I have the file "global power plants" with a column "capacity_in_mw" (with numbers 30, 100, 45, ...) and another column is "primary_fuel" (Coal, Hydro, Oil, Solar, Nuclear, Wind, Coal).
I can generate a map in function of "capacity_in_mw" by setting the condition
plotdata = data.query('capacity_in_mw > 50')
Now, I would like to generate a map in function of "primary_fuel". Because data is alphanumeric, how do I set up the condition?
Furthermore, when making the map, to assign color='black' for Coal, color='green' for Wind, color='yellow' for Solar, ... etc.
Python.
I am a novice, but I think I found the solution. It is more of an issue of syntax, to use the == in the query to identify the alphanumeric value.
plotdata = data.query('primary_fuel == "Hydro"')
Also, lesson learned in the future to dig more before posting a question.
Apparently Guava's ImmutableRangeSet cannot store overlapping ranges. This makes sense, but is there an interface to resolve/merge overlapping ranges and then put the resultant ranges into an ImmutableRangeSet?
Currently I'm building a TreeRangeSet, which automatically merges overlapping ranges, and passing this as an argument to ImmutableRangeSet.builder().addAll(). This process works, but it seems a little too indirect just to resolve overlapping ranges.
Can you be more specific about your use case? I guess you have a collection of ranges and you're trying to create ImmutableRangeSet using copyOf method, which throws IAE in case of overlapping ranges. Let's see this test case:
#Test
public void shouldHandleOverlappingRanges()
{
//given
ImmutableList<Range<Integer>> ranges = ImmutableList.of(
Range.closed(0, 2),
Range.closed(1, 4),
Range.closed(9, 10)
);
//when
ImmutableRangeSet<Integer> rangeSet = ImmutableRangeSet.copyOf(ranges);
//then
assertThat(rangeSet.asSet(DiscreteDomain.integers()))
.containsOnly(0, 1, 2, 3, 4, 9, 10);
}
which fails with
java.lang.IllegalArgumentException:
Overlapping ranges not permitted but found [0..2] overlapping [1..4]
In this case you should use unionOf instead of copyOf and it'd pass:
//when
ImmutableRangeSet<Integer> rangeSet = ImmutableRangeSet.unionOf(ranges);
I'm trying to figure out why the following works on the first string cluster (character) but not on a second one. Perhaps the endIndex cannot be applied on another String?
let part = "A"
let full = "ABC"
print(full[part.startIndex ... part.startIndex]) // "A"
print(full[part.endIndex ... part.endIndex]) // "" <- ???
print(full[part.endIndex ... full.index(after: part.endIndex)]) // "B"
bSecond should hold "B", but instead is empty. But the proof that one string index works on another is that the last statement works.
EDIT:
Assuming full.hasPrefix(part) is true.
Swift puzzles.
You cannot use the indices of one string to subscript a different
string. That may work by chance (in your first example) or not
(in your second example), or crash at runtime.
In this particular case, part.endIndex (which is the "one past the end position" for the part string) returns
String.UnicodeScalarView.Index(_position: 1), _countUTF16: 0)
with _countUTF16: (which is the "count of this extended grapheme cluster in UTF-16 code units") being zero, i.e. it describes
a position (in the unicode scalar view) with no extent. Then
full[part.endIndex ... part.endIndex]
returns an empty string. But that is an implementation detail
(compare StringCharacterView.swift). The real answer is just "you can't do that".
A safe way to obtain the intended (?) result is
let part = "A"
let full = "ABC"
if let range = full.range(of: part) {
print(full[range]) // A
if range.upperBound != full.endIndex {
print(full[range.upperBound...range.upperBound]) // B
}
}
First of all, this question is not about "what does $0 mean". I learnt in swift document that $0 is like index.
My question is "How numbers.sort { $0 > $1 } can be used to implement a sort function". I searched for this syntax numbers.sort { $0 > $1 } in some other websites, for example this one. It is apparently not the current version. So I still can't understand what the meaning of it.
print(numbers) //[20, 19, 1, 12]
let sortedNumbers = numbers.sort { $0 > $1 }
print(sortedNumbers) //[20, 19, 12, 1]
Can someone explain this simple piece of code above for me? Like how this simple code $0 > $1 implement the sort function, sorting the numbers from big to small.
I know some about index, and this $0 looks like index, but it only has $0 and $1 two indices. So how can it be used into 4 numbers? According to my knowledge in C++ before, I can't understand the principle in this.
Please make your answer as specific as possible. Thank you!
----------------- Below is edited extra part -------------------
I don't know whether stackoverflow would allow me to edit my question like this, but this extra part is too long, so I can't add it in the comment.
