this recursive definition of sum thrumped me, is this some sort of
gotcha or am I just braindead today?
and yes i know this is easy a a for x in xs acc += x or just using the
builtin.
def suma(xs, acc=0):
if len(xs) == 0:
acc
else:
suma(xs[1:], acc+xs[0])
it returns none.
def summa(xs):
if not xs:
0
else:
xs[0]+summa(xs[1:])
Traceback (most recent call last):
File "<pyshell#6 >", line 1, in <module>
summa([1,2,3,4,5])
File "<pyshell#5 >", line 5, in summa
xs[0]+summa(xs[1:])
File "<pyshell#5 >", line 5, in summa
xs[0]+summa(xs[1:])
File "<pyshell#5 >", line 5, in summa
xs[0]+summa(xs[1:])
File "<pyshell#5 >", line 5, in summa
xs[0]+summa(xs[1:])
File "<pyshell#5 >", line 5, in summa
xs[0]+summa(xs[1:])
TypeError: unsupported operand type(s) for +: 'int' and 'NoneType' 5 1574
On Sep 14, 9:01*am, cnb <circularf...@y ahoo.sewrote:
def suma(xs, acc=0):
* * * * if len(xs) == 0:
* * * * * * * * acc
* * * * else:
* * * * * * * * suma(xs[1:], acc+xs[0])
it returns none.
Yep, that's because there is no "return" statement anywhere. Python
doesn't return expressions "by default", like functional languages do,
so where you say "suma(xs[1:], acc+xs[0])" this just calls itself and
returns nothing.
Try this:
def suma(xs, acc=0):
if len(xs) == 0:
return acc
else:
return suma(xs[1:], acc+xs[0])
print suma([1, 2, 3, 4, 5])
(prints 15)
Roman
On Sun, Sep 14, 2008 at 10:01 AM, cnb <ci**********@y ahoo.sewrote:
this recursive definition of sum thrumped me, is this some sort of
gotcha or am I just braindead today?
and yes i know this is easy a a for x in xs acc += x or just using the
builtin.
def suma(xs, acc=0):
if len(xs) == 0:
acc
else:
suma(xs[1:], acc+xs[0])
You're just missing the "return" statements?
def suma(xs, acc=0):
if len(xs) == 0:
return acc
else:
return suma(xs[1:], acc+xs[0])
Regards
Marco
--
Marco Bizzarri http://notenotturne.blogspot.com/ http://iliveinpisa.blogspot.com/
On Sun, Sep 14, 2008 at 10:08 AM, Marco Bizzarri
<ma************ @gmail.comwrote :
On Sun, Sep 14, 2008 at 10:01 AM, cnb <ci**********@y ahoo.sewrote:
>this recursive definition of sum thrumped me, is this some sort of gotcha or am I just braindead today? and yes i know this is easy a a for x in xs acc += x or just using the builtin.
def suma(xs, acc=0): if len(xs) == 0: acc else: suma(xs[1:], acc+xs[0])
You're just missing the "return" statements?
def suma(xs, acc=0):
if len(xs) == 0:
return acc
else:
return suma(xs[1:], acc+xs[0])
Besides: you can avoid the "acc" parameter:
def suma(xs):
if len(xs) == 0:
return 0
else:
return xs[0] + suma(xs[1:])
Regards
Marco
--
Marco Bizzarri http://notenotturne.blogspot.com/ http://iliveinpisa.blogspot.com/
On Sep 14, 9:44*am, "Marco Bizzarri" <marco.bizza... @gmail.comwrote :
On Sun, Sep 14, 2008 at 10:08 AM, Marco Bizzarri
<marco.bizza... @gmail.comwrote :
On Sun, Sep 14, 2008 at 10:01 AM, cnb <circularf...@y ahoo.sewrote:
this recursive definition of sum thrumped me, is this some sort of
gotcha or am I just braindead today?
and yes i know this is easy a a for x in xs acc += x or just using the
builtin.
def suma(xs, acc=0):
* * * *if len(xs) == 0:
* * * * * * * *acc
* * * *else:
* * * * * * * *suma(xs[1:], acc+xs[0])
You're just missing the "return" statements?
def suma(xs, acc=0):
* * * if len(xs) == 0:
* * * * * * *return acc
* * * else:
* * * * * * *return suma(xs[1:], acc+xs[0])
Besides: you can avoid the "acc" parameter:
def suma(xs):
* * if len(xs) == 0:
* * * * return 0
* * else:
* * * * return xs[0] + suma(xs[1:])
I think the OP tried to make it tail-recursive, which of course has no
benefit in Python. In fact it looks like a Scheme implementation of
sum translated literally to Python.
In Python this algorithm is expressed naturally as:
def suma(xs):
acc = 0
for x in xs:
acc += x
return acc
--
Arnaud
cnb wrote:
this recursive definition of sum thrumped me, is this some sort of
gotcha or am I just braindead today?
and yes i know this is easy a a for x in xs acc += x or just using the
builtin.
def suma(xs, acc=0):
if len(xs) == 0:
acc
else:
suma(xs[1:], acc+xs[0])
it returns none.
Without return statement, the only recursive solution is a lambda expr :
>>suma = lambda xs : xs[0]+suma(xs[1:]) if xs else 0
>>suma(range(10 1))
5050
Note that suma(xs[1:]) implies copying the remainder of xs, what in turn makes
the time grow quadratically with the length of xs. So instead of passing a
superfluous acc second variable, you could pass an index into the list, eg
def suma(xs,pos=0) :
if pos>=len(xs) :
return 0
else :
return xs[pos]+suma(xs,pos+1)
Cheers, BB This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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