function with variable arguments

[Xah Lee]

i wanted to define a function where the number of argument matters.
Example:

def Range(n):
return range(n+1)

def Range(n,m):
return range(n,m+1)

def Range(n,m,step):
return range(n,m+1,step)

this obvious doesn't work. The default argument like
Range(n=1,m,step=1) obviously isn't a solution.

can this be done in Python?

or, must the args be changed to a list?

Xah
xa*@xahlee.org
∑ http://xahlee.org/

[Steve]

On 5/13/05, Wolfram Kriesing <wo**************@gmail.com> wrote:

using default args does actually solve it
what about
def Range(n, m=None, step=None)
if step==None:
if m==None:
range(n)
else:

<...snip...>
.....or better still :

def Range(*args):
return range(*args)

Regards
Steve

PS: but what do I know, I'm a F'ing imcompetent ass

[Dan Sommers]

On 13 May 2005 02:52:34 -0700,
"Xah Lee" <xa*@xahlee.org> wrote:

i wanted to define a function where the number of argument matters.
Example: def Range(n):
return range(n+1) def Range(n,m):
return range(n,m+1) def Range(n,m,step):
return range(n,m+1,step) this obvious doesn't work. The default argument like
Range(n=1,m,step=1) obviously isn't a solution. can this be done in Python?

Assuming you're doing something more interesting than wrapping range:

def Range( start, stop = None, step = 1 ):
if stop == None: # i,e., we only got one argument
stop = start
start = 1
# rest of function goes here....

HTH,
Dan

--
Dan Sommers
<http://www.tombstonezero.net/dan/>

[Peter Dembinski]

On Fri, 13 May 2005 11:52:34 +0200, Xah Lee <xa*@xahlee.org> wrote:

i wanted to define a function where the number of argument matters.
Example:

def Range(n):
return range(n+1)

def Range(n,m):
return range(n,m+1)

def Range(n,m,step):
return range(n,m+1,step)

this obvious doesn't work. The default argument like
Range(n=1,m,step=1) obviously isn't a solution.

can this be done in Python?

or, must the args be changed to a list?

It can be written this way:

def Range_3args(n, m, step):
return range(n, m + 1, step)

def Range_2args(n, m):
return range(n, m + 1)

def Range(n, m = None, step = None):
if (m is None) and (step is None):
return range(n + 1)

if (not (m is None)) and (step is None):
return Range_2args(n, m)

if (not (m is None)) and (not (step is None)):
return Return_3args(n, m, step)

return []
--
http://www.peter.dembinski.prv.pl

[PoD]

On Fri, 13 May 2005 02:52:34 -0700, Xah Lee wrote:

i wanted to define a function where the number of argument matters.
Example:

def Range(n):
return range(n+1)

def Range(n,m):
return range(n,m+1)

def Range(n,m,step):
return range(n,m+1,step)

this obvious doesn't work. The default argument like
Range(n=1,m,step=1) obviously isn't a solution.

can this be done in Python?

or, must the args be changed to a list?

def Range(n,m=None,step=1):
if m is None:
n,m = 0,n+1
else:
n,m = n,m+1
return range(n,m,step)

[Wolfram Kriesing]

> def Range(n,m=None,step=1):

if m is None:
n,m = 0,n+1
else:
n,m = n,m+1
return range(n,m,step)

i like this one.

coming from php (just a couple weeks ago) its always again interesting
to see how i have to start thinking to program differently, it can be
so much easier with python. i dont want to go back to php!

--
cu

Wolfram

[Xah Lee]

Thanks to all for the reply. (i should've known better)

on a related topic,
I think it would be a improvement for the built-in range() so that step
needs not be an integer.
Further, it'd be better to support decreasing range. e.g.

Range( 5, 7, 0.3); # returns [5, 5.3, 5.6, 5.9, 6.2, 6.5, 6.8]
Range( 5, -4, -2); # returns [5,3,1,-1,-3]

Xah
xa*@xahlee.org
∑ http://xahlee.org/

[tiissa]

Xah Lee wrote:

on a related topic,
I think it would be a improvement for the built-in range() so that step
needs not be an integer.
There are easy workarounds but I'd find it useful as well.
Further, it'd be better to support decreasing range. e.g.

Range( 5, 7, 0.3); # returns [5, 5.3, 5.6, 5.9, 6.2, 6.5, 6.8]
Range( 5, -4, -2); # returns [5,3,1,-1,-3]

The last one already works:

range(5,-4,-2)

[5, 3, 1, -1, -3]

[Harald Schmidt]

Xah Lee wrote:

I think it would be a improvement for the built-in range() so that step
needs not be an integer. [...]

Range( 5, 7, 0.3); # returns [5, 5.3, 5.6, 5.9, 6.2, 6.5, 6.8]

This may not return what you expect it to return.

For example let's use a naive implementation like this:

def Range(start, stop, step):
values = []
while start < stop:
values.append(start)
start += step
return values

The result is:

Range(5, 7, 0.3)

[5, 5.2999999999999998, 5.5999999999999996, 5.8999999999999995,
6.1999999999999993, 6.4999999999999991, 6.7999999999999989]

Worse: Range(5, 7.1, 0.3) would return 8 values, not 7 as expected from e.g.
range(50, 71, 3).

Welcome to the interesting world of floating point numbers.

Harald