Velocity Reviews > Recursive function

# Recursive function

Ana Dionísio
Guest
Posts: n/a

 03-05-2013
Hello!

I have to make a script that calculates temperature, but one of the
parameters is the temperature in the iteration before, for example:
temp = (temp_-1)+1

it = 0
temp = 3

it = 1
temp = 3+1

it = 2
temp = 4+1

How can I do this in a simple way?

Thanks a lot!

Dave Angel
Guest
Posts: n/a

 03-05-2013
On 03/05/2013 10:32 AM, Ana Dionísio wrote:
> Hello!
>
> I have to make a script that calculates temperature, but one of the
> parameters is the temperature in the iteration before, for example:
> temp = (temp_-1)+1
>
> it = 0
> temp = 3
>
> it = 1
> temp = 3+1
>
> it = 2
> temp = 4+1
>
> How can I do this in a simple way?
>
> Thanks a lot!
>

Recursion only works when you have a termination condition. Otherwise,
it just loops endlessly (or more specifically until the stack runs out
of room). Still, you don't have a problem that even implies recursion,
just iteration.

For your simple case, there's a simple formula:

def find_temp(it):
return it+3

Methinks you have simplified the problem too far to make any sense.

import itertools
temp = 3
for it in itertools.count():
temp += 1
print it, temp

Warning: that loop will never terminate.

Is this what you were looking for? It uses recursion, and I used <=0
for the termination condition.

def find_temp2(it):
if it <=0: return 3
return find_temp(it-1)

--
DaveA

Vlastimil Brom
Guest
Posts: n/a

 03-05-2013
2013/3/5 Ana Dionísio <(E-Mail Removed)>:
> Hello!
>
> I have to make a script that calculates temperature, but one of the
> parameters is the temperature in the iteration before, for example:
> temp = (temp_-1)+1
>
> it = 0
> temp = 3
>
> it = 1
> temp = 3+1
>
> it = 2
> temp = 4+1
>
> How can I do this in a simple way?
>
> Thanks a lot!
> --
> http://mail.python.org/mailman/listinfo/python-list

Hi,
it is not quite clear from the examples, what should be achieved (I
guess, the actual computation is probably mor complex).
I'd probably approach an iterative computation iteratively, rather
than recursively;
e.g. simply:

def compute_iteratively(starting_value, number_of_iterations):
tmp = starting_value
for i in range(number_of_iterations):
tmp = tmp + 1
return tmp

print(compute_iteratively(starting_value=7, number_of_iterations=3))

hth,
vbr

Ana Dionísio
Guest
Posts: n/a

 03-05-2013
Yes, I simplified it a lot.

I need to run it 24 times. What I don't really understand is how to put the final temperature (result) in it = 0 in temp_-1 in it =1

Ana Dionísio
Guest
Posts: n/a

 03-05-2013
Yes, I simplified it a lot.

I need to run it 24 times. What I don't really understand is how to put the final temperature (result) in it = 0 in temp_-1 in it =1

Neil Cerutti
Guest
Posts: n/a

 03-05-2013
On 2013-03-05, Ana Dion?sio <(E-Mail Removed)> wrote:
> Yes, I simplified it a lot.
>
> I need to run it 24 times. What I don't really understand is
> how to put the final temperature (result) in it = 0 in temp_-1
> in it =1

One way to compose a function that recurses by calling itself is
to first write the simplest version of the function you can think
of, the version that solves the most trivial form for the
problem.

For example, lets say I want to write a recursive function to
reverse a string (you would never do this except as an exercise),
I'd first write a version that can successfully reverse an empty
string.

def reverse0(s):
if len(s) == 0:
return s

The next most complex problem is a string with one character.
I'll write that next. They key is to write it in terms of the

def reverse1(s):
if len(s) == 1:
return s + reverse0('')

At this point it becomes clear that reverse0 is not needed. I'll
just use a new version of reverse1 instead:

def reverse1(s):
if len(s) <= 1:
return s

The next vesion will be able to handle two characters. I'll write
it to use my previously composed reverse1 function.

def reverse2(s):
if len(s) == 2:
return s[-1] + reverse1(s[:-1])

And so on:

def reverse3(s):
if len(s) == 3:
return s[-1] + reverse2(s[:-1])

def reverse4(s):
if len(s) == 4:
return s[-1] + reverse3(s[:-1])

By this time a pattern has emerged; every version of reverse
after reverse1 is exactly the same. I can now write my recursive
reverse function.

def reverse_any(s):
if len(s) <= 1:
return s
else:
return s[-1] + reverse_any(s[:-1])

Try this exercise with your conversion problem and see if you can
make progress.

--
Neil Cerutti