#pbodsk #Paul Richter
So the sort() syntax in swift uses quick sort to deal with sort function?
Actually my question is more about "what is the operating principle of sort{$0 > $1}". I know what you mean above, and I think it's similar with what swift 2.1 document says, but your answer is not what I really want to know. Sorry, my English expression is not very good. Let me try another way.
When I learnt C++ before, there are always some documents to explain what a function's operating principle is or how this function (like sort() here) operate in background. Sort() here needs to compare first and second interchange. In C++, it's like
if numbers[1] < numbers[2]{ //just consider this pseudocode
int k;
k = numbers[1];
numbers[1] = numbers[2];
numbers[2] = k;
}
We can see this process is obvious. In swift, it's like
numbers.sort({(val1: Int, val2: Int) -> Bool in
return val1 > val2
})
Where is it compared? And how is it interchanged? Does return val1 > val2 automatically compare and interchange these two values and return them? Just this one syntax implement these all 3 processes? How? This is what I really want to know. Sorry again for my poor English expression.
#the_UB and #moonvader are both right, but I just thought that I would extend the example from #moonvader a bit, just to show you how we end up with $0 > $1
If you look at the example in "The Swift Programming Language" about Closure Expressions you can see that to sort an array you call the sort method which can then take a function as a parameter.
This function must take two parameters and compare them, and then return a boolean.
So if we have this array:
let numbers = [4, 6, 8, 1, 3]
and this method
func sortBackwards(val1: Int, val2: Int) -> Bool {
print("val1: \(val1) - val2: \(val2)" )
return val1 > val2
}
We can sort the elements like so:
numbers.sort(sortBackwards) //gives us [8, 6, 4, 3, 1]
The sort method will use our sortBackwards method on each of the elements in the array and compare them.
Here's the output of the print
val1: 6 - val2: 4
val1: 8 - val2: 4
val1: 8 - val2: 6
val1: 1 - val2: 4
val1: 3 - val2: 1
val1: 3 - val2: 4
OK, let's reduce that.
Instead of defining a function, we can add that directly as a parameter to the sort method like so:
numbers.sort({(val1: Int, val2: Int) -> Bool in
return val1 > val2
})
And we still end up with [8, 6, 4, 3, 1] (how fortunate!)
OK, the next thing we can do is what in "The Swift Programming Language" (the link above) is called "Infering Type From Context". As we call this method on an array of Ints, Swift can figure out that our val1 and val2 parameters must be Ints too, there's no need for us to tell it. So, lets remove the types. That leaves us with:
numbers.sort({val1, val2 in
return val1 > val2
})
And still the same result.
OK, getting there. The next thing we can do is what in the book is called "Implicit Returns from Single-Expression Closures"
As our comparison can be done in one line there's no need for us to use return. So:
numbers.sort({val1, val2 in val1 > val2})
Still gives us [8, 6, 4, 3, 1]
Finally we're getting to what #moonvader used much much less words to explain :-) namely "Shorthand Argument Names"
As it says in the book:
Swift automatically provides shorthand argument names to inline closures, which can be used to refer to the values of the closure’s arguments by the names $0, $1, $2, and so on.
So, in our example, val1 can be replaced by $0 and val2 can be replaced by $1
Which gives us:
numbers.sort({$0 > $1})
And still we get [8, 6, 4, 3, 1]
We can then continue to use a "Trailing Closure", which means that if the last parameter of a function is a closure, we can add that parameter "outside" the function.
So we end up with:
numbers.sort{$0 > $1}
And the outcome is still [8, 6, 4, 3, 1]
Hope that helps to clarify things.
Here is what all need to know: Sort and Sorted.
To be more specific, Sorting can be two type : Ascending and Descending.
Q - So to do sorting, what do we need?
A - We need two variables to hold two variable(I don't know if it is the correct word)
Hence in this case we have two variable $0 and $1. These both are shorthands to represent left and right variable. Both will help to sort.
">" will do descending.
"<" will do ascending.
From developer.apple.com
Shorthand Argument Names
Swift automatically provides shorthand argument names to inline closures, which can be used to refer to the values of the closure’s arguments by the names $0, $1, $2, and so on.
If you use these shorthand argument names within your closure expression, you can omit the closure’s argument list from its definition, and the number and type of the shorthand argument names will be inferred from the expected function type. The in keyword can also be omitted, because the closure expression is made up entirely of its body:
reversed = names.sort( { $0 > $1 } )
Here, $0 and $1 refer to the closure’s first and second String arguments.
The process of sorting a list consists of repeatedly reordering its elements until nothing remains to be reordered. Now there are many sorting algorithms, but they all do this, in different ways. So then how are elements reordered? By comparing two given elements, and deciding which comes first, and swapping them if needed.
