# Find the x for y in chart

Hi.

If i know the y value how can i find the x value and vice versa?

Example for (Chart1.SeriesList.Series[0].MaxYValue) how can i find the x ?
LVL 16
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Assuming that the x values and y values are unique, the following functions could be used against a given chart series.

Russell

---

function FindXForY(YValue: Double; Series: TChartSeries): Double;
var  dwIndex:       Integer;
begin

// Set default result
result:=0;

// Find the index of the y value
for dwIndex:=0 to Pred(Series.YValues.Count) do
begin
if (Series.YValue[dwIndex] = YValue) then
begin
// Return the x value
result:=Series.XValue[dwIndex];
// Done
break;
end;
end;

end;

function FindYForX(XValue: Double; Series: TChartSeries): Double;
var  dwIndex:       Integer;
begin

// Set default result
result:=0;

// Find the index of the y value
for dwIndex:=0 to Pred(Series.XValues.Count) do
begin
if (Series.XValue[dwIndex] = XValue) then
begin
// Return the y value
result:=Series.YValue[dwIndex];
// Done
break;
end;
end;

end;

0

The MaxYValue property returns the largest Y value in the chart series, it does not return you the index of the data point (which would be needed in order to get the corresponding X value). You could walk the series YValues[index] until you found the yvalue that = MaxYValue, which would then give you the index, but if there are 2 data points where the YValue = MaxYValue, then you would not be able to determine what data point you needed to get. Eg, given the 6 sample data points in the format of index = (x, y)

0 = (1, 5)
1 = (2, 7)
2 = (10, 20)
3 = (11, 14)
4 = (15, 12)
5 = (8, 20)

The series MaxYValue would return 20. But because there are 2 data points with Y=20, you would be unable to determine if the corresponding X value was supposed to be 10 or 8. So, only if EVERY y value and EVERY x value is unique, will you be able to do what you are asking.

Russell

0

Author Commented:
Thank you Russell.

But how can find the x. Given that y is unique.
Can you tell me the commands?
0

Author Commented:
Thank you very much Russell :)
0

No problem,
Russell
0
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