How to set period for vertical ticks?
I use d3 library version 5.7.0. I make simple graph. Vertical axis have several ticks? these ticks have period 5000.
JSFIDDLE
But i need period 10000. Please help me.
js code:
xScale = d3.scaleLinear()
.domain([0, 100000])
.range([paddingYaxis, w - padding]);
yScale = d3.scaleLinear()
.domain([0, d3.max(data, d => +d['price'] + yPeriod)])
.range([h - padding, tickSize]);
xAxis = d3.axisBottom(xScale)
.tickSizeInner(0)
.tickSizeOuter(tickSize)
.tickPadding(tickPadding);
yAxis = d3.axisLeft(yScale)
.tickSizeInner(-(w - paddingYaxis - tickSize * 2))
.tickSizeOuter(0)
.tickPadding(tickPadding)
.tickFormat(d3.format(' '));
javascript d3.js
asked Nov 24 '18 at 21:47
ptortyr45ptortyr45
576
add a comment |
I use d3 library version 5.7.0. I make simple graph. Vertical axis have several ticks? these ticks have period 5000.
JSFIDDLE
But i need period 10000. Please help me.
js code:
xScale = d3.scaleLinear()
.domain([0, 100000])
.range([paddingYaxis, w - padding]);
yScale = d3.scaleLinear()
.domain([0, d3.max(data, d => +d['price'] + yPeriod)])
.range([h - padding, tickSize]);
xAxis = d3.axisBottom(xScale)
.tickSizeInner(0)
.tickSizeOuter(tickSize)
.tickPadding(tickPadding);
yAxis = d3.axisLeft(yScale)
.tickSizeInner(-(w - paddingYaxis - tickSize * 2))
.tickSizeOuter(0)
.tickPadding(tickPadding)
.tickFormat(d3.format(' '));
javascript d3.js
asked Nov 24 '18 at 21:47
ptortyr45ptortyr45
576
1
you can use.ticks(5)or use.tickValues([0,1000,2000,3000,4000,5000])
– rioV8
Nov 24 '18 at 23:05
add a comment |
I use d3 library version 5.7.0. I make simple graph. Vertical axis have several ticks? these ticks have period 5000.
JSFIDDLE
But i need period 10000. Please help me.
js code:
xScale = d3.scaleLinear()
.domain([0, 100000])
.range([paddingYaxis, w - padding]);
yScale = d3.scaleLinear()
.domain([0, d3.max(data, d => +d['price'] + yPeriod)])
.range([h - padding, tickSize]);
xAxis = d3.axisBottom(xScale)
.tickSizeInner(0)
.tickSizeOuter(tickSize)
.tickPadding(tickPadding);
yAxis = d3.axisLeft(yScale)
.tickSizeInner(-(w - paddingYaxis - tickSize * 2))
.tickSizeOuter(0)
.tickPadding(tickPadding)
.tickFormat(d3.format(' '));
javascript d3.js
asked Nov 24 '18 at 21:47
ptortyr45ptortyr45
576
I use d3 library version 5.7.0. I make simple graph. Vertical axis have several ticks? these ticks have period 5000.
JSFIDDLE
But i need period 10000. Please help me.
js code:
xScale = d3.scaleLinear()
.domain([0, 100000])
.range([paddingYaxis, w - padding]);
yScale = d3.scaleLinear()
.domain([0, d3.max(data, d => +d['price'] + yPeriod)])
.range([h - padding, tickSize]);
xAxis = d3.axisBottom(xScale)
.tickSizeInner(0)
.tickSizeOuter(tickSize)
.tickPadding(tickPadding);
yAxis = d3.axisLeft(yScale)
.tickSizeInner(-(w - paddingYaxis - tickSize * 2))
.tickSizeOuter(0)
.tickPadding(tickPadding)
.tickFormat(d3.format(' '));
javascript d3.js
javascript d3.js
asked Nov 24 '18 at 21:47
ptortyr45ptortyr45
576
asked Nov 24 '18 at 21:47
ptortyr45ptortyr45
576
asked Nov 24 '18 at 21:47
ptortyr45ptortyr45
576
asked Nov 24 '18 at 21:47
ptortyr45ptortyr45
576
asked Nov 24 '18 at 21:47
ptortyr45ptortyr45
576
576
1
you can use.ticks(5)or use.tickValues([0,1000,2000,3000,4000,5000])
– rioV8
Nov 24 '18 at 23:05
add a comment |
1
you can use.ticks(5)or use.tickValues([0,1000,2000,3000,4000,5000])
– rioV8
Nov 24 '18 at 23:05
1
1
you can use
.ticks(5) or use .tickValues([0,1000,2000,3000,4000,5000])– rioV8
Nov 24 '18 at 23:05
you can use
.ticks(5) or use .tickValues([0,1000,2000,3000,4000,5000])– rioV8
Nov 24 '18 at 23:05
add a comment |
1 Answer
1
active
oldest
votes
This should provide a generic solution for various ranges (not just up to 50000). The ticks function allows you to choose a certain amount of ticks, dividing by the period size, gives us the wanted result.
yAxis = d3.axisLeft(yScale)
.tickSizeInner(-(w - paddingYaxis - tickSize * 2))
.tickSizeOuter(0)
.tickPadding(tickPadding)
.tickFormat(d3.format(' '))
.ticks(d3.max(data, d => +d['price'] + yPeriod) / 10000);
answered Nov 25 '18 at 11:11
ShushanShushan
1,1171814
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
votes
active
oldest
votes
This should provide a generic solution for various ranges (not just up to 50000). The ticks function allows you to choose a certain amount of ticks, dividing by the period size, gives us the wanted result.
yAxis = d3.axisLeft(yScale)
.tickSizeInner(-(w - paddingYaxis - tickSize * 2))
.tickSizeOuter(0)
.tickPadding(tickPadding)
.tickFormat(d3.format(' '))
.ticks(d3.max(data, d => +d['price'] + yPeriod) / 10000);
answered Nov 25 '18 at 11:11
ShushanShushan
1,1171814
add a comment |
This should provide a generic solution for various ranges (not just up to 50000). The ticks function allows you to choose a certain amount of ticks, dividing by the period size, gives us the wanted result.
yAxis = d3.axisLeft(yScale)
.tickSizeInner(-(w - paddingYaxis - tickSize * 2))
.tickSizeOuter(0)
.tickPadding(tickPadding)
.tickFormat(d3.format(' '))
.ticks(d3.max(data, d => +d['price'] + yPeriod) / 10000);
answered Nov 25 '18 at 11:11
ShushanShushan
1,1171814
add a comment |
This should provide a generic solution for various ranges (not just up to 50000). The ticks function allows you to choose a certain amount of ticks, dividing by the period size, gives us the wanted result.
yAxis = d3.axisLeft(yScale)
.tickSizeInner(-(w - paddingYaxis - tickSize * 2))
.tickSizeOuter(0)
.tickPadding(tickPadding)
.tickFormat(d3.format(' '))
.ticks(d3.max(data, d => +d['price'] + yPeriod) / 10000);
answered Nov 25 '18 at 11:11
ShushanShushan
1,1171814
This should provide a generic solution for various ranges (not just up to 50000). The ticks function allows you to choose a certain amount of ticks, dividing by the period size, gives us the wanted result.
yAxis = d3.axisLeft(yScale)
.tickSizeInner(-(w - paddingYaxis - tickSize * 2))
.tickSizeOuter(0)
.tickPadding(tickPadding)
.tickFormat(d3.format(' '))
.ticks(d3.max(data, d => +d['price'] + yPeriod) / 10000);
answered Nov 25 '18 at 11:11
ShushanShushan
1,1171814
answered Nov 25 '18 at 11:11
ShushanShushan
1,1171814
answered Nov 25 '18 at 11:11
ShushanShushan
1,1171814
answered Nov 25 '18 at 11:11
ShushanShushan
1,1171814
1,1171814
add a comment |
add a comment |
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1
you can use
.ticks(5)or use.tickValues([0,1000,2000,3000,4000,5000])– rioV8
Nov 24 '18 at 23:05