I have a somewhat weird issue. I'm supposed to take an inputted number, and if it's even divide it until it's odd. But. The code works inside the function, and the while loop, but it doesn't work when I try to return that value. It literally just returns the inputted value.
Code for reference:
num = int(input("Enter a number."))
def makeodd(num):
while num % 2 == 0:
num = num / 2
print(num)
return num
makeodd(num)
print(num)
I apologize that this is a bit basic. I'm looking for more of an explanation then a real solution. Although a solution would be nice. This is python 3.4 for reference. I looked around but couldn't actually find anything that was like this.
Change makeodd(num) to:
num = makeodd(num)
That puts the return value into num.
You could also use the return value directly in the print statement:
print(makeodd(num))
Related
Does anyone know what is going on with the following R code?
library(inline)
increment <- cfunction(c(a = "integer"), "
INTEGER(a)[0]++;
return R_NilValue;
")
print_one = function(){
one = 0L
increment(one)
print(one)
}
print_one() # prints 1
print_one() # prints 2
The printed results are 1, 2. Replacing one = 0L with one = integer(1) gives result 1, 1.
Just to clarify. I know I'm passing the variable one by reference to the C function, so its value should change (become 1). What I don't understand is that, how come resetting one = 0L seem to have no effect at all after the first call to print_one (the second call to print_one prints 2 instead of 1).
This really (as the last comment hinted) have little to do with Rcpp. It's really mostly about the .C() interface used by the old inline package and here by cfunction approach in the question.
That leads to two answers.
First, the consensus among R developers is that .C() is deprecated and should no longer be used. Statements to that effect can be found on the r-devel and r-package-devel lists. .C() uses plain old types as pointers in the interface so here the integer value is passed as int* by reference and can be altered.
If we switch to Rcpp uses, and hence to the underlying .Call() interface using only SEXP types for input and output, then an int no passes by reference. So the code behaves and prints only 0:
Rcpp::cppFunction("void increment2(int a) { a++; }")
print_two <- function(){
two <- 0L
increment2(two)
print(two)
}
print_two() # prints 0
print_two() # prints 0
Lastly, Rcpp (capital R) is of course not the "sucessor" to inline (as it does a whole lot more than inline but it among all its functionality is (since around 2013) a quasi-replacement for inline in Rcpp Attributes. So with Rcpp 'as-is' since about 2013 you no longer need the examples and approach from inline.
dictOne = {'a':1, 'b':2}
dictTwo = {'aa':dictOne['a'], 'bb':dictOne['b']}
print(dictTwo['aa']
# returns 1
If I make a change to dictOne like:
dictOne['a'] = 2
print(dictOne['a'])
# returns 2
but the second dict that refers to the first still returns the original value.
print(dictTwo['aa'])
# returns 1
What is happening here? I'm sure this is somehow an inappropriate usage of dict but I need to resolve this in the immediate. Thanks.
You're extracting the value from the key 'a' inside dictOne with the below piece of code
dictTwo = {'aa':dictOne['a']}
You may find some value in reading the python FAQ on pass by assignment
Without knowing more about the problem, it's difficult to say exactly how you can solve this. If you need to create a mapping between different sets of keys, there's the option to do something like:
dictTwo = {'aa' : 'a', 'bb' : 'b'}
dictOne[dictTwo['aa']]
Although maybe what you're looking for is a multi key dict
The line here:
dictTwo = {'aa':dictOne['a'], 'bb':dictOne['b']}
Is equivalent to:
dictTwo = {'aa':1, 'bb':2}
Since dictOne['a'] and dictOne['b'] both return immutable values (integers), they are passed by copy, and not by reference. See How do I pass a variable by reference?
Had you done dictTwo = dictOne, updating dictOne would also update dictTwo, however they would have the same key values.
I'm not getting the output I want. I don't understand why the result is duplicated. Can someone help me?
for $i in 1 to 2
let $rng:=random-number-generator()
let $rng1:=$rng('permute')(1 to 10)
let $rng:=$rng('next')()
let $rng2:=$rng('permute')(1 to 10)
let $rng:=$rng('next')()
let $rng3:=$rng('permute')(1 to 10)
return (string-join($rng1),string-join($rng2),string-join($rng3),",")
result:
23496815107
31018674529
31017684259
23496815107
31018674529
31017684259
The result is duplicated because of the initial for $i in 1 to 2, and because the variable $i is not actually used anywhere.
