linear.nice should use d3.tickIncrement for precise rounding.

[pavlin99th]

Say you want a nice linear scale and ticks for your [5.83, 6.2] domain.
Due to float calculations .nice() sets the domain to:

d3.scaleLinear().domain([5.83, 6.2]).nice().domain()
[5.800000000000001, 6.2]

and .ticks() returns:

d3.scaleLinear().domain([5.83, 6.2]).nice().ticks()
[5.8500000000000005, 5.9, 5.95, 6.000000000000001, 6.050000000000001, 6.1000000000000005, 6.15, 6.2]

Now you are missing the first 5.8 tick.

I know this happens because of float calculations (0.05 * 116), what is the best way to deal with the situation? Should it be solved by d3 or by a user?

I ended up with rounding and resetting the domain with the help of d3.format, but looking for a more general solution.

[mbostock]

Thanks for the report. I think we can fix this without too much trouble. This will affect both d3-array and d3-scale, since the core tick methods are in d3-array, but the implementation of linear.nice will be affected. Rather than tracking this in both places, let’s just use d3/d3-array#45.

[pmalouin]

Just a heads up that I can still reproduce this issue with the latest version of d3-array (1.2.0). I expected d3/d3-array@04cf783 would solve the problem.

Here are my dependencies:

└─┬ d3-scale@1.0.5
  ├── d3-array@1.2.0
  ├── d3-collection@1.0.3
  ├── d3-color@1.0.3
  ├── d3-format@1.2.0
  ├── d3-interpolate@1.1.4
  ├── d3-time@1.0.6
  └── d3-time-format@2.0.5

> var d3Scale = require('d3-scale');
undefined
> d3Scale.scaleLinear().domain([5.83, 6.2]).nice().domain()
[ 5.800000000000001, 6.2 ]
> d3Scale.scaleLinear().domain([5.83, 6.2]).nice().ticks()
[ 5.85, 5.9, 5.95, 6, 6.05, 6.1, 6.15, 6.2 ]

[mbostock]

No; linear scales still use d3.tickStep, not the new d3.tickIncrement, and so there’s no protection against rounding errors. See linear.js.

However, I don’t see any reason why we couldn’t update linear.nice to use d3.tickIncrement and thus guarantee exact behavior (within the limits of IEEE 754). So I’ll reopen and retitle this issue accordingly.