Fork/Join calculation - groovy

I have this example of a fork/join calculation. Could somebody please describe me shortly how it works here?
def quicksortForkJoin(numbers) {
withPool {
runForkJoin(0, numbers) {index, list ->
def groups = list.groupBy {it <=> list[list.size().intdiv(2)]}
if ((list.size() < 2) || (groups.size() == 1)) {
return [index: index, list: list.clone()]
}
(-1..1).each { forkOffChild(it, groups[it] ?: [])}
return [index: index, list: childrenResults.sort {it.index}.sum {it.list}]
}.list
}
}

Any good?
def quicksortForkJoin(numbers) {
// Create a pool of workers the default size
withPool {
// Run a fork with index 0 and the numbers
runForkJoin(0, numbers) {index, list -> // [1]
// Split numbers into 3 groups:
// -1: those less than the "middle" number
// 0: those equal to the "middle" number
// 1: those greater than the "middle" number
def groups = list.groupBy {it <=> list[list.size().intdiv(2)]}
// If there are less than 2 numbers to sort, or all numbers are equal
if ((list.size() < 2) || (groups.size() == 1)) {
// return the index and a clone of the current list
return [index: index, list: list.clone()]
}
// Otherwise, fork off a child process for each of the
// groups above (less than, equal and greater than)
// forkOffChild will not block, and will effectively go back
// to line [1] with the new index and list
(-1..1).each { forkOffChild(it, groups[it] ?: [])}
// Block waiting for all 3 children to finish, then sort the
// results so the indexes are [ -1, 0, 1 ] and then join the
// lists of numbers back together
return [ index: index,
list: childrenResults.sort {it.index}.sum {it.list}]
// when done, return the `list` property from the final map
}.list
}
}

Related

Why is the following a memory leak? [duplicate]

