Re: retain values between fun calls
Gary Wessle wrote:
Hi
the second argument in the functions below suppose to retain its value
between function calls, the first does, the second does not and I
would like to know why it doesn't?
Fisrt thing to remember is that function's default args are eval'd only
once - when the def block is eval'd, which is usually at import time.
# it does
def f(a, L=[]):
L.append(a)
Here, you are modifying the (list) object bound to local name L.
return L
print f('a')
print f('b')
# it does not
def f(a, b=1):
b = a + b
And here, you are *rebinding* the local name b to another object.
Understand that modifying an object in-place and rebinding a name are
two really different operations...
return b
print f(1)
print f(2)
and how to make it so it does?
The QD solution is to wrap :
def f(a, b = [1])
b[0] = b[0] + a
return b[0]
But this is really a hack - it's ok for a throw-away script, but may not
be the best thing to do in a more serious piece of software.
The clean solution to maintain state between calls is to use a custom
class - hopefully, Python is OO enough to let you write your own callables:
class MyFunc(object):
def __init__(self, initstate=1):
self._state = default
def __call__(self, a):
self._state += a
return self._state
f = MyFunc(1)
print f(1)
print f(2)
HTH
--
bruno desthuilliers
python -c print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for
p in '[EMAIL PROTECTED]'.split('@')])
--
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Re: retain values between fun calls
George Sakkis wrote:
Gary Wessle wrote:
Hi
the second argument in the functions below suppose to retain its value
between function calls, the first does, the second does not and I
would like to know why it doesn't? and how to make it so it does?
thanks
# it does
def f(a, L=[]):
L.append(a)
return L
print f('a')
print f('b')
# it does not
def f(a, b=1):
b = a + b
return b
print f(1)
print f(2)
It's a FAQ:
http://www.python.org/doc/faq/general/#why-are-default-values-shared-between-objects.
Whenever you want to control one or more objects beyond the lifetime of
a single function call, your first thought should be to use a class to
couple behaviour with state:
class SomeFancyClassName(object):
def __init__(self, b=1):
self.b = b
def f(self, a):
self.b += a
return self.b
x = SomeFancyClassName()
print x.f(1)
print x.f(2)
If you make the class callable you can match the original syntax:
In [40]: class F(object):
: b=1
: def __call__(self, a):
: F.b += a
: return F.b
:
:
In [41]: f=F()
In [42]: f(1)
Out[42]: 2
In [43]: f(2)
Out[43]: 4
Alternately you can use an attribute of the function to save the state:
In [35]: def f(a):
: f.b += a
: return f.b
:
In [36]: f.b=1
In [37]: f(1)
Out[37]: 2
In [38]: f(2)
Out[38]: 4
Kent
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retain values between fun calls
Hi
the second argument in the functions below suppose to retain its value
between function calls, the first does, the second does not and I
would like to know why it doesn't? and how to make it so it does?
thanks
# it does
def f(a, L=[]):
L.append(a)
return L
print f('a')
print f('b')
# it does not
def f(a, b=1):
b = a + b
return b
print f(1)
print f(2)
--
http://mail.python.org/mailman/listinfo/python-list
Re: retain values between fun calls
Gary Wessle wrote:
Hi
the second argument in the functions below suppose to retain its value
between function calls, the first does, the second does not and I
would like to know why it doesn't? and how to make it so it does?
thanks
# it does
def f(a, L=[]):
L.append(a)
return L
print f('a')
print f('b')
# it does not
def f(a, b=1):
b = a + b
return b
print f(1)
print f(2)
It's a FAQ:
http://www.python.org/doc/faq/general/#why-are-default-values-shared-between-objects.
Whenever you want to control one or more objects beyond the lifetime of
a single function call, your first thought should be to use a class to
couple behaviour with state:
class SomeFancyClassName(object):
def __init__(self, b=1):
self.b = b
def f(self, a):
self.b += a
return self.b
x = SomeFancyClassName()
print x.f(1)
print x.f(2)
HTH,
George
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