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I would like to use a function that takes an enumerable and a function and does the same thing as Python's itertools.accumulate. For example,
iex> accumulate([1,3,7], &Kernel.+)
[1, 4, 11]
For an explanation, it's equal to [1, 1+2, 1+4+7]. Does such a function exist in Elixir's standard library?
Enum.scan/2 does the same thing.
Enum.scan([1, 3, 7], &+/2)
#⇒ [1, 4, 11]
I have an interesting situation. I'm looping over a list several times and I don't know how to produce the list I want. I'm essentially trying to order the second tuple in a list of tuples according to the order of an outside list.
aclist = [{2,4},{2,6},{4,1},{4,8},{1, 2},{1,5},{3,3},{3,7}]
plist = [1,2,3,4]
newplist =
for pid <- plist do
Enum.map(aclist, fn({p_id,c_id}) ->
if p_id == pid do
c_id
end
end)
end
the output from this code is:
[[2, 5, nil, nil, nil, nil, nil, nil], [nil, nil, 4, 6, nil, nil, nil, nil],
[nil, nil, nil, nil, 3, 7, nil, nil], [nil, nil, nil, nil, nil, nil, 1, 8]]
I need the output to be [2,5,4,6,3,7,1,8] but that would require me to loop over it again in a nested loop to pull those numbers out. So obviously I'm missing something, how do I loop over it and pull out the correct data the first time?
for is the perfect thing to use here. It allows iterating over multiple lists while producing a flat list as result and also allows filtering:
iex(1)> aclist = [{2,4},{2,6},{4,1},{4,8},{1, 2},{1,5},{3,3},{3,7}]
[{2, 4}, {2, 6}, {4, 1}, {4, 8}, {1, 2}, {1, 5}, {3, 3}, {3, 7}]
iex(2)> plist = [1,2,3,4]
[1, 2, 3, 4]
iex(3)> for pid <- plist, {p_id, c_id} <- aclist, p_id == pid, do: c_id
[2, 5, 4, 6, 3, 7, 1, 8]
I am new to Prolog and came across this practice excercise. The question asks to define a predicate
zipper([[List1,List2]], Zippered). //this is two lists within one list.
This predicate should interleave elements of List1 with elements of List2.
For example,
zipper([[1,3,5,7], [2,4,6,8]], Zippered) -> Zippered = [1,2,3,4,5,6,7,8].
zipper([[1,3,5], [2,4,6,7,8]], Zippered) -> Zippered = [1,2,3,4,5,6,7,8].
So far I have a solution for two different list:
zipper ([],[],Z).
zipper([X],[],[X]).
zipper([],[Y],[Y]).
zipper([X|List1],[Y|List2],[X,Y|List]) :- zipper(List1,List2,List).
I am not sure how I can translate this solution for one list. Any suggestion on where I can start would be greatly helpful!
Firstly you should change zipper ([],[],Z). to zipper ([],[],[]).. Then to make it work for one list you could do what mat recommended in the comment or you could change it a little. So my version is:
zipper([],[],[]).
zipper([X,[]],X).
zipper([[],Y],Y).
zipper([[X|List1],[Y|List2]],[X,Y|List]) :- zipper([List1,List2],List).
And for your examples:
?- zipper([[1,3,5,7], [2,4,6,8]], Zippered).
Zippered = [1, 2, 3, 4, 5, 6, 7, 8] ;
Zippered = [1, 2, 3, 4, 5, 6, 7, 8] ;
false.
?- zipper([[1,3,5],[2,4,6,7,8]],Zippered).
Zippered = [1, 2, 3, 4, 5, 6, 7, 8] ;
false.
I am a little confused on the answer that Xcode is giving me to this experiment in the Swift Programming Language Guide:
// Use a for-in to iterate through a dictionary (experiment)
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25]
]
var largest = 0
for (kind, numbers) in interestingNumbers {
for number in numbers {
if number > largest {
largest = number
}
}
}
largest
I understand that as the dictionary is being transversed, the largest number is being set to the variable, largest. However, I am confused as to why Xcode is saying that largest is being set 5 times, or 1 time, or 3 times, depending on each test.
When looking through the code, I see that it should be set 6 times in "Prime" alone (2, 3, 5, 7, 11, 13). Then it should skip over any numbers in "Fibonacci" since those are all less than the largest, which is currently set to 13 from "Prime". Then, it should be set to 16, and finally 25 in "Square", yielding a total of 8 times.
