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