n a+1
add = reduce(myfunc, nums)
reduce would call your lambda function succesively with arguments
(1,2) # which would return 1+1 i.e. 2
(2,3) # use previous result (2) and next number in the sequence
(3); returns 2+1 i.e. 3
(3,4) # use previous result (3) and next number in
Shivlal Sharma wrote:
> from functools import*
> nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]
> add = reduce(lambda a : a + 1, nums)
> print(add)
>
> error: -
> TypeError Traceback (most recent call
> last) in ()
> 1 from functools import*
> 2 nums = [1, 2, 3, 4
from functools import*
nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]
add = reduce(lambda a : a + 1, nums)
print(add)
error: -
TypeError Traceback (most recent call last)
in ()
1 from functools import*
2 nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]
> 3 add = reduce(lambd
"fl" a écrit dans le message de
news:323866d1-b117-4785-ae24-7d04c49bc...@googlegroups.com...
Hi,
def make_incrementor(n):
... return lambda x: x + n
...
f = make_incrementor(42)
f(0)
42
f(1)
43
make_incrementor is a fonction which return a function !
and the returned function jus
On Tue, Jun 2, 2015 at 11:14 AM, fl wrote:
> Hi,
>
> I see the description of lambda at the online tutorial, but I cannot
> understand it. '42' is transferred to the function. What 'x' value should
> be? I do not see it says that it is '0'. And, what is 'x'?
>
The lambda keyword is merely anothe
Hi,
I see the description of lambda at the online tutorial, but I cannot
understand it. '42' is transferred to the function. What 'x' value should
be? I do not see it says that it is '0'. And, what is 'x'?
>>> def make_incrementor(n):
... return lambda x: x + n
...
>>> f = make_increm
This is the second part of my posting on the Turing completeness of
Python's lambda expressions. This time I am going to define a recursive
function as a lambda expression (I use lambda when I am talking about
Python's lambda expressions, and λ for the theory – λ calculus.)
Now of course it is eas
Musical Notation writes:
> Is it possible to write a Turing-complete lambda function (which does
> not depend on named functions) in Python?
The wording of this question is questionable. Turing completeness is not
an attribute of a function, but of a system (for example a programming
la
On Wed, Jul 31, 2013 at 11:55 PM, Steven D'Aprano
wrote:
> On Wed, 31 Jul 2013 13:53:26 +0700, Musical Notation wrote:
>
>> Is it possible to write a Turing-complete lambda function (which does
>> not depend on named functions) in Python?
>
>
> lambda s: e
On Wed, 31 Jul 2013 13:53:26 +0700, Musical Notation wrote:
> Is it possible to write a Turing-complete lambda function (which does
> not depend on named functions) in Python?
lambda s: eval(s)
--
Steven
--
http://mail.python.org/mailman/listinfo/python-list
On Wed, Jul 31, 2013 at 12:53 AM, Musical Notation
wrote:
> Is it possible to write a Turing-complete lambda function (which does not
> depend on named functions) in Python?
Yes, lambda functions are Turing-complete. You can get anonymous
recursion by defining the function to take a rec
On Wed 31 Jul 2013 08:53:26 AM CEST, Musical Notation wrote:
Is it possible to write a Turing-complete lambda function (which does not
depend on named functions) in Python?
what should a sinlge Turing-complete lambda function be?
For me, a programming language can be Turing-complete or a
Is it possible to write a Turing-complete lambda function (which does not
depend on named functions) in Python?
--
http://mail.python.org/mailman/listinfo/python-list
On Aug 11, 3:42 pm, Duncan Booth wrote:
> Terry wrote:
> > I'm trying to implement something like:
>
> > remote_map(fun, list)
>
> > to execute the function on a remove machine. But the problem is I
> > cannot pickle a lambda function and send it to
Terry wrote:
> I'm trying to implement something like:
>
> remote_map(fun, list)
>
> to execute the function on a remove machine. But the problem is I
> cannot pickle a lambda function and send it to the remote machine.
>
> Is there any possible way to pickle (o
Hi,
I'm trying to implement something like:
remote_map(fun, list)
to execute the function on a remove machine. But the problem is I
cannot pickle a lambda function and send it to the remote machine.
Is there any possible way to pickle (or other method) any functions
including lambda
En Wed, 25 Feb 2009 12:42:32 -0200, Albert Hopkins
escribió:
On Wed, 2009-02-25 at 17:56 +0530, aditya saurabh wrote:
I defined two functions - lets say
fa = lambda x: 2*x
fb = lambda x: 3*x
Now I would like to use fa*fb in terms of x
is there a way?
Thanks in advance
I'm not sure what "us
On Wed, 2009-02-25 at 17:56 +0530, aditya saurabh wrote:
> I defined two functions - lets say
> fa = lambda x: 2*x
> fb = lambda x: 3*x
> Now I would like to use fa*fb in terms of x
> is there a way?
> Thanks in advance
I'm not sure what "use fa*fb in terms of x" means.
But if you mean fa(x) * fb
I defined two functions - lets say
fa = lambda x: 2*x
fb = lambda x: 3*x
Now I would like to use fa*fb in terms of x
is there a way?
Thanks in advance
--
http://mail.python.org/mailman/listinfo/python-list
unction'). FWIW and WWAI, Python's
methods are thin callable wrappers around the function, the class and
(for bound methods) the instance.
I figured an easy way to do it would be to
create a lambda function that calls the correct method, but this is
proving more difficult tha
nter.
s/pointer/object/
There's nothing like a pointer in Python, and Python's functions are
plain objects (instances of class 'function'). FWIW and WWAI, Python's
methods are thin callable wrappers around the function, the class and
(for bound methods) the instance.
>
Justin Delegard wrote:
> So I am trying to pass an object's method call to a function that
> requires a function pointer. I figured an easy way to do it would be to
> create a lambda function that calls the correct method, but this is
> proving more difficult than I imagined
So I am trying to pass an object's method call to a function that
requires a function pointer. I figured an easy way to do it would be to
create a lambda function that calls the correct method, but this is
proving more difficult than I imagined.
Here is the function I'm u
j (which is declear with the lambda
> function), I can get expected answer [...]
'j' is not declared with the lambda function. Inside the lambda, 'j' is a
global name. Global names inside a function body are looked up when the
function is called, not when the function is def
result is very strange:
print f[k](1.),
4.0 4.0 4.0 4.0 4.0 expect 0. 1. 2. 3. 4.
print f[j](1.),
0.0 1.0 2.0 3.0 4.0 expect 0. 1. 2. 3. 4.
print f[0](1.), f[1](1.), ...
4.0 4.0 4.0 4.0 4.0
expect 0. 1. 2. 3. 4.
It seems only when I use the index j (which is declear with the lambda
funct
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