Am I missing something entirely obvious?
Dictionaries in Swift (and other languages) are not ordered. When you iterate through the dictionary, there's no guarantee that the order will match the initialization order. In this example, Swift processes the "Square" key before the others. You can see this by adding a print statement to the loop. 25 is the 5th element of Square so largest would be set 5 times for the 5 elements in Square and then would stay at 25.
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25]
]
var largest = 0
for (kind, numbers) in interestingNumbers {
println("kind: \(kind)")
for number in numbers {
if number > largest {
largest = number
}
}
}
largest
This prints:
kind: Square
kind: Prime
kind: Fibonacci
let dict : [String : Any] = ["FirstName" : "Maninder" , "LastName" : "Singh" , "Address" : "Chandigarh"]
dict.forEach { print($0) }
Result would be
("FirstName", "Maninder")
("LastName", "Singh")
("Address", "Chandigarh")
This is a user-defined function to iterate through a dictionary:
func findDic(dict: [String: String]) {
for (key, value) in dict {
print("\(key) : \(value)")
}
}
findDic(dict: ["Animal": "Lion", "Bird": "Sparrow"])
// prints…
// Animal : Lion
// Bird : Sparrow
If you want to iterate over all the values:
dict.values.forEach { value in
// print(value)
}
Here is an alternative for that experiment (Swift 3.0). This tells you exactly which kind of number was the largest.
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25],
]
var largest = 0
var whichKind: String? = nil
for (kind, numbers) in interestingNumbers {
for number in numbers {
if number > largest {
whichKind = kind
largest = number
}
}
}
print(whichKind)
print(largest)
OUTPUT:
Optional("Square")
25
You can also use values.makeIterator() to iterate over dict values, like this:
for sb in sbItems.values.makeIterator(){
// do something with your sb item..
print(sb)
}
You can also do the iteration like this, in a more swifty style:
sbItems.values.makeIterator().forEach{
// $0 is your dict value..
print($0)
}
sbItems is dict of type [String : NSManagedObject]
I have a set of datapoints such as (THIS IS AN EXAMPLE)
val=4; (*this value is calculated before in the program, so it not known a priori*)
x={0,1,2,3};
data=Table[0, {val}];
data[[1]] = {1,5,6,8};
data[[2]] = {9,7,1,3};
data[[3]] = {3,4,5,6};
data[[4]] = {2,2,4,6};
Now I can plot each of these data with ListPlot as
ListPlot[Transpose[{x,data[[1]]}]]
and if I want to plot more than one I can do
ListPlot[{Transpose[{x, data[[1]]}], Transpose[{x, data[[2]]}]}]
but how can I plot all of them in one code single line, by considering that val is calculated before in the program?
Is there a way to do something like
For[i = 1, i < val + 1, i++, ListPlot[Transpose[{x,data[i]}]]......]
having a single graph with all x-y curves?
Indeed I would like a static picture of
Manipulate[ListPlot[Transpose[{x, data[[i]]}]], {i, 1, val,1}]
Thanks
Virgilio
You want to "do the same thing" to every element of a list. That should tell you to think of using Map. Your list is named data and each element is your four element sublist. If you look at the help page for Map it shows you need to think up a function that does what you need to do to each individual sublist. You have already understood that you need to use Transpose with x and your sublist so that tells you your function and you are almost there. The result of Map will be a list of all those results. So
In[1]:= x = {0, 1, 2, 3};
data = {{1, 5, 6, 8}, {9, 7, 1, 3}, {3, 4, 5, 6}, {2, 2, 4, 6}};
ListPlot[Map[Transpose[{x, #}] &, data], Joined -> True]
Out[3]= ...FourOverlaidPlotsSnipped...
Go through that a word at a time until you can really understand the thinking that was done to be able to write that. You will use this idea again and again if you keep using Mathematica.
For the example you give the cleanest method is to use DataRange:
data = {{1, 5, 6, 8}, {9, 7, 1, 3}, {3, 4, 5, 6}, {2, 2, 4, 6}};
ListLinePlot[data, DataRange -> {0, 3}]
Please ask your future questions on the dedicated Mathematica StackExchange site: