how to modify path's order according to the x axis when using d3.js
I want to draw a stacked area graph which can show the tendency and the value.
For example:
data:
year,orange,apple
1976,30,20
1980,20,40
1988,50,30
related code:
stack = d3.stack()
.order(d3.stackOrderAscending)
// .order(order)
.keys(keys);
series = stack(data);
area = d3
.area()
.x(function(d) {
return xScale(new Date(d.data.year, 0, 1));
})
.y0(function(d) {
return yScale(d[0]);
})
.y1(function(d) {
return yScale(d[1]);
});
$ele
.selectAll('path')
.data(series)
.enter()
.append('path')
.attr('class', 'area')
.attr('transform', 'translate(' + template.padding + ',0)')
.attr('d', area)
.attr('fill', function(d, i) {
return get_colors(i);
});
It shows like this:
The current result
But what I want to do is to put the max value on the top (according to the x axis's value)of the graph
And therefore the order of the categories will change with the x axis.
I tried to define my own order function,but it only return an array which seems to define only a order, which means the order can not change.
I hope you can understand what I said.Thanks！！！
d3.js stacked-chart
edited Nov 22 '18 at 12:21
rioV8
4,5742311
asked Nov 22 '18 at 12:17
朱融晨朱融晨
82
add a comment |
I want to draw a stacked area graph which can show the tendency and the value.
For example:
data:
year,orange,apple
1976,30,20
1980,20,40
1988,50,30
related code:
stack = d3.stack()
.order(d3.stackOrderAscending)
// .order(order)
.keys(keys);
series = stack(data);
area = d3
.area()
.x(function(d) {
return xScale(new Date(d.data.year, 0, 1));
})
.y0(function(d) {
return yScale(d[0]);
})
.y1(function(d) {
return yScale(d[1]);
});
$ele
.selectAll('path')
.data(series)
.enter()
.append('path')
.attr('class', 'area')
.attr('transform', 'translate(' + template.padding + ',0)')
.attr('d', area)
.attr('fill', function(d, i) {
return get_colors(i);
});
It shows like this:
The current result
But what I want to do is to put the max value on the top (according to the x axis's value)of the graph
And therefore the order of the categories will change with the x axis.
I tried to define my own order function,but it only return an array which seems to define only a order, which means the order can not change.
I hope you can understand what I said.Thanks！！！
d3.js stacked-chart
edited Nov 22 '18 at 12:21
rioV8
4,5742311
asked Nov 22 '18 at 12:17
朱融晨朱融晨
82
you have to write the stacking yourself, start by using the series as created bystack(). Result: i.imgur.com/TjKwy9Z.png
– rioV8
Nov 22 '18 at 14:28
Yeah! I succeeded it as you said! Thanks！
– 朱融晨
Nov 23 '18 at 3:27
add a comment |
I want to draw a stacked area graph which can show the tendency and the value.
For example:
data:
year,orange,apple
1976,30,20
1980,20,40
1988,50,30
related code:
stack = d3.stack()
.order(d3.stackOrderAscending)
// .order(order)
.keys(keys);
series = stack(data);
area = d3
.area()
.x(function(d) {
return xScale(new Date(d.data.year, 0, 1));
})
.y0(function(d) {
return yScale(d[0]);
})
.y1(function(d) {
return yScale(d[1]);
});
$ele
.selectAll('path')
.data(series)
.enter()
.append('path')
.attr('class', 'area')
.attr('transform', 'translate(' + template.padding + ',0)')
.attr('d', area)
.attr('fill', function(d, i) {
return get_colors(i);
});
It shows like this:
The current result
But what I want to do is to put the max value on the top (according to the x axis's value)of the graph
And therefore the order of the categories will change with the x axis.
I tried to define my own order function,but it only return an array which seems to define only a order, which means the order can not change.
I hope you can understand what I said.Thanks！！！
d3.js stacked-chart
edited Nov 22 '18 at 12:21
rioV8
4,5742311
asked Nov 22 '18 at 12:17
朱融晨朱融晨
82
I want to draw a stacked area graph which can show the tendency and the value.
For example:
data:
year,orange,apple
1976,30,20
1980,20,40
1988,50,30
related code:
stack = d3.stack()
.order(d3.stackOrderAscending)
// .order(order)
.keys(keys);
series = stack(data);
area = d3
.area()
.x(function(d) {
return xScale(new Date(d.data.year, 0, 1));
})
.y0(function(d) {
return yScale(d[0]);
})
.y1(function(d) {
return yScale(d[1]);
});
$ele
.selectAll('path')
.data(series)
.enter()
.append('path')
.attr('class', 'area')
.attr('transform', 'translate(' + template.padding + ',0)')
.attr('d', area)
.attr('fill', function(d, i) {
return get_colors(i);
});
It shows like this:
The current result
But what I want to do is to put the max value on the top (according to the x axis's value)of the graph
And therefore the order of the categories will change with the x axis.
I tried to define my own order function,but it only return an array which seems to define only a order, which means the order can not change.
I hope you can understand what I said.Thanks！！！
d3.js stacked-chart
d3.js stacked-chart
edited Nov 22 '18 at 12:21
rioV8
4,5742311
asked Nov 22 '18 at 12:17
朱融晨朱融晨
82
edited Nov 22 '18 at 12:21
rioV8
4,5742311
asked Nov 22 '18 at 12:17
朱融晨朱融晨
82
edited Nov 22 '18 at 12:21
rioV8
4,5742311
edited Nov 22 '18 at 12:21
rioV8
4,5742311
edited Nov 22 '18 at 12:21
rioV8
4,5742311
4,5742311
asked Nov 22 '18 at 12:17
朱融晨朱融晨
82
asked Nov 22 '18 at 12:17
朱融晨朱融晨
82
asked Nov 22 '18 at 12:17
朱融晨朱融晨
82
82
you have to write the stacking yourself, start by using the series as created bystack(). Result: i.imgur.com/TjKwy9Z.png
– rioV8
Nov 22 '18 at 14:28
Yeah! I succeeded it as you said! Thanks！
– 朱融晨
Nov 23 '18 at 3:27
add a comment |
you have to write the stacking yourself, start by using the series as created bystack(). Result: i.imgur.com/TjKwy9Z.png
– rioV8
Nov 22 '18 at 14:28
Yeah! I succeeded it as you said! Thanks！
– 朱融晨
Nov 23 '18 at 3:27
you have to write the stacking yourself, start by using the series as created by
stack(). Result: i.imgur.com/TjKwy9Z.png– rioV8
Nov 22 '18 at 14:28
you have to write the stacking yourself, start by using the series as created by
stack(). Result: i.imgur.com/TjKwy9Z.png– rioV8
Nov 22 '18 at 14:28
Yeah! I succeeded it as you said! Thanks！
– 朱融晨
Nov 23 '18 at 3:27
Yeah! I succeeded it as you said! Thanks！
– 朱融晨
Nov 23 '18 at 3:27
add a comment |
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you have to write the stacking yourself, start by using the series as created by
stack(). Result: i.imgur.com/TjKwy9Z.png– rioV8
Nov 22 '18 at 14:28
Yeah! I succeeded it as you said! Thanks！
– 朱融晨
Nov 23 '18 at 3:27