I have issue i'm using MpChart to show radar chart.
The chart render by Mpchart
Now, I want to show outermost line same
The chart i want to show
i have found solution. I override RadarRenderer class, fun drawExtra
for (int j = 0; j < labelCount; j++) {
if (j < labelCount - 1) continue;
for (int i = 0; i < mChart.getData().getEntryCount(); i++) {
float r = (mChart.getYAxis().mEntries[j] - mChart.getYChartMin()) * factor;
Utils.getPosition(center, r, sliceangle * i + rotationangle, p1out);
Utils.getPosition(center, r, sliceangle * (i + 1) + rotationangle, p2out);
c.drawLine(p1out.x, p1out.y, p2out.x, p2out.y, mWebPaint);
}
}
Related
I'm tring to access the values of a dictionary property in a grid,such as Fluvial facies or lithologies etc.I have read the coursebook and help docs, but didn't find anything relevant.The coursebook only has examples of creating properties, but not accessing properties.Below is the code I tried:
Grid grid = arguments.Input_Grid;
if (grid == null)
{
PetrelLogger.ErrorStatus("HelloGrid: Arguments cannot be empty.");
return;
}
Index3 currentCell = new Index3();
int maxI = grid.NumCellsIJK.I;
int maxJ = grid.NumCellsIJK.J;
int maxK = grid.NumCellsIJK.K;
for (int i = 0; i < maxI; i++)
{
for (int j = 0; j < maxJ; j++)
{
for (int k = 0; k < maxK; k++)
{
currentCell.I = i; currentCell.J = j; currentCell.K = k;
if (grid.IsCellDefined(currentCell) && grid.HasCellVolume(currentCell))
{
//DictionaryProperty p = ???
//int val = p[currentCell] ???
}
}
}
}
You need to use the "FastDictionaryPropertyIndexer" or "FastPropertyIndexer" for regular properties.
foreach (var dictProp in grid.DictionaryProperties)
{
int numCellsI = dictProp.NumCellsIJK[0];
int numCellsJ = dictProp.NumCellsIJK[1];
int numCellsK = dictProp.NumCellsIJK[2];
float[] values = new float[dictProp.NumCells];
var dpsa = dictProp.SpecializedAccess;
using (var fdpi = dpsa.OpenFastDictionaryPropertyIndexer())
{
int index = 0;
for (int k = 0; k < numCellsK; k++)
{
for (int j = 0; j < numCellsJ; j++)
{
for (int i = 0; i < numCellsI; i++)
{
values[index] = fdpi[i, j, k];
index++;
}
}
}
}
}
You also need to be careful about the indexing since it varies by project. For instance, you may need to reverse the order of traversal in the J direction or you could end up with some strange results.
I tried to find the time complexity of the following two functions:
the first one
public static int myMethod1(int[] arr) {
int x = 0;
for (int i = 0; i < arr.length / 2; i++) {
for (int j = 0; j < arr.length; j++) {
for (int k = 0; k < arr.length; k++) {
x++;
if (k == arr.length / 2) {
break;
}
}
}
}
return x;
}
So with this one i am thinking.
The method contains 3 loops, and the loops are iterating over variable i, j and k…
i and j, and k are both incremented by 1 for each passing… this gives us as N For each LOOP which leaves us with three N’s.., which gives is O(N^3)
The next one is:
public static int myMethod(int N) {
int x = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N / 2; j++) {
for (int k = 1; k < N;) {
x++;
k *= 2;
}
}
}
return x;
}
With this i am thinking.
The method contains 3 loops, and the loops are iterating over variable i, j and k… i and j are both incremented by 1 for each passing… this gives us as N For each LOOP which leaves us with two N’s.. The last loop k doubles, which gives is log(n).
The result of the this problem is therefore O(N^2· log (N))
is this correct? and if it is not, why?
You are right. In both of the questions
nevertheless i change all cells of sheets in epplus to double sheets,but the error operand += happen in my code.this is my code.
sheet.Cells["C1:C"+sheet.Dimension.Rows].Style.Numberformat.Format = "0";
//sheet.Cells["C1:C" + sheet.Dimension.Rows].Value = s;
for (int x = 1; x <= sheet2.Dimension.Rows; x++)
{
for (int y = 1; y <= sheet.Dimension.Rows; y++)
{
if (sheet.Cells[y, 1].Value.ToString() != null)
{
if (sheet2.Cells[x, 1].Value.Equals(sheet.Cells[y, 1].Value.ToString()))
{
sheet2.Cells[x, 4].Value.GetType<Double>() += sheet.Cells[y, 3].Value.GetType<Double>();
}
}
}
}
thank you for your help
your resolve help me to solve my code tnx. this is the resolution of code.
for (int x = 1; x <= 604; x++)
{
for (int y = 1; y <= sheet.Dimension.Rows; y++)
{
if (sheet.Cells[y, 1].Value.ToString() != null)
{
if (sheet.Cells[y, 1].Value.Equals(sheet2.Cells[x, 1].Value.ToString()))
{
double cellValue = Convert.ToDouble(sheet2.Cells[x, 4].Value);
cellValue += Convert.ToDouble(sheet.Cells[y, 3].Value);
sheet2.Cells[x, 4].Value = cellValue;
}
}
}
}
The reason of the error is because you are trying to += a getter function. You can only do that operation in objects that can be assigned values.