We can separate the overall reordering and swapping parts from the comparison part, and write a sort function that will take care of all the repeated reordering stuff, and just require the caller to specify how to compare two elements. If the list consists of numbers, it's almost always the case that the way to compare them is to just take their value. But suppose the list consists of things a little more complicated, like cars. How do you compare two cars? Well, you could compare them by numerically comparing their top speed. Or their gas mileage. Or price.
But the comparison doesn't have to be numerical. We could compare two cars by actually racing them. We could compare two cars by just saying if one is blue and the other isn't, the blue one is ordered first, and if neither or both are blue they are ordered as they already are.
We could come up with all sorts of ways to compare two cars. And the sorting algorithm could then sort a list of cars, without knowing anything about cars, as long as we the caller just tell it how to compare cars - any two given cars. We just have to express that comparison as an expression that returns a boolean, where if it's true, the first car is ordered before the second one, and if it's false, the first car is ordered after the second one.
Returning to numbers, that's what sort { $0 > $1 } means, in Swift's very concise syntax: "Sort, where if the first element is > the second one, order the first one before the second one."
You asked how it can sort four numbers with only two indices. $0 and $1 are not bound to the four specific elements in the list [20, 19, 1, 12], they are bound to any two given numbers that need to be compared, because the sorting algorithm repeately needs to do this.
There are a few things to note. First, the operator > has to be defined for the kinds of elements you are sorting. In the example the elements are numbers, and > is indeed defined. Second, the sort function specifies that the boolean true orders the first one before the second rather than the other way around, so the comparison function follows that specification. Third, the last evaluated expression is taken as the boolean value to be used. Having these two assumptions beforehand allows the comparison function to be written so concisely.
So if we wanted to sort those cars by racing them, we could write it like this:
cars.sort {
winner_of_race_between($0, $1) == $0
// if the first car beats the second, it is sorted ahead
}
Or exclusive blueness:
cars.sort { //not guaranteed to be valid Swift, just consider this pseudocode
if(($0.color != Color.blue) && ($1.color == Color.blue) {
$1
} else if (($0.color == Color.blue) && ($1.color != Color.blue)) {
$0
} else { //leave them in same order
$0
}
}
extension Array {
public func mySorted(by areInIncreasingOrder: (Element, Element) throws -> Bool) rethrows -> [Element] {
var newArray: [Element] = self
if newArray.count <= 1 {
/// nothing to do
} else if newArray.count <= 32 { /// 32 ?? 64
for l in 1..<count {
for r in (0..<l).reversed() {
if try areInIncreasingOrder(newArray[r + 1], newArray[r]) {
(newArray[r + 1], newArray[r]) = (newArray[r], newArray[r + 1])
} else {
break
}
}
}
} else {
/// others sort
}
return newArray
}
}
var array: [Int] = [4, 6, 8, 1, 3]
let a1 = array.sorted {
print("\($0) \($1)")
return $0 > $1
}
print("---------")
let a2 = array.mySorted {
print("\($0) \($1)")
return $0 > $1
}
print("==========")
let a1 = array.sorted {
print("\($0) \($1)")
return $0 < $1
}
print("+++++++")
let a2 = array.mySorted {
print("\($0) \($1)")
return $0 < $1
}
I am making a timetabling program which does one to one matches from SubjectTeacherPeriod (planning entity) to Period. There comes a case when I need to: "for y periods, atleast x of the SubjectTeacherPeriod must match a match_condition"
For example, I want to constrain 3 particular periods, atleast two of them to be taught by teachers who match to asst prof.
Here is the data structure holding such a constraint:
Class XOfYPeriods
SomeType match_condition
int x
List<Period> Periods //problem
SubjectTeacherPeriod has a Period, of course
class SubjectTeacherPeriod
int id
SomeType attrib
Period period
How do I write a rule that evaluates individual Periods from a list to check if x number of SubjectTeacherPeriods that are allocated those Periods meet the match condition?
Do correct me if I am defining my classes in bad form.
For the sake of example, here is a statement to be evaluated to determine a match: eval(matches($stp_attrib,$match_condition))
Sorry for the use of Pseudocode if it confused more than clarified. The SomeType is actually List< String> and thus the match condition is checked with a Collections.disjoint
I will give it a try, but not sure I completely understand your problem statement:
rule "X of Y Periods"
when
$c : XOfYPeriods( )
$list : List( size > $c.x ) from
accumulate( $stp : SubjectTeacherPeriod( matches(attrib, $c.match_condition),
period memberOf $c.periods ),
collectList( $stp ) )
then
// $list of STP that match the condition and
// whose period matches one of the periods in the list
end
Hope it helps.