What is the right way to make 1 from the string "one"

Ruby
1

Hello,

Im trying to understand currying.
I understand how you can add, substract two numbers by using currying.

But I want to make it a step harder.

How can I translate the word “one” by the number 1 so I can use it on a
add function.

I tried :

def one 1
Proc.new do |x|
x = 1
end
end

but that one does not give the 1 back.

Roelof

2

On 7 June 2014 17:48, Roelof W. [email protected] wrote:

Hello,

Im trying to understand currying.
I understand how you can add, substract two numbers by using currying.

​I think maybe you don’t actually understand what currying is.

Currying basically​ means: taking a function that takes N arguments, and
turning it into a nested sequence of N functions that each take 1
argument.
The practical use is “partial application”, where you assign values to
some
of the functions.

For example, take this lambda function:

mean = ->(a, b, c) do
(a + b + c) / 3
end
mean[1, 2, 6] # => 3

​It takes three parameters, and returns their geometric mean. It doesn’t
make sense to call it with only one parameter.

If I curry it, I get a new lambda function which takes one parameter,
and
returns a new lambda function which remembers that parameter, and
accepts
the next parameter, etc.:

curried = mean.curry # ~= mean(?, ?, ?)
mean_1 = curried[1] # ~= mean(1, ?, ?)
mean_1_2 = mean_1[2] # ~= mean(1, 2, ?)
mean_1_2[6] # ~= mean(1, 2, 3) => 3
​ mean_1_2[9] # ~= mean(1, 2, 9) => 4​
mean_1_3 = mean_1[3] # ~= mean(1, 3, ?)
mean_1_3[11] # ~= mean(1, 3, 11) => 5

But I want to make it a step harder.

end

but that one does not give the 1 back.

I can’t even work out the intent of that function, to give advice on how
to
make it work. Does this not do what you want?

def one
1
end
3 + one # => 4

​Cheers

3

On Saturday, June 07, 2014 11:37:32 AM Roelof W. wrote:

Roelof

Can you please ask your questions using plain text ? I am not able to
read
your lines. All lines are wrapped withing HTML.

–

Regards,
Arup R.

Debugging is twice as hard as writing the code in the first place.
Therefore,
if you write the code as cleverly as possible, you are, by definition,
not
smart enough to debug it.

–Brian Kernighan

4
Matthew K. schreef op 7-6-2014 11:33:
On 7 June 2014 17:48, Roelof W. <[email protected]> wrote:
Hello,

Im trying to understand currying.
I understand how you can add, substract two numbers by using currying.


???I think maybe you don't actually understand what currying is.

Currying basically??? means: taking a function that takes N arguments, and turning it into a nested sequence of N functions that each take 1 argument. The practical use is "partial application", where you assign values to some of the functions.

For example, take this lambda function:??

?? mean = ->(a, b, c) do
?? ?? (a + b + c) / 3
?? end
?? mean[1, 2, 6] # => 3

???It takes three parameters, and returns their geometric mean. It doesn't make sense to call it with only one parameter.

If I curry it, I get a new lambda function which takes one parameter, and returns a new lambda function which remembers that parameter, and accepts the next parameter, etc.:

?? curried = mean.curry??# ~= mean(?, ?, ?)
?? mean_1 = curried[1] ??# ~= mean(1, ?, ?)
?? mean_1_2 = mean_1[2] # ~= mean(1, 2, ?)
?? mean_1_2[6] ?? ?? ?? ?? ??# ~= mean(1, 2, 3) => 3
??? ??mean_1_2[9] ?? ?? ?? ?? ??# ~= mean(1, 2, 9) => 4???
?? mean_1_3 = mean_1[3] # ~= mean(1, 3, ?)
?? mean_1_3[11] ?? ?? ?? ?? # ~= mean(1, 3, 11) => 5


But I want to make it a step harder.

How can I ??translate the word "one" by the number 1 so I can use it on a add function.

I tried :

def one 1
?? ??Proc.new do ?? |x|
?? ?? ?? x = 1
?? ??end
end

but that one does not give the 1 back.

I can't even work out the intent of that function, to give advice on how to make it work. Does this not do what you want?

?? def one
?? ?? 1
?? end
?? 3 + one # => 4

??? Cheers
--

The exercise I try to do it this one

Suppose you have this

one(add(one)) which should provide 2

So I have two functions

def one
end

def add
end

Roelof

5

On 7 June 2014 19:37, Roelof W. [email protected] wrote:

end

def add
end

Ah, I begin to understand. I’d approach it from the other direction:

add = ->(a,b) { a + b }
one_plus = add.curry[1] # ~= add(1, ?)

one_plus[43] # => 44

Or, if you want to be obtuse:

one = 1
def one fn
fn[1]
end
def add x
->(y) { y + x }
end

one(add(one)) # => 2

However I’d argue that this is using a simple closure (the lambda
function
returned from #add) rather than currying. It also relies on the fact
that
one and one() are unambiguously a local variable and a function,
respectively; however if you were to use Ruby’s currying, you’d have a
proc
object rather than a function, and so couldn’t use the () syntax to
invoke
it, so the two definitions of one would be ambiguous.

6
Matthew K. schreef op 7-6-2014 12:06:
On 7 June 2014 19:37, Roelof W. <[email protected]> wrote:

The exercise I try to do it this one

Suppose you have this

one(add(one)) which should provide 2

So I have two functions

def one
end

def add
end


Ah, I begin to understand. I'd approach it from the other direction:

?? add = ->(a,b) { a + b }
?? one_plus = add.curry[1] # ~= add(1, ?)
????
?? one_plus[43] # => 44

Or, if you want to be obtuse:

?? one = 1
?? def one fn
?? ?? fn[1]
?? end
?? def add x
?? ?? ->(y) { y + x }
?? end
????
?? one(add(one)) # => 2

However I'd argue that this is using a simple closure (the lambda function returned from #add) rather than currying. It also relies on the fact that one and one() are unambiguously a local variable and a function, respectively; however if you were to use Ruby's currying, you'd have a proc object rather than a function, and so couldn't use the () syntax to invoke it, so the two definitions of `one` would be ambiguous.

--


oke, I already thoug you need a proc that why I tried

def one.
?? Proc.new do |x|
?????????? x = 1
???? end
end

But as I said this is not working.
And I think I need ruby currying to do the job.

Roelof

7

On Jun 8, 2014 3:13 AM, “Roelof W.” [email protected] wrote:

Currying is harder then I thought.

What you’re doing is not currying.

I gave you an example using currying, and an example that correctly
evaluates ´one(add(one))´

Here’s one without shadowing:

def one fn=nil
if fn
fn[1]
else
1
end
end
def add x
->(y) { y + x }
end

*I made this up off the top of my head, on my phone, in bed. It probably
works, though. #add is a partially applied, curried version of ´y + x´,
but
#one is just intentional obscurity

8
Roelof W. schreef op 7-6-2014 13:48:
Matthew K. schreef op 7-6-2014 12:06:
On 7 June 2014 19:37, Roelof W. <[email protected]> wrote:

The exercise I try to do it this one

Suppose you have this

one(add(one)) which should provide 2

So I have two functions

def one
end

def add
end


Ah, I begin to understand. I'd approach it from the other direction:

?? add = ->(a,b) { a + b }
?? one_plus = add.curry[1] # ~= add(1, ?)
????
?? one_plus[43] # => 44

Or, if you want to be obtuse:

?? one = 1
?? def one fn
?? ?? fn[1]
?? end
?? def add x
?? ?? ->(y) { y + x }
?? end
????
?? one(add(one)) # => 2

However 
            I'd argue that this is using a simple closure (the
            lambda function returned from #add) rather than
            currying. It also relies on the fact that </span><font
            face="courier new, monospace"

color=“#000000”>one and
one() are
unambiguously a local variable and a function,
respectively; however if you were to use Ruby’s
currying, you’d have a proc object rather than a
function, and so couldn’t use the () syntax to invoke
it, so the two definitions of one would be
ambiguous.





–









oke, I already thoug you need a proc that why I tried



def one.

?? Proc.new do |x|

??? x = 1

??? end

end



But as I said this is not working.

And I think I need ruby currying to do the job.



Roelof






I think I know how to solve this.

Can I also yield to a specific block and can I save the outcome like
this : x = yield



Roelof


9
Matthew K. schreef op 8-6-2014 1:37:


On Jun 8, 2014 3:13 AM, "Roelof W." <[email protected]> wrote:
>
> Currying is harder then I thought.
>

What you're doing is not currying.

I gave you an example using currying, *and* an example that correctly evaluates ??one(add(one))??

Here's one without shadowing:

?? def one fn=nil
?????? if fn
?????????? fn[1]
?????? else
?????????? 1
?????? end
?? end
?? def add x
?????? ->(y) { y + x }
?? end

*I made this up off the top of my head, on my phone, in bed. It probably works, though. #add is a partially applied, curried version of ??y + x??, but #one is just intentional obscurity



Thanks,

This works and I have studie this and the rest of this topic.
If I understand everything right it works like this :

one(add(one))??
function one will be executed.
if fn = right??
?? fn[1] will be add[1, one]
add will?? be executed
(y)?? { 1+ one}
so one will be executed again.
if fn = false
1
(y) { 1+1}
y = 2

Do I understand everything well ?

Roelof


10

On 8 June 2014 18:57, Roelof W. [email protected] wrote:

add will be executed
(y) { 1+ one}
so one will be executed again.
if fn = false
1
(y) { 1+1}
y = 2

Do I understand everything well ?

Pretty much, but the execution order is the opposite. ​The inner-most
parameters have to be evaluated before they can be passed to the
function,
so first is executed the inner one, then add(...), then the outer
one(...).​

11
Roelof W. schreef op 7-6-2014 14:07:
Roelof W. schreef op 7-6-2014 13:48:
Matthew K. schreef op 7-6-2014 12:06:
On 7 June 2014 19:37, Roelof W. <[email protected]> wrote:

The exercise I try to do it this one

Suppose you have this

one(add(one)) which should provide 2

So I have two functions

def one
end

def add
end


Ah, I begin to understand. I'd approach it from the other direction:

?? add = ->(a,b) { a + b }
?? one_plus = add.curry[1] # ~= add(1, ?)
????
?? one_plus[43] # => 44

Or, if you want to be obtuse:

?? one = 1
?? def one fn
?? ?? fn[1]
?? end
?? def add x
?? ?? ->(y) { y + x }
?? end
????
?? one(add(one)) # => 2

However 
              I'd argue that this is using a simple closure (the
              lambda function returned from #add) rather than
              currying. It also relies on the fact that </span><font
              face="courier new, monospace"

color=“#000000”>one and
one() are
unambiguously a local variable and a function,
respectively; however if you were to use Ruby’s
currying, you’d have a proc object rather than a
function, and so couldn’t use the () syntax to invoke
it, so the two definitions of one would be
ambiguous.





–









oke, I already thoug you need a proc that why I tried



def one.

?? Proc.new do |x|

??? x = 1

??? end

end



But as I said this is not working.

And I think I need ruby currying to do the job.



Roelof






I think I know how to solve this.

Can I also yield to a specific block and can I save the outcome
like this : x = yield



Roelof








Chips, still not working.



When I try this :

??

def one

??? 1

end



def add(x,y)

??Proc.new do |x,y|

??? y + x

?? end

end



puts one(add(one))



I see these errors :



main.rb:8: warning: shadowing outer local variable - x
main.rb:8: warning: shadowing outer local variable - y
main.rb:7:in add': wrong number of arguments (1 for 2) (ArgumentError) from main.rb:14:in



Currying is harder then I thought.



Roelof






12
Matthew K. schreef op 8-6-2014 13:13:
On 8 June 2014 18:57, Roelof W. <[email protected]> wrote:

Thanks,

This works and I have studie this and the rest of this topic.
If I understand everything right it works like this :

one(add(one))??
function one will be executed.
if fn = right??
?? fn[1] will be add[1, one]
add will?? be executed
(y)?? { 1+ one}
so one will be executed again.
if fn = false
1
(y) { 1+1}
y = 2

Do I understand everything well ?

Pretty much, but the execution order is the opposite. ???The inner-most parameters have to be evaluated before they can be passed to the function, so first is executed the inner `one`, then `add(...)`, then the outer `one(...)`.???

--

Oke,
Thanks for the help, explantion andt the patience with me.
Still I find it a wierd order.

Roelof

13

On Mon, Jun 9, 2014 at 11:09 AM, Roelof W. [email protected] wrote:

If I understand everything right it works like this :
(y) { 1+1}
–

end
if fn
3

fn[6]

end
def nine fn=nil

end

four(plus(nine))

and now I see this error :

main.rb:83:in plus': wrong number of arguments (1 for 0) (ArgumentError) from main.rb:99:in ’
Roelof

Your methods need to receive an argument:

def plus x
->(y) {y + x}
end

The argument they receive is the one that is curried:

2.0.0p195 :023 > %w{one two three four five six seven eight
nine}.each_with_index do |x, i|
2.0.0p195 :024 > define_method(x) do |fn=nil|
2.0.0p195 :025 > fn.nil? ? i+1 : fn[i+1]
2.0.0p195 :026?> end
2.0.0p195 :027?> end

2.0.0p195 :030 > def plus x
2.0.0p195 :031?> ->(y) {y + x}
2.0.0p195 :032?> end
=> nil
2.0.0p195 :033 > one(plus(one))
=> 2
2.0.0p195 :034 > def minus x
2.0.0p195 :035?> ->(y) {y - x}
2.0.0p195 :036?> end
=> nil
2.0.0p195 :037 > def times x
2.0.0p195 :038?> ->(y) {y * x}
2.0.0p195 :039?> end

2.0.0p195 :042 > three(times(one(plus(one))))
=> 6

Jesus.