You could for example do the following
if (sheet2.Cells[x, 1].Value.Equals(sheet.Cells[y, 1].Value.ToString()))
{
double cellValue = (double)sheet2.Cells[x, 4].Value;
cellValue += (double)sheet.Cells[y, 3].Value;
sheet2.Cells[x, 4].Value = cellvalue;
}
Edit: The following is untested, but could also work: ((double)sheet2.Cells[x, 4].Value) += (double)sheet.Cells[y, 3].Value
Here's a sample solution for Sliding Window Maximum problem in Java.
Given an array nums, there is a sliding window of size k which is
moving from the very left of the array to the very right. You can only
see the k numbers in the window. Each time the sliding window moves
right by one position.
I want to get the time and space complexity of this function. Here's what I think would be the answer:
Time: O((n-k)(k * logk)) == O(nklogk)
Space (auxiliary): O(n) for return int[] and O(k) for pq. Total of O(n).
Is this correct?
private static int[] maxSlidingWindow(int[] a, int k) {
if(a == null || a.length == 0) return new int[] {};
PriorityQueue<Integer> pq = new PriorityQueue<Integer>(k, new Comparator<Integer>() {
// max heap
public int compare(Integer o1, Integer o2) {
return o2 - o1;
}
});
int[] result = new int[a.length - k + 1];
int count = 0;
// time: n - k times
for (int i = 0; i < a.length - k + 1; i++) {
for (int j = i; j < i + k; j++) {
// time k*logk (the part I'm not sure about)
pq.offer(a[j]);
}
// logk
result[count] = pq.poll();
count = count + 1;
pq.clear();
}
return result;
}
You're right in most of the part except -
for (int j = i; j < i + k; j++) {
// time k*logk (the part I'm not sure about)
pq.offer(a[j]);
}
Here total number of executions is log1 + log2 + log3 + log4 + ... + logk. The summation of this series -
log1 + log2 + log3 + log4 + ... + logk = log(k!)
And second thought is, you can do it better than your linearithmic time solution using double-ended queue property which will be O(n). Here is my solution -
public int[] maxSlidingWindow(int[] nums, int k) {
if (nums == null || k <= 0) {
return new int[0];
}
int n = nums.length;
int[] result = new int[n - k + 1];
int indx = 0;
Deque<Integer> q = new ArrayDeque<>();
for (int i = 0; i < n; i++) {
// remove numbers out of range k
while (!q.isEmpty() && q.peek() < i - k + 1) {
q.poll();
}
// remove smaller numbers in k range as they are useless
while (!q.isEmpty() && nums[q.peekLast()] < nums[i]) {
q.pollLast();
}
q.offer(i);
if (i >= k - 1) {
result[indx++] = nums[q.peek()];
}
}
return result;
}
HTH.
I am trying to print out arrays incrementally like this;
TractMultBox->Text = rows[0] + newline;
TractMultBox->Text += rows[1] + rows[0] + newline;
TractMultBox->Text += rows[2] + rows[1] + rows[0] + newline;
which would give an output like this
3
43
543
I can do fine with this code, however. It would like to use a for loop, that would make it easier, since I would like it to output all arrays until max is reached automatically.
I'm assuming you want to concatenate and not sum.
string text;
for (int i = 0; i < rows.count; ++i)
{
text = rows[i] + text;
TractMultBox->Text = text + newline;
}
for less lines of code.
string text = newline;
for (int i = 0; i < rows.count; ++i)
{
TractMultBox->Text = (text = rows[i] + text);
}
but that's a little hard to read.
Sounds like a job for a for loop indeed perhaps something like this:
#include <iostream>
int main()
{
int rows[3] = {3, 4, 5};
for (int i(0); i < 3; ++i)
{
for (int j(i); j >= 0; --j)
std::cout << rows[j];
std::cout << "\n";
}
std::cin.get();
return 0;
}
If rows contained 345 this would give you the following output:
3
43
543
Not sure if that's what you wanted but you can adjust the loops accordingly. The key is to have 2 for loops.
Edit: Changed to self contained example you can play with
What about a double loop like:
for (int i = 0; i < maxNRows; ++i)
{
for (int j = 0; j < i; ++j)
{
TractMultBox->Text += rows[j];
}
TractMultBox->Text += newline;
}