You have to change a few things in your code. First, the viewColumns needs
to contain both columns 0 and 1 to start if you want Column1 and Column2 to
both be in the output. Then, you need to adjust the columns you add to the
viewColumns: each one needs a calc parameter which calculates the value in
the column. In this case, you want to compare the value of the column to
distinctValues[i] and return 1 when they match and 0 when they don't. In
the aggregation function for groupColumns, use sum instead of count:
var distinctValues = data.getDistinctValues(2);
var viewColumns = [0, 1];
var groupColumns = [];
// build column arrays for the view and grouping
for (var i = 0; i < distinctValues.length; i++) {
viewColumns.push({
type: 'number',
label: distinctValues[i],
calc: (function (x) {
return function (dt, row) {
return (dt.getValue(row, 2) == x) ? 1 : 0;
}
})(distinctValues[i])
});
groupColumns.push({
column: i+2,
type: 'number',
//label: distinctValues[i],
aggregation: google.visualization.data.sum
});
}
Then, in the grouping function, pass columns 0 and 1 in the first array:
var pivotedData = google.visualization.data.group(view, [0, 1],
groupColumns);
See these changes working here: http://jsfiddle.net/asgallant/DUn6B/1/
On Monday, December 16, 2013 1:02:35 AM UTC-5, Pradeep Shankar M wrote:
>
> I excited with this fiddle <http://jsfiddle.net/asgallant/HkjDe/> and I
> tried to create the same kind with reference to that fiddle. My modified
> sample is given in here <http://jsfiddle.net/DUn6B/> and i'm trying to
> create a view as follows.
>
> var distinctValues = data.getDistinctValues(2);
>
> var viewColumns = [1];
> var groupColumns = [];
> // build column arrays for the view and grouping
> for (var i = 0; i < distinctValues.length; i++) {
> viewColumns.push({
> type: 'number',
> label: distinctValues[i],
> aggregation: google.visualization.data.count
> });
> groupColumns.push({
> column: i+1,
> type: 'number',
> //label: distinctValues[i],
> aggregation: google.visualization.data.sum
> });
> }
>
> But my aim is to create a pivot table something like as follows.
>
> ['Column1', 'Column2', 100, 200, 300, 400],['A', 'bar', 0, 1, 1, 0],['A',
> 'baz', 0, 0, 1, 0],['A', 'foo', 3, 1, 0, 0],['B', 'baz', 0, 1, 0, 0],['B',
> 'cad', 1, 0, 1, 1],['B', 'qud', 1, 1, 1, 2]
>
> How can I proceed?
>
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