14

Jes??s Gabriel y Gal??n schreef op 9-6-2014 12:06:

If I understand everything right it works like this :
(y) { 1+1}
Matthew K.

 fn[2]

end
def five fn=nil
else
end
if fn
def minus

2.0.0p195 :026?> end
2.0.0p195 :036?> end
=> nil
2.0.0p195 :037 > def times x
2.0.0p195 :038?> ->(y) {y * x}
2.0.0p195 :039?> end

2.0.0p195 :042 > three(times(one(plus(one))))
=> 6

Jesus.

Thanks, still I wonder why on the simple case the x is not needed .

Roelof

15
Roelof W. schreef op 8-6-2014 22:39:
Matthew K. schreef op 8-6-2014 13:13:
On 8 June 2014 18:57, Roelof W. <[email protected]> wrote:

Thanks,

This works and I have studie this and the rest of this topic.
If I understand everything right it works like this :

one(add(one))??
function one will be executed.
if fn = right??
?? fn[1] will be add[1, one]
add will?? be executed
(y)?? { 1+ one}
so one will be executed again.
if fn = false
1
(y) { 1+1}
y = 2

Do I understand everything well ?

Pretty 
          much, but the execution order is the opposite. ???The
          inner-most parameters have to be evaluated before they can
          be passed to the function, so first is executed the inner
          `one`, then `add(...)`, then the outer

one(...).???





–







Oke,

Thanks for the help, explantion andt the patience with me.

Still I find it a wierd order.



Roelof








Still a wierd problem.

I did make this with your example.



def zero fn=nil

?? if fn

??? fn[0]

?? else

??? 0

?? end

end



def one fn=nil

?? if fn

??? fn[1]

?? else

??? 1

?? end

end



def two fn=nil

?? if fn

??? fn[2]

?? else

??? 2

?? end

end



def three fn=nil

?? if fn

??? fn[3]

?? else

??? 3

?? end

end



def four fn=nil

?? if fn

??? fn[4]

?? else

??? 4

?? end

end



def five fn=nil

?? if fn

??? fn[5]

?? else

??? 5

?? end

end



def six fn=nil

?? if fn

??? fn[6]

?? else

??? 6

?? end

end



def seven fn=nil

?? if fn

??? fn[7]

?? else

??? 7

?? end

end



def eight fn=nil

?? if fn

??? fn[8]

?? else

??? 8

?? end

end



def nine fn=nil

?? if fn

??? fn[9]

?? else

??? 9

?? end

end



def plus

?? ->(y) { y + x }

end



def minus

?? ->(y) { y - x }

end



def times

?? ->(y) { y * x }

end



def divided_by

?? ->(y) { y / x }

end



four(plus(nine))



and now I see this error :



main.rb:83:in plus': wrong number of arguments (1 for 0) (ArgumentError) from main.rb:99:in
'

Roelof





16

On Mon, Jun 9, 2014 at 12:29 PM, Roelof W. [email protected] wrote:

Thanks, still I wonder why on the simple case the x is not needed .

What do you mean by ‘the simple case’?

Jesus.

17

Jes??s Gabriel y Gal??n schreef op 9-6-2014 12:37:

On Mon, Jun 9, 2014 at 12:29 PM, Roelof W. [email protected] wrote:

Thanks, still I wonder why on the simple case the x is not needed .
What do you mean by ‘the simple case’?

Jesus.

this one :

def one fn=nil
if fn
fn[1]
else
1
end
end

def add x
->(y) { y + x }
end

18

Jes??s Gabriel y Gal??n schreef op 9-6-2014 12:37:

On Mon, Jun 9, 2014 at 12:29 PM, Roelof W. [email protected] wrote:

Thanks, still I wonder why on the simple case the x is not needed .
What do you mean by ‘the simple case’?

Jesus.

This one :

19

On Mon, Jun 9, 2014 at 12:50 PM, Roelof W. [email protected] wrote:

def add x

         ^^^

you do have the parameter there.

->(y) { y + x }

end

Jesus.