Recipe 1.19. Checking a String for Any of Multiple Endings
Credit: Michele
Simionato
Problem
For a certain string s, you must check
whether s has any of several endings; in
other words, you need a handy, elegant equivalent of
s.endswith(end1) or s.endswith(end2) or
s.endswith(end3) and so on.
Solution
The itertools.imap function is just as handy for
this task as for many of a similar nature:
import itertools
def anyTrue(predicate, sequence):
return True in itertools.imap(predicate, sequence)
def endsWith(s, *endings):
return anyTrue(s.endswith, endings)
Discussion
A typical use for endsWith might be to print all
names of image files in the current directory:
import os
for filename in os.listdir('.'):
if endsWith(filename, '.jpg', '.jpeg', '.gif'):
print filename
The same general idea shown in this recipe's
Solution is easily applied to other tasks related to checking a
string for any of several possibilities. The auxiliary function
anyTrue is general and fast, and you can pass it as
its first argument (the predicate) other bound
methods, such as s.startswith or s._
_contains_ _. Indeed, perhaps it would be better to do
without the helper function endsWithafter
all, directly coding
if anyTrue(filename.endswith, (".jpg", ".gif", ".png")):
seems to be already readable enough.
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Whenever a Python object supplies a method,
you can get the method, already bound to the
object, by just accessing the method on the
object. (For example, you can assign it, pass it as an argument,
return it as a function's result, etc.) For example:
L = ['fee', 'fie', 'foo']
x = L.append
Now, name x refers to a bound
method of list object L.
Calling, say, x('fum') is the same as calling
L.append('fum'): either call mutates object
L into ['fee', 'fie',
'foo', 'fum'].
If
you access a method on a type or class, rather than an instance of
the type or class, you get an unbound method,
not "attached" to any particular
instance of the type or class: when you call it, you need to pass as
its first argument an instance of that type or class. For example, if
you set y = list.append, you cannot just call
y('I')Python couldn't
possibly guess which list you want to append
I to! You can, however, call y(L,
'I'), and that is just the same as calling
L.append('I') (as long as isinstance(L,
list)).
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This recipe originates from a discussion on news:comp.lang.python. and summarizes inputs
from many people, including Raymond Hettinger, Chris Perkins, Bengt
Richter and others.
See Also
Library Reference and Python in a
Nutshell docs for itertools and string
methods.
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