Enumerable.Range(1, 999).Select((n,i) =>{ return n*i;})
what does the "i" get in every time?
Enumerable.Range(1, 999).Select((n,i,j) =>{ return n*i*j;})
why cant I add "j"?
The Select overload accepting a lambda with two parameters will take the first parameter from the sequence, and the second is the index of the element.
In your example i will always be n-1, so there is not much use of the second parameter. When working with non-trivial sequences or sequences of non-numeric types it can sometimes be an advantage to have the order number of the element available in the select expression.
There is no three parameter version. That's why (n,i,j) doesn't work.
The first argument to the Select() extension method on IEnumerable has two forms. One takes one argument (the current element of the enumeration) and the second two arguments (the current element and the index). There is no version that takes three arguments. See http://msdn.microsoft.com/en-us/library/bb548891.aspx for more information.
The i is the name you've used for the index of the element you are currently projecting. You can't specify a third parameter because there is no overload of Select which defines a delegate that takes three parameters.
You can either build new elements on the basis of value (Select(n)) or on the basis of value and index (Select(n, i)). Select() with three parameters is not defined.
Compare: http://msdn.microsoft.com/en-us/library/system.linq.enumerable.select
Related
Suppose I have several arrays in Ruby which I add/subtract values and afterwards I limit their range, like so:
array[x][y]=array[x][y]+1
array[x][y]=array[x][y].clamp (0..99)
Since I have many different arrays with rather long (index) names - and in order not to repeat those names twice in one line, I'd like to achieve something like
array[x][y]+=1.clamp (0..99)
Which is accepted by the interpreter, but doesn't work. It adds, but the value in the array does not get clamped.
Splitting it in at least two lines
array[x][y]+=1
array[x][y].clamp(0..99)
does also add, but doesn't clamp.
Is there any solution for this to fit the entire command in one line?
Many thanks!
The #clamp method doesn't take a range as a single argument for Ruby versions before 2.7, but rather two arguments representing the min and max, and #clamp does not mutate the object it's called on.
array[x][y] = (array[x][y] + 1).clamp(0, 99)
Note that because it's valid to call a method without parentheses, if parentheses are used around an argument list, there should not be any space between the method name and the parentheses. E.g. 1.clamp(0..4) rather than 1.clamp (0..4).
In the problem that I want to solve a well defined matrix has no empty rows or columns.
For example the matrix [[],[]] is not valid.
When I call the function first_column, how do I prevent to execute it if the matrix that I send as an argument is not valid as defined before?
first_column([],[],[]).
first_column([[H|T]|Tail],[H|Col],[T|Rows]):- first_column(Tail,Col,Rows).
Technically, what you're asking can be done by testing for an end-condition of a list with one element, rather than an empty list, based on the specs you gave.
first_column([[H|T]],[H],[T]).
first_column([[H|T]|Tail],[H|Col],[T|Rows]):- first_column(Tail,Col,Rows).
However, beyond your specs, I suspect that you'll also need to "transfer" your final Col,Rows to end variables, something like:
first_column([[H|T]],C,R,[H|C],[T|R]).
first_column([[H|T]|Tail],[H|C],[T|R],Col,Rows):-
first_column(Tail,C,R,Col,Rows).
The modified predicate would be called with initial conditions, like
first_column(List,[],[],Col,Rows).
I have this parameter as an array. The array is big, 100 cells. It is a parameter that can be initiated in omnet.ini file. The cells with even numbers should get value A and odd numbers should get value B. How can I do this in an automated manner?
Is there a way besides having all odd and even indices initiated one by one manually?
Wildcards can be useful but I do not know how to use them to separate odd and even indices.
Thanks.
You can access the actual module index with the index operator. Combining this with the conditional operator ?: you can easily define the value:
**.myModule[*].myParameter = index % 2 == 0 ? "A" : "B"
I'm not aware of any feature like this. There are a number of work-arounds you could use:
Provide two parameters and select the correct one in code
Use the volatile keyword (probably not appropriate here)
Put the entire thing in your .ini file
I'd personally implement the first approach, that way you can use the wildcard to pass both parameters ([*].myNode.parameterEven and [*].myNode.parameterUneven) and then set the correct values in your array in a for loop.
However, you could also use the volatile keyword in your NED file, see the manual for more details. However, this approach mostly works well if you have different parameters depending on which node you are assigning it to. For this case I think the first approach is better.
The last alternative is just putting the entire thing in your .ini file, which may be useful if you want to parameterize the array later.
My rule requires me to apply them only to methods without 'get' as part of their name. In another words, my rules need to apply to only non-getter methods in the class. I know to get a hold of all the non-getter methods, I can use
//MethodDeclarator[not(contains(#Image,'get'))]
However, I don't know the syntax about where I insert my logic for the rules. Is it like
//MethodDeclarator[
not(contains(#Image,'get'))
'Some Rule Statements'
]
I saw the use of . in the beginning of statement inside [] in some example code. what are they used for?
In my particular case, I need to combine following pieces together but so far I am unable to accomplish it yet.
Piece 1:
//PrimaryExpression[not(PrimarySuffix/Arguments)]
Piece 2:
//MethodDeclarator[not(contains(#Image,'get'))]
Piece 3:
//PrimaryExpression[PrimaryPrefix/#Label='this']
You need to have at least some basic knowledge/understanding of XPath.
I saw the use of . in the beginning of statement inside [] in some
example code. what are they used for?
[] is called predicate. It must contain a boolean expression. It must immediately follow a node-test. This specifies an additional condition for a node that satisfies the node-test to be selected.
For example:
/*/num
selects all elements named num that are children of the top element of the XML document.
However, if we want to select only such num elements, whose value is an odd integer, we add this additional condition inside a predicate:
/*/num[. mod 2 = 1]
Now this last expression selects all elements named num that are children of the top element of the XML document and whose string value represents an odd integer.
. denotes the context node -- this is the node that has been selected so-far (or the starting node off which the complete XPath expression is evaluated).
In my particular case, I need to combine following pieces together ...
You forgot to say in what way / how the three expressions should be combined. In XPath some of the frequently used "combinators" are the operators and, or, and the function not().
For example, if you want to select elements that are selected by all three provided XPath expressions, you can use the and operator:
//PrimaryExpression
[not(PrimarySuffix/Arguments)
and
PrimaryPrefix/#Label='this'
]
suppose I have a list ListSum, and I want to append a new list to ListSum recursively, like
appList(ListSum):-
%%generate a list: ListTemp,
append(ListTemp,ListSum,ListSum),
appList(ListSum).
but append(ListTemp,ListSum,ListSum) didn't work in the way i wanted.
Can anyone help me out?
Cheers
You have to understand the concept of unification (or actually "matching" as implemented in Prolog). You can't bind two or more values to the same variable. Variables in Prolog once matched persisted its value until the final goal achieved, or fails somewhere. After that, if there're more possibilities then the variable is re-instantiated with another value and so on.
For example, if I query appList([]), then the append would be tested to match as:
append(ListTemp,[],[])
If ListTemp isn't empty list, this clause would fail because the semantic of append is "append the first argument with second, both are lists, resulting in the third". The recursive call to appList(ListSum) would be called as appList([]) since ListSum is matched with [] previously, resulting in infinite recursion (fortunately, if ListTemp isn't [], this won't be reached).
You must have two arguments in the clause, where one is the original list, and the other is the resulting list. The first two argument of append is then ListSum and ListTemp (depends on the append order you want), while the third is the resulting list. Done, no recursion required.
here's a non-recursive solution, not sure why you even need recursion:
appself(L,X) :- append(L,L,X).