I've got code that looks like this:
for (std::list<item*>::iterator i=items.begin();i!=items.end();i++)
{
bool isActive = (*i)->update();
//if (!isActive)
// items.remove(*i);
//else
other_code_involving(*i);
}
items.remove_if(CheckItemNotActive);
I'd like remove inactive items immediately after update them, inorder to avoid walking the list again. But if I add the commented-out lines, I get an error when I get to i++: "List iterator not incrementable". I tried some alternates which didn't increment in the for statement, but I couldn't get anything to work.
What's the best way to remove items as you are walking a std::list?
You have to increment the iterator first (with i++) and then remove the previous element (e.g., by using the returned value from i++). You can change the code to a while loop like so:
std::list<item*>::iterator i = items.begin();
while (i != items.end())
{
bool isActive = (*i)->update();
if (!isActive)
{
items.erase(i++); // alternatively, i = items.erase(i);
}
else
{
other_code_involving(*i);
++i;
}
}
You want to do:
i= items.erase(i);
That will correctly update the iterator to point to the location after the iterator you removed.
You need to do the combination of Kristo's answer and MSN's:
// Note: Using the pre-increment operator is preferred for iterators because
// there can be a performance gain.
//
// Note: As long as you are iterating from beginning to end, without inserting
// along the way you can safely save end once; otherwise get it at the
// top of each loop.
std::list< item * >::iterator iter = items.begin();
std::list< item * >::iterator end = items.end();
while (iter != end)
{
item * pItem = *iter;
if (pItem->update() == true)
{
other_code_involving(pItem);
++iter;
}
else
{
// BTW, who is deleting pItem, a.k.a. (*iter)?
iter = items.erase(iter);
}
}
Of course, the most efficient and SuperCool® STL savy thing would be something like this:
// This implementation of update executes other_code_involving(Item *) if
// this instance needs updating.
//
// This method returns true if this still needs future updates.
//
bool Item::update(void)
{
if (m_needsUpdates == true)
{
m_needsUpdates = other_code_involving(this);
}
return (m_needsUpdates);
}
// This call does everything the previous loop did!!! (Including the fact
// that it isn't deleting the items that are erased!)
items.remove_if(std::not1(std::mem_fun(&Item::update)));
I have sumup it, here is the three method with example:
1. using while loop
list<int> lst{4, 1, 2, 3, 5};
auto it = lst.begin();
while (it != lst.end()){
if((*it % 2) == 1){
it = lst.erase(it);// erase and go to next
} else{
++it; // go to next
}
}
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
2. using remove_if member funtion in list:
list<int> lst{4, 1, 2, 3, 5};
lst.remove_if([](int a){return a % 2 == 1;});
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
3. using std::remove_if funtion combining with erase member function:
list<int> lst{4, 1, 2, 3, 5};
lst.erase(std::remove_if(lst.begin(), lst.end(), [](int a){
return a % 2 == 1;
}), lst.end());
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
4. using for loop , should note update the iterator:
list<int> lst{4, 1, 2, 3, 5};
for(auto it = lst.begin(); it != lst.end();++it){
if ((*it % 2) == 1){
it = lst.erase(it); erase and go to next(erase will return the next iterator)
--it; // as it will be add again in for, so we go back one step
}
}
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
Use std::remove_if algorithm.
Edit:
Work with collections should be like:
prepare collection.
process collection.
Life will be easier if you won't mix this steps.
std::remove_if. or list::remove_if ( if you know that you work with list and not with the TCollection )
std::for_each
The alternative for loop version to Kristo's answer.
You lose some efficiency, you go backwards and then forward again when deleting but in exchange for the extra iterator increment you can have the iterator declared in the loop scope and the code looking a bit cleaner. What to choose depends on priorities of the moment.
The answer was totally out of time, I know...
typedef std::list<item*>::iterator item_iterator;
for(item_iterator i = items.begin(); i != items.end(); ++i)
{
bool isActive = (*i)->update();
if (!isActive)
{
items.erase(i--);
}
else
{
other_code_involving(*i);
}
}
Here's an example using a for loop that iterates the list and increments or revalidates the iterator in the event of an item being removed during traversal of the list.
for(auto i = items.begin(); i != items.end();)
{
if(bool isActive = (*i)->update())
{
other_code_involving(*i);
++i;
}
else
{
i = items.erase(i);
}
}
items.remove_if(CheckItemNotActive);
Removal invalidates only the iterators that point to the elements that are removed.
So in this case after removing *i , i is invalidated and you cannot do increment on it.
What you can do is first save the iterator of element that is to be removed , then increment the iterator and then remove the saved one.
If you think of the std::list like a queue, then you can dequeue and enqueue all the items that you want to keep, but only dequeue (and not enqueue) the item you want to remove. Here's an example where I want to remove 5 from a list containing the numbers 1-10...
std::list<int> myList;
int size = myList.size(); // The size needs to be saved to iterate through the whole thing
for (int i = 0; i < size; ++i)
{
int val = myList.back()
myList.pop_back() // dequeue
if (val != 5)
{
myList.push_front(val) // enqueue if not 5
}
}
myList will now only have numbers 1-4 and 6-10.
Iterating backwards avoids the effect of erasing an element on the remaining elements to be traversed:
typedef list<item*> list_t;
for ( list_t::iterator it = items.end() ; it != items.begin() ; ) {
--it;
bool remove = <determine whether to remove>
if ( remove ) {
items.erase( it );
}
}
PS: see this, e.g., regarding backward iteration.
PS2: I did not thoroughly tested if it handles well erasing elements at the ends.
You can write
std::list<item*>::iterator i = items.begin();
while (i != items.end())
{
bool isActive = (*i)->update();
if (!isActive) {
i = items.erase(i);
} else {
other_code_involving(*i);
i++;
}
}
You can write equivalent code with std::list::remove_if, which is less verbose and more explicit
items.remove_if([] (item*i) {
bool isActive = (*i)->update();
if (!isActive)
return true;
other_code_involving(*i);
return false;
});
The std::vector::erase std::remove_if idiom should be used when items is a vector instead of a list to keep compexity at O(n) - or in case you write generic code and items might be a container with no effective way to erase single items (like a vector)
items.erase(std::remove_if(begin(items), end(items), [] (item*i) {
bool isActive = (*i)->update();
if (!isActive)
return true;
other_code_involving(*i);
return false;
}));
do while loop, it's flexable and fast and easy to read and write.
auto textRegion = m_pdfTextRegions.begin();
while(textRegion != m_pdfTextRegions.end())
{
if ((*textRegion)->glyphs.empty())
{
m_pdfTextRegions.erase(textRegion);
textRegion = m_pdfTextRegions.begin();
}
else
textRegion++;
}
I'd like to share my method. This method also allows the insertion of the element to the back of the list during iteration
#include <iostream>
#include <list>
int main(int argc, char **argv) {
std::list<int> d;
for (int i = 0; i < 12; ++i) {
d.push_back(i);
}
auto it = d.begin();
int nelem = d.size(); // number of current elements
for (int ielem = 0; ielem < nelem; ++ielem) {
auto &i = *it;
if (i % 2 == 0) {
it = d.erase(it);
} else {
if (i % 3 == 0) {
d.push_back(3*i);
}
++it;
}
}
for (auto i : d) {
std::cout << i << ", ";
}
std::cout << std::endl;
// result should be: 1, 3, 5, 7, 9, 11, 9, 27,
return 0;
}
I think you have a bug there, I code this way:
for (std::list<CAudioChannel *>::iterator itAudioChannel = audioChannels.begin();
itAudioChannel != audioChannels.end(); )
{
CAudioChannel *audioChannel = *itAudioChannel;
std::list<CAudioChannel *>::iterator itCurrentAudioChannel = itAudioChannel;
itAudioChannel++;
if (audioChannel->destroyMe)
{
audioChannels.erase(itCurrentAudioChannel);
delete audioChannel;
continue;
}
audioChannel->Mix(outBuffer, numSamples);
}

