Is there a better way to calculate a moving sum of a list?
List<double?> rollingSum({int window = 3, List data = const []}) {
List<double?> sum = [];
int i = 0;
int maxLength = data.length - window + 1;
while (i < maxLength) {
List tmpData = data.getRange(i, i + window).toList();
double tmpSum = tmpData.reduce((a, b) => a + b);
sum.add(tmpSum);
i++;
}
// filling the first n values with null
i = 0;
while (i < window - 1) {
sum.insert(0, null);
i++;
}
return sum;
}
Well, the code is already clean for what you need. Maybe just some improvements like:
Use a for loop
You can use the method sublist which creates a "view" of a list, which is more efficient
To insert some values in the left/right of a list, there is a specific Dart method called padLeft, where you specify the lenght of the list which you want it to become (first parameter), then the value you want to use to fill it (second parameter). For example, if you have an array of N elements, and you want to fill it with X "null"s to the left, use padLeft(N+X, null).
List<double?> rollingSum({int window = 3, List data = const []}) {
List<double?> sum = [];
for (int i = 0; i < data.length - window + 1; i++) {
List tmpData = data.sublist(i, i + window);
double tmpSum = tmpData.reduce((a, b) => a + b);
sum.add(tmpSum);
}
sum.padLeft(window - 1, null);
return sum;
}
if I understand your problem correctly you can just calculate the window one time and in one loop you can for each iteration you can add the current element to the sum and subtract i - (window - 1)
so for an input like this
data = [1,2,3,4,5,6]
window = 3
the below code will result in [6,9,12,15]
int sum = 0;
List<double> res = [];
for (int i = 0;i<data.length;i++) {
sum += data[i];
if (i < window - 1) {
continue;
}
res.add(sum);
sum -= data[i - (window - 1)]; // remove element that got out of the window size
}
this way you won't have to use getRange nor sublist nor reduce as all of those are expensive functions in terms of time and space complexity
In Copper Nickel Mine (Cloud) Simulation, In MinePanel Agent, there is a function Called setupTunnelLayout.
The original code in the above function is as following:
//create corridor of already empty rooms
`RoomBlock emptyRoom;
for ( int j = 0; j < nColumns; j++ ) {
emptyRoom = add_roomBlocks();
emptyRoom.jumpToCell( 0, j );
if ( j == nColumns - 1 )
emptyRoom.isStartBlock = true;
emptyRoom.isTunnel = true;
}`
But in my scenario all the cells are accessible initially, so all can be tunnel (path to move), not only the (0, j) row as the above example!
I was thinking I can change it as the followings; (1) or (2);
(1)
//create corridor of already empty rooms
RoomBlock emptyRoom;
for ( int j = 0; j < nColumns; j++ )
for (int i = 0; i < nRows; i++){
emptyRoom = add_roomBlocks();
emptyRoom.jumpToCell( 0, j );
emptyRoom.jumpToCell( i, 0 );
if ( j == nColumns - 1 )
emptyRoom.isStartBlock = true;
emptyRoom.isTunnel = true;
}
Or it can be like this;
(2)
//create corridor of already empty rooms
`RoomBlock emptyRoom;
for ( int j = 0; j < nColumns; j++ )
for (int i = 0; i < nRows; i++){
emptyRoom = add_roomBlocks();
emptyRoom.jumpToCell( i, j );
if ( j == nColumns - 1 )
emptyRoom.isStartBlock = true;
emptyRoom.isTunnel = true;
}`
'Can you please let me know if (1) or (2) are correct? Which one is preferred?
Thank you so much,
Neda.'
The second is preferable; there is no point jumping the agent to the column and then the row when you can directly move it to the exact cell coordinates.
In both cases though you are missing a closing } for your extra i loop (and you're missing an opening one in your second case).
//create corridor of already empty rooms
RoomBlock emptyRoom;
for (int j = 0; j < nColumns; j++) {
for (int i = 0; i < nRows; i++) {
emptyRoom = add_roomBlocks();
emptyRoom.jumpToCell(i, j);
if (j == nColumns - 1) {
emptyRoom.isStartBlock = true;
}
emptyRoom.isTunnel = true;
}
}
[I used the StackOverflow formatting to mark the full block as Java code. It's also best-practice (although AnyLogic doesn't do it many of their example models) to always use curly brackets for loops, etc. even if their body only has one line.]
Plus whether the change will do what you want or not obviously depends on the rest of the model and how it handles the grid of cells and tunnels.
I using netbeans 8.
I need to loop to collect all employee ID from first column of jtable and store those IDs into an arraylist.
if (jTabledetail.getRowCount() > 0) {
String ecode = "";
int ishasRow = jTabledetail.getRowCount();// total 1 row
for (int r = 0; r <= ishasRow; r++) {// loop twice. First loop is gone, return to second loop or final loop for 1 row exists giving error bellow.
ecode = jTabledetail.getValueAt(r, 0).toString();
arrempcode.add(ecode);
}
}
I also tried changing to ==>> for (int r = 0; r < ishasRow; r++) but not worked.
Exception in thread "AWT-EventQueue-0" java.lang.ArrayIndexOutOfBoundsException: 1 >= 1
at java.util.Vector.elementAt(Vector.java:474)
at javax.swing.table.DefaultTableModel.getValueAt(DefaultTableModel.java:648)
at javax.swing.JTable.getValueAt(JTable.java:2717)
I don't understand the error. I known that the error comes from loop expression. I am not sure for this error.
Now my jtable named "jTabledetail" has 1 row exists.
