Rule of order for dot operators?

[Albert van der Horst]

In article <mailman.118.1431989304.17265.python-list at python.org>,
Cameron Simpson  <cs at zip.com.au> wrote:
>On 16May2015 12:20, C.D. Reimer <chris at cdreimer.com> wrote:
>>title = slug.replace('-',' ').title()
>>This line also works if I switched the dot operators around.
>>title = slug.title().replace('-',' ')
>>
>>I'm reading the first example as character replacement first and title
>>capitalization second, and the second example as title capitalization
>>first and character replacement second.
>>
>>Does python perform the dot operators from left to right or according
>>to a rule of order (i.e., multiplication/division before add/subtract)?
>
>I've been thinking about the mindset that asks this question.
>
>I think the reason you needed to ask it was that you were thinking in terms of
>each .foo() operation acting on the original "slug". They do not.
>
>"slug" is an expression returning, of course, itself.
>
>"slug.title()" is an expression returning a new string which is a titlecased
>version of "slug".

Why is "slug.title" a valid decomposition of the total string>
(Or is it?)
What is the ()-brackets doing? Does it force the execution of title,
which gives something to be dotted onto slug etc. See below.

>
>"slug.title().replace(...)" is an expression which _operates on_ that new
>string, _not_ on "slug".
>
>In particular, each .foo() need not return a string - it might return anything,
>and the following .bah() will work on that anything.

I interpreted the question as about the associative of the
dot operator.

title = slug.title().replace('-',' ')

Does that mean
title = slug.( title().replace('-',' ') )
or

title = ( slug.( title()) .replace('-',' ')

or is it even    <slot>.<slot>.<slot> a ternary operator with
special syntax like we have in
    <slot> <= <slot> <= <slot>
lately.

It seems to me that you presuppose the answer.

>
>Cheers,
>Cameron Simpson <cs at zip.com.au>
-- 
Albert van der Horst, UTRECHT,THE NETHERLANDS
Economic growth -- being exponential -- ultimately falters.
albert at spe&ar&c.xs4all.nl &=n http://home.hccnet.nl/a.w.m.van.der.horst