Inline Conditional Map Literal in Groovy

Working on some translation / mapping functionality using Maps/JsonBuilder in Groovy.
Is is possible (without creating extra code outside of the map literal creation) .. to conditionally include/exclude certain key/value pairs ? Some thing along the lines of the following ..
def someConditional = true
def mapResult =
[
"id":123,
"somethingElse":[],
if(someConditional){ return ["onlyIfConditionalTrue":true]}
]
Expected results:
If someConditional if false, only 2 key/value pairs will exist in mapResult.
If someConditional if true, all 3 key/value pairs will exist.
Note that I'm sure it could be done if I create methods / and split things up.. for to keep things concise I would want to keep things inside of the map creation.
You can help yourself with with:
[a:1, b:2].with{
if (false) {
c = 1
}
it
}
With a small helper:
Map newMap(m=[:], Closure c) {
m.with c
m
}
E.g.:
def m = newMap {
a = 1
b = 1
if (true) {
c = 1
}
if (false) {
d = 1
}
}
assert m.a == 1
assert m.b == 1
assert m.c == 1
assert !m.containsKey('d')
Or pass an initial map:
newMap(a:1, b:2) {
if (true) {
c = 1
}
if (false) {
d = 1
}
}
edit
Since Groovy 2.5, there is an alternative for with called tap. It
works like with but does not return the return value from the closure,
but the delegate. So this can be written as:
[a:1, b:2].tap{
if (false) {
c = 1
}
}
You could potentially map all false conditions to a common key (e.g. "/dev/null", "", etc) and then remove that key afterwards as part of a contract. Consider the following:
def condA = true
def condB = false
def condC = false
def mapResult =
[
"id":123,
"somethingElse":[],
(condA ? "condA" : "") : "hello",
(condB ? "condB" : "") : "abc",
(condB ? "condC" : "") : "ijk",
]
// mandatory, arguably reasonable
mapResult.remove("")
assert 3 == mapResult.keySet().size()
assert 123 == mapResult["id"]
assert [] == mapResult["somethingElse"]
assert "hello" == mapResult["condA"]
There is no such syntax, the best you can do is
def someConditional = true
def mapResult = [
"id":123,
"somethingElse":[]
]
if (someConditional) {
mapResult.onlyIfConditionalTrue = true
}
I agree with Donal, without code outside of map creation it is difficult.
At least you would have to implement your own ConditionalMap, it is a little work but perfectly doable.
Each element could have it's own condition like
map["a"] = "A"
map["b"] = "B"
map.put("c","C", true)
map.put("d","D", { myCondition })
etc...
Here an incomplete example (I did only put, get, keySet, values and size to illustrate, and not typed - but you probably don't need types here?), you will probably have to implement few others (isEmpty, containsKey etc...).
class ConditionalMap extends HashMap {
/** Default condition can be a closure */
def defaultCondition = true
/** Put an elemtn with default condition */
def put(key, value) {
super.put(key, new Tuple(defaultCondition, value))
}
/** Put an elemetn with specific condition */
def put(key, value, condition) {
super.put(key, new Tuple(condition, value))
}
/** Get visible element only */
def get(key) {
def tuple = super.get(key)
tuple[0] == true ? tuple[1] : null
}
/** Not part of Map , just to know the real size*/
def int realSize() {
super.keySet().size()
}
/** Includes only the "visible" elements keys */
def Set keySet() {
super.keySet().inject(new HashSet(),
{ result, key
->
def tuple = super.get(key)
if (tuple[0])
result.add(key)
result
})
}
/** Includes only the "visible" elements keys */
def Collection values() {
this.keySet().asCollection().collect({ k -> this[k] })
}
/** Includes only the "visible" elements keys */
def int size() {
this.keySet().size()
}
}
/** default condition that do not accept elements */
def map = new ConditionalMap(defaultCondition: false)
/** condition can be a closure too */
// def map = new ConditionalMap(defaultCondition : {-> true == false })
map["a"] = "A"
map["b"] = "B"
map.put("c","C", true)
map.put("d","D", false)
assert map.size() == 1
assert map.realSize() == 4
println map["a"]
println map["b"]
println map["c"]
println map["d"]
println "size: ${map.size()}"
println "realSize: ${map.realSize()}"
println "keySet: ${map.keySet()}"
println "values: ${map.values()}"
/** end of script */
You can use the spread operator to do this for for both maps and lists:
def t = true
def map = [
a:5,
*:(t ? [b:6] : [:])
]
println(map)
[a:5, b:6]
This works in v3, haven't tried in prior versions.

