purist I'd actually prefer it if Python raised a
TypeErr
Yes (unless I was testing the assertion that the second envelope did
not contain counterfeits of the first)
Scott David Daniels wrote:
> Would you say that envelope containing five $100 bills is equal to
> an envelope containing five $100 bills with different serial numbers?
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Truthfully, I wouldn't mind it at all. In Python, I frequently write
things like
i == int(f)
or vice versa just to avoid subtle bugs that sometimes creep in when
later modifications to code change the original assumptions.
When working in C, I always set the compiler for maximum warnings and
d
I believe that 'is' tests equality of reference, such that
>>> a = range(1,3)
>>> b = range(1,3)
>>> a is b
False
The 'is' operator tells you whether a and b refer to the same object.
What I've been discussing is whether == should test for "structural"
equality so that a and b remain equivalent u
'. I haven't used such perjoratives in
any of my posts and would appreciate the same courtesy.
Cheers,
Mike
'''
StrongEquality -- a first cut at the definition proposed by M. Ellis.
Author: Michael F. Ellis, Ellis & Grant, Inc.
'''
def indices(ite
Hey Alex, lighten up! Python is a programming language -- not your
family, religion, or civil rights.
Cheers,
Mike
Alex Martelli wrote:
> <[EMAIL PROTECTED]> wrote:
>...
> > (As an aside, may I point out that Python In A Nutshell states on page
> > 46 "The result of S*n or n*S is the concatena
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