Do I need to change something for this case of error? I am not sure that the loop expression is wrong.
Thank you very much.
DefaultTableModel tableModel = (DefaultTableModel) TableName.getModel();
Get the row count of the table
int rowCount = tableModel.getRowCount();
Declare ArrayList
ArrayList<Object> list = new ArrayList<Object>();
Traversing table and adding values into arraylist
for(int i=0; i<rowCount; i++){
for(int j=0; j<tableModel.getColumnCount(); j++){
if(j==0){
list.add(tableModel.getValueAt(i,j));
}
}
}
For any array of length greater than 10, is it safe to say that merge sort performs fewer comparisons among the array's elements than does insertion sort on the same array because the best case for the run time of merge sort is O(N log N) while for insertion sort, its O(N)?
My take on this. First off, you are talking about comparisons, but there are swaps as well that matter.
In insertion sort in the worst case (an array sorted in opposite direction) you have to do n^2 - n comparisons and swaps (11^2 - 11 = 121 - 11 = 110 for 11 elements, for example). But if the array is even partially sorted in needed order (I mean many elements already stay at correct positions or even not far from them), the number of swaps&comparisons may significantly drop. The right position for the element will be found pretty soon and there will be no need for performing as many actions as in case of an array sorted in opposite order. So, as you can see for arr2, which is almost sorted, the number of actions will become linear (in relation to the input size) - 6.
var arr1 = [11,10,9,8,7,6,5,4,3,2,1];
var arr2 = [1,2,3,4,5,6,7,8,11,10,9];
function InsertionSort(arr) {
var arr = arr, compNum = 0, swapNum = 0;
for(var i = 1; i < arr.length; i++) {
var temp = arr[i], j = i - 1;
while(j >= 0) {
if(temp < arr[j]) { arr[j + 1] = arr[j]; swapNum++; } else break;
j--;
compNum++;
}
arr[j + 1] = temp;
}
console.log(arr, "Number of comparisons: " + compNum, "Number of swaps: " + swapNum);
}
InsertionSort(arr1); // worst case, 11^2 - 11 = 110 actions
InsertionSort(arr2); // almost sorted array, few actions
In merge sort we always do aprox. n*log n actions - the properties of the input array don't matter. So, as you can see in both cases we will get both of our arrays sorted in 39 actions:
var arr1 = [11,10,9,8,7,6,5,4,3,2,1];
var arr2 = [1,2,3,4,5,6,7,8,11,10,9];
var actions = 0;
function mergesort(arr, left, right) {
if(left >= right) return;
var middle = Math.floor((left + right)/2);
mergesort(arr, left, middle);
mergesort(arr, middle + 1, right);
merge(arr, left, middle, right);
}
function merge(arr, left, middle, right) {
var l = middle - left + 1, r = right - middle, temp_l = [], temp_r = [];
for(var i = 0; i < l; i++) temp_l[i] = arr[left + i];
for(var i = 0; i < r; i++) temp_r[i] = arr[middle + i + 1];
var i = 0, j = 0, k = left;
while(i < l && j < r) {
if(temp_l[i] <= temp_r[j]) {
arr[k] = temp_l[i]; i++;
} else {
arr[k] = temp_r[j]; j++;
}
k++; actions++;
}
while(i < l) { arr[k] = temp_l[i]; i++; k++; actions++;}
while(j < r) { arr[k] = temp_r[j]; j++; k++; actions++;}
}
mergesort(arr1, 0, arr1.length - 1);
console.log(arr1, "Number of actions: " + actions); // 11*log11 = 39 (aprox.)
actions = 0;
mergesort(arr2, 0, arr2.length - 1);
console.log(arr2, "Number of actions: " + actions); // 11*log11 = 39 (aprox.)
So, answering your question:
For any array of length greater than 10, is it safe to say that merge sort performs fewer comparisons among the array's elements than does insertion sort on the same array
I would say that no, it isn't safe to say so. Merge sort can perform more actions compared to insertion sort in some cases. The size of an array isn't important here. What is important in this particular case of comparing insertion sort vs. merge sort is how far from the sorted state is your array. I hope it helps :)
BTW, merge sort and insertion sort have been united in a hybrid stable sorting algorithm called Timsort to get the best from both of them. Check it out if interested.
I am having problems with bottoms-up mergesort. I have problems sorting/merging. Current code includes:
public void mergeSort(long[] a, int len) {
long[] temp = new long[a.length];
int length = 1;
while (length < len) {
mergepass(a, temp, length, len);
length *= 2;
}
}
public void mergepass(long[] a, long[] temp, int blocksize, int len) {
int k = 0;
int i = 1;
while(i <= (len/blocksize)){
if(blocksize == 1){break;}
int min = a.length;
for(int j = 0; j < blocksize; j++){
if(a[i*j] < min){
temp[k++] = a[i*j];
count++;
}
else{
temp[k++] = a[(i*j)+1];
count++;
}
}
for(int n = 0; n < this.a.length; n++){
a[n] = temp[n];
}
}
}
Obvious problems:
i is never incremented.
At no point do you compare two elements in the array. (Is that what if(a[i*j] < min) is supposed to be doing? I can't tell.)
Why are you multiplying i and j?
What's this.a.length?
Style problems:
mergeSort() takes len as an argument, even though arrays have an implicit length. To make matters worse, the function also uses a.length and length.
Generally poor variable names.
Nitpicks:
If you're going to make a second array of the same size, it is common to make one the "source" and the other the "destination" and swap them between passes, instead of sorting into a temporary array and copying them back again.