How to implement Haskell's splitEvery in Swift?

PROBLEM
let x = (0..<10).splitEvery( 3 )
XCTAssertEqual( x, [(0...2),(3...5),(6...8),(9)], "implementation broken" )
COMMENTS
I am running into problems calculating number of elements in the Range, etc...
extension Range
{
func splitEvery( nInEach: Int ) -> [Range]
{
let n = self.endIndex - self.startIndex // ERROR - cannot invoke '-' with an argument list of type (T,T)
}
}
The values in a range are of ForwardIndexType, so you can only advance() them,
or compute the distance(), but the subtraction - is not defined. The advance amount has to be of the corresponding
type T.Distance. So this would be a possible implementation:
extension Range {
func splitEvery(nInEach: T.Distance) -> [Range] {
var result = [Range]() // Start with empty array
var from = self.startIndex
while from != self.endIndex {
// Advance position, but not beyond the end index:
let to = advance(from, nInEach, self.endIndex)
result.append(from ..< to)
// Continue with next interval:
from = to
}
return result
}
}
Example:
println( (0 ..< 10).splitEvery(3) )
// Output: [0..<3, 3..<6, 6..<9, 9..<10]
Note however that 0 ..< 10 is not a list (or array) of integers. To split an array into subarrays you could define a similar extension:
extension Array {
func splitEvery(nInEach: Int) -> [[T]] {
var result = [[T]]()
for from in stride(from: 0, to: self.count, by: nInEach) {
let to = advance(from, nInEach, self.count)
result.append(Array(self[from ..< to]))
}
return result
}
}
Example:
println( [1, 1, 2, 3, 5, 8, 13].splitEvery(3) )
// Output: [[1, 1, 2], [3, 5, 8], [13]]
A more general approach could be to split all sliceable objects. But Sliceable
is protocol and protocols cannot be extended. What you can do instead is to
define a function that takes the sliceable object as the first argument:
func splitEvery<S : Sliceable>(seq : S, nInEach : S.Index.Distance) -> [S.SubSlice] {
var result : [S.SubSlice] = []
var from = seq.startIndex
while from != seq.endIndex {
let to = advance(from, nInEach, seq.endIndex)
result.append(seq[from ..< to])
from = to
}
return result
}
(Note that this function is completely unrelated to the (extension) methods
defined above.)
Example:
println( splitEvery("abcdefg", 2) )
// Output: [ab, cd, ef, g]
println( splitEvery([3.1, 4.1, 5.9, 2.6, 5.3], 2) )
// Output: [[3.1, 4.1], [5.9, 2.6], [5.3]]
Ranges are not sliceable, but you could define a separate function that takes a
range argument:
func splitEvery<T>(range : Range<T>, nInEach : T.Distance) -> [Range<T>] {
var result : [Range<T>] = []
var from = range.startIndex
while from != range.endIndex {
let to = advance(from, nInEach, range.endIndex)
result.append(from ..< to)
from = to
}
return result
}
Example:
println( splitEvery(0 ..< 10, 3) )
// Output: [0..<3, 3..<6, 6..<9, 9..<10]

