Streamline FAQ entry about the ternary operator, and suggest using io.StringIO for a mutable unicode container.

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Antoine Pitrou 2011-12-03 22:11:11 +01:00
parent 9cb41dfbaa
commit c5b266efb5
1 changed files with 20 additions and 54 deletions

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@ -679,61 +679,21 @@ are not truly operators but syntactic delimiters in assignment statements.
Is there an equivalent of C's "?:" ternary operator?
----------------------------------------------------
Yes, this feature was added in Python 2.5. The syntax would be as follows::
Yes, there is. The syntax is as follows::
[on_true] if [expression] else [on_false]
x, y = 50, 25
small = x if x < y else y
For versions previous to 2.5 the answer would be 'No'.
Before this syntax was introduced in Python 2.5, a common idiom was to use
logical operators::
.. XXX remove rest?
[expression] and [on_true] or [on_false]
In many cases you can mimic ``a ? b : c`` with ``a and b or c``, but there's a
flaw: if *b* is zero (or empty, or ``None`` -- anything that tests false) then
*c* will be selected instead. In many cases you can prove by looking at the
code that this can't happen (e.g. because *b* is a constant or has a type that
can never be false), but in general this can be a problem.
Tim Peters (who wishes it was Steve Majewski) suggested the following solution:
``(a and [b] or [c])[0]``. Because ``[b]`` is a singleton list it is never
false, so the wrong path is never taken; then applying ``[0]`` to the whole
thing gets the *b* or *c* that you really wanted. Ugly, but it gets you there
in the rare cases where it is really inconvenient to rewrite your code using
'if'.
The best course is usually to write a simple ``if...else`` statement. Another
solution is to implement the ``?:`` operator as a function::
def q(cond, on_true, on_false):
if cond:
if not isfunction(on_true):
return on_true
else:
return on_true()
else:
if not isfunction(on_false):
return on_false
else:
return on_false()
In most cases you'll pass b and c directly: ``q(a, b, c)``. To avoid evaluating
b or c when they shouldn't be, encapsulate them within a lambda function, e.g.:
``q(a, lambda: b, lambda: c)``.
It has been asked *why* Python has no if-then-else expression. There are
several answers: many languages do just fine without one; it can easily lead to
less readable code; no sufficiently "Pythonic" syntax has been discovered; a
search of the standard library found remarkably few places where using an
if-then-else expression would make the code more understandable.
In 2002, :pep:`308` was written proposing several possible syntaxes and the
community was asked to vote on the issue. The vote was inconclusive. Most
people liked one of the syntaxes, but also hated other syntaxes; many votes
implied that people preferred no ternary operator rather than having a syntax
they hated.
However, this idiom is unsafe, as it can give wrong results when *on_true*
has a false boolean value. Therefore, it is always better to use
the ``... if ... else ...`` form.
Is it possible to write obfuscated one-liners in Python?
@ -852,15 +812,21 @@ the :ref:`string-formatting` section, e.g. ``"{:04d}".format(144)`` yields
How do I modify a string in place?
----------------------------------
You can't, because strings are immutable. If you need an object with this
ability, try converting the string to a list or use the array module::
You can't, because strings are immutable. In most situations, you should
simply construct a new string from the various parts you want to assemble
it from. However, if you need an object with the ability to modify in-place
unicode data, try using a :class:`io.StringIO` object or the :mod:`array`
module::
>>> s = "Hello, world"
>>> a = list(s)
>>> print(a)
['H', 'e', 'l', 'l', 'o', ',', ' ', 'w', 'o', 'r', 'l', 'd']
>>> a[7:] = list("there!")
>>> ''.join(a)
>>> sio = io.StringIO(s)
>>> sio.getvalue()
'Hello, world'
>>> sio.seek(7)
7
>>> sio.write("there!")
6
>>> sio.getvalue()
'Hello, there!'
>>> import array