I edited the query based on your comment (getting 10 numbers). From what I understand, the difficulty here is to chain the calls (alternating between 'next' and 'permute'). Chaining calls can be done with a tail recursion.
declare function local:multiple-calls(
$rng as function(*),
$number-of-times as xs:integer) as item()* {
if($number-of-times le 0)
then ()
else
let $rng := $rng('next')
return ($rng('permute')(1 to 10),
local:multiple-calls($rng, $number-of-times - 1))
};
local:multiple-calls(random-number-generator(), 10)
Note: I am not sure if (1 to 10) is what needs to actually be passed to the call to $rng('permute'), or if it was an attempt to output ten numbers. In doubt, I haven't changed it.
The specification is here:
http://www.w3.org/TR/xpath-functions-31/#func-random-number-generator
It says:
Both forms of the function are ·deterministic·: calling the function
twice with the same arguments, within a single ·execution scope·,
produces the same results.
If you supply $i as the $seed argument to random-number-generator then the two sequences should be different.
I think I now understand what confuses you in this original query. One could indeed expect the random numbers to be generated differently for each iteration of $i.
However, XQuery is (to put it simply, with a few exceptions) deterministic. This means that the random generator probably gets initialized in each iteration with the same, default seed.
Thus, I have a second potential answer:
If you have a way to pass a different seed to $rng, you could slightly modify your initial query by constructing a seed based on $i and maybe current-dateTime() in each iteration before generating the numbers. But it will still be the same if you execute the query several times unless you involve the current date/time.
Is it possible to return multiple values from a function?
I want to pass the return values into another function, and I wonder if I can avoid having to explode the array into multiple values
My problem?
I am upgrading Capybara for my project, and I realized, thanks to CSS 'contains' selector & upgrade of Capybara, that the statement below will no longer work
has_selector?(:css, "#rightCol:contains(\"#{page_name}\")")
I want to get it working with minimum effort (there are a lot of such cases), So I came up with the idea of using Nokogiri to convert the css to xpath. I wanted to write it so that the above function can become
has_selector? xpath(:css, "#rightCol:contains(\"#{page_name}\")")
But since xpath has to return an array, I need to actually write this
has_selector?(*xpath(:css, "#rightCol:contains(\"#{page_name}\")"))
Is there a way to get the former behavior?
It can be assumed that right now xpath func is like the below, for brevity.
def xpath(*a)
[1, 2]
end
You cannot let a method return multiple values. In order to do what you want, you have to change has_selector?, maybe something like this:
alias old_has_selector? :has_selector?
def has_selector? arg
case arg
when Array then old_has_selector?(*arg)
else old_has_selector?(arg)
end
end
Ruby has limited support for returning multiple values from a function. In particular a returned Array will get "destructured" when assigning to multiple variables:
def foo
[1, 2]
end
a, b = foo
a #=> 1
b #=> 2
However in your case you need the splat (*) to make it clear you're not just passing the array as the first argument.
If you want a cleaner syntax, why not just write your own wrapper:
def has_xpath?(xp)
has_selector?(*xpath(:css, xp))
end
i would love to have functionality like this:
print(randomParameter(1,2,3))
-- prints 1 2 or 3... randomly picks a parameter
i have tried using the func(...) argument but i cant seem to use the table ARG when i pass multiple parameters. I tried this:
function hsv(...)
return arg[math.random(1,#arg)] -- also tried: return arg[math.random(#arg)]
end
print(hsv(5,32,7))
i have even tried putting the #arg into a variable using the rand function, also making a for loop with it sequentially adding a variable to count the table. still nothing works.
i remember doing this a while back, amd it looked different then this. can anyone Help with this? THANKS!
To elaborate a bit on #EgorSkriptunoff's answer (who needs to change his habit of providing answers in comments ;)): return (select(math.random(select('#',...)),...)).
... provides access to vararg parameter in the function
select('#', ...) returns the number of parameters passed in that vararg
math.random(select('#',...)) gives you a random number between 1 and the number of passed parameters
select(math.random(select('#',...)),...) gives you the element with the index specified by that random number from the passed parameters.
The other solution that is using arg = {...} gives you almost the same result with one subtle difference related to the number of arguments when nil is included as one of the parameters:
> function f(...) print(#{...}, select('#', ...)) end
> f(1,2,3)
3 3
> f(1,2,nil)
2 3
> f(1,2,nil,3)
2 4
As you can see select('#',...) produces more accurate results (this is running LuaJIT, but as far as I remember, Lua 5.1 produces similar results).
function randomNumber(...)
t = {...}
return t[math.random(1,#t)]
end
print(randomNumber(1, 5, 2, 9))
> 1 or 5 or 2 or 9