Verify if two lists share values in C#

I'd like to know if two lists share values before applying an intersection. Something like bool DoIntersect(listA, listB) would be fabulous!
This is the code I came up with:
// Person is a class with Id and Name properties
List<Person> people1;
List<Person> people2;
// Populate people1 and people2...
// My current solution (pseudocode obviously)...
if (DoIntersect(people1, people2))
{
people1 = people1.Intersect(people2)
}
else
{
/* No shared people */
throw exception;
}
// Continue with the process...
It depends on exactly what you want:
// are there any common values between a and b?
public static bool SharesAnyValueWith<T>(this IEnumerable<T> a, IEnumerable<T> b)
{
return a.Intersect(b).Any();
}
For lists that don't overlap, this will iterate through a and b each once. For lists that overlap, this will iterate all the way through a, then through b until the first overlapping element is found.
// does a contain all of b? (ignores duplicates)
public static bool ContainsAllFrom<T>(this IEnumerable<T> a, IEnumerable<T> b)
{
return !b.Except(a).Any();
}
This will iterate through a once, then will iterate through b, stopping on the first element in b not in a.
// does a contain all of b? (considers duplicates)
public static bool ContainsAllFrom<T>(this IEnumerable<T> a, IEnumerable<T> b)
{
// get the count of each distinct element in a
var counts = a.GroupBy(t => t).ToDictionary(g => g.Key, g => g.Count());
foreach (var t in b) {
int count;
// if t isn't in a or has too few occurrences return false. Otherwise, reduce
// the count by 1
if (!counts.TryGetValue(t, out count) || count == 0) { return false; }
counts[t] = count - 1;
}
return true;
}
Similarly, this will iterate through a once, then will iterate through b, stopping on the first element in b not in a.
I believe without altering the fact that you're using a List you can't get better performance.
However, if you would have 2 sorted lists to begin with (requires overhead when creating them), then you could iterate through them with complexity of O(n) in order to find out if you have shared values.
Edit:
Although original OP doesn't have 2 sorted lists, in case someone will need it, here is the implementation for checking Intersection at O(n):
public Boolean DoIntersect(SortedList<int,String> listA,SortedList<int,String> listB )
{
if (listA == null || listA.Count == 0 || listB == null || listB.Count == 0)
{
return false;
}
var keysA = listA.Keys;
var keysB = listB.Keys;
int i = 0, j = 0;
while (i < listA.Count && j < listB.Count)
{
if (keysA[i] < keysB[j])
{
i++;
}else if (keysA[i] > keysB[j])
{
j++;
}
else
{
return true;
}
}
The above approach can be used also with IEnumerable lists, given that they are sorted, with slight variation - using GetEnumerator and iterating with it.

groovy.lang.MissingMethodException: No signature of method

I am getting the following error -
groovy.lang.MissingMethodException: No signature of method: Script64$_run_closure5_closure7_closure8_closure9_closure10_closure11.doCall() is applicable for argument types: (java.lang.String) values: Possible solutions: doCall(java.lang.Object, java.lang.Object), isCase(java.lang.Object), isCase(java.lang.Object) error at line:
Code - EDIT
import groovy.xml.*
List tempList = []
List listgenerated = []
def count = 0
for (a in 0..totalCount-1)
{
//nameList and valueList lists will have all the contents added as below commented pseudo code
/*for (b in 0..50)
{
nameList.add(b,number) // number is some calculated value
valueList.add(b,number)
e.g. nameList=[name1, name2, name3,name4, name5]
valueList =[val1, val2, val3, , val5]
listgenerated should be = [[name1:val1, name2:val2], [name3:val3, name4: , name5:val5]]
} */
tempList = []
for (j in count..nameList.size())
{
count = j
def nameKey = nameList[j]
def value
if (nameKey != null)
{
value = valueList[j]
tempList << [(nameKey) : value]
}
}
count = count
listgenerated.putAt(a,tempList)
number = number +1
}
def process = { binding, element, name ->
if( element[ name ] instanceof Collection ) {
element[ name ].each { n ->
binding."$name"( n )
}
}
else if( element[ name ] ) {
binding."$name"( element[ name ] )
}
}
class Form {
List fields
}
def list = [[ name:'a', val:'1' ], [ name:'b', val :'2', name2:4, xyz:'abc', pqr:'']] //Edited list
f = new Form( fields: list ) //Works fine
f = new Form( fields: listgenerated ) //Gives the above error
String xml = XmlUtil.serialize( new StreamingMarkupBuilder().with { builder ->
builder.bind { binding ->
data {
f.fields.each { fields ->
item {
fields.each { name, value ->
process( binding, fields, name )
}
}
}
}
}
} )
If while creating the "listgenerated" single quotes are added around values it takes it as character and while printing both lists seem different.
I am unable to figure it out what exactly is going wrong. Any help is appreciated. Thanks.
Ref - Groovy: dynamically create XML for collection of objects with collections of properties
I believe, where you do:
//some loop to add multiple values to the list
listgenerated << name+":"+value
You need to do:
//some loop to add multiple values to the list
listgenerated << [ (name): value ]
And add a map to the list rather than a String. It's hard to say though as your code example doesn't run without alteration, and I don't know if it's the alterations that are solving the problem

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