What's up with UnitRange?

question
#1

Consider in matlab:

>> 1.12*100:0.02*100:1.2*100

ans =

          112          114          116          118          120

In julia I get

julia> 1.12*100:0.02*100:1.2*100
112.00000000000001:2.0:118.00000000000001

which when I collect is definitely not what I expected

julia> collect(1.12*100:0.02*100:1.2*100)
4-element Array{Float64,1}:
 112.00000000000001
 114.00000000000001
 116.00000000000001
 118.00000000000001
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#2

In Matlab everything is a Complex Float, but it rounds the answer when possible. In Julia you explicitly asked for a Float, and so you got one.

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#3

Makes sense, but I still find it very unintuitive. Is there anything I can do to match the functionality of matlab in this case?

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#4

Is this not just printing though? Matlab just truncates it to avoid that question.

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#5

Yes, Matlab may be lying to you and truncating its display. It’s also returning a floating point number — it’s just formatted like Julia formats integers. I no longer have a Matlab installed, but try

r = 1.12*100:0.02*100:1.2*100
fprintf('%.17f\n',r(1)) 

I do know that Matlab also does some “snapping” to integral endpoints within its colon syntax, so that may (also) be happening here.

Floating point arithmetic doesn’t behave how you might expect — across all languages. Indeed, just check out the result of 1.12*100 in both Matlab and Julia. Again, Matlab will likely obscure the “real” answer via its printing.

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#6

collect(round(Int, 1.12*100):round(Int, 0.02*100):round(Int, 1.2*100)) would work.

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#7

I tried the snippet you wrote, and yeah Matlab is just hiding the ugly in the print statements. But what I’m more concerned about is properly computing the number of elements in the range (the “snapping” as you say). 112.00000000000001421 is pretty much 112, but a range with 4 elements where it’s supposed to have 5 is a much bigger problem.

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#8

That would work if I expected all the elements in the UnitRange to be integers, but I don’t. For example, if I do this in matlab, I get what I expect:

>> 1.12*87:0.02*87:1.2*87

ans =

        97.44        99.18       100.92       102.66        104.4

But in Julia, I’m still losing an element because of the small +ϵ tacked on

julia> collect(1.12*87:0.02*87:1.2*87)
4-element Array{Float64,1}:
  97.44000000000001
  99.18
 100.92000000000002
 102.66000000000001
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#9

Ah, I missed that. That’s the other part of my answer — the slop for snapping. We don’t do slop, but it’s still something we try very hard to get right. It’s a challenge because of floating point issues like this. In general, if you write the values directly as floating point literals (instead of an multiplication) we’ll get them almost always right.

As such, a nice workaround is to do the multiplication outside the range. It also hits your intended values much more accurately.

julia> (1.12:0.02:1.2) .* 87
97.44:1.74:104.4

julia> collect(ans)
5-element Array{Float64,1}:
  97.44
  99.18
 100.92
 102.66
 104.4
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#10

You can use range with keyword length:

julia> collect(range(1.12*100,1.20*100, length=5))
5-element Array{Float64,1}:
 112.00000000000001
 114.00000000000001
 116.0
 118.0
 120.0

Here the step size is implicit, which might be inconvenient in some cases, but this always guarantees the end point to be included.

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#11

Thanks, I appreciate the lack of slop with Julia but it can be frustrating at times. This is a simple fix that’ll work for me.

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#12

How does Julia calculate (1.12:0.02:1.2)*87 to get the crisp values?

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#13

It infers more precision than 1.12 and 1.2 specify since taken at face value, those end-points and a length of 5 are incompatible since 1.12 + 4*0.02 != 1.2. If you call dump on the range object, you can see the higher precision values that are inferred upon construction:

julia> dump((1.12:0.02:1.2))
StepRangeLen{Float64,Base.TwicePrecision{Float64},Base.TwicePrecision{Float64}}
  ref: Base.TwicePrecision{Float64}
    hi: Float64 1.12
    lo: Float64 -1.0658141036401502e-16
  step: Base.TwicePrecision{Float64}
    hi: Float64 0.01999999999999999
    lo: Float64 9.992007221626408e-18
  len: Int64 5
  offset: Int64 1

Indeed, the true value 1.12 is about 1.0658141036401502e-16 smaller than the floating-point value 1.12.

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#14

The end-point is always included but it may not be what you wanted. In this case, 1.12*100 == 112.00000000000001 != 112 whereas:

julia> collect((1.12:0.02:1.2)*100)
5-element Array{Float64,1}:
 112.0
 114.0
 116.0
 118.0
 120.0

The values are all exact integer values.

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#15

So range object triggers high precision multiplication. Very neat!

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#16

Julia is doing two things: it first realizes that the floating point value 0.02 — that is, 0.0200000000000000004163336342344337026588618755340576171875 — is the closest representable value to 1//50. And that the start and stop can also be expressed in terms of a common denominator with the step. Then, secondly, it finds the closest twice-precision (128-bit) floating point representation for this exact fraction:

julia> big(0.02)
0.0200000000000000004163336342344337026588618755340576171875

julia> big((1.12:0.02:1.2).step)
0.019999999999999999999999999999999630221450677650716213252235023693954840684483542645466513931751251220703125

It’s not just high precision; it’s also a rationalization of the number you gave it. I should note that we are snapping, but we’re only changing the exact values by less than eps(x)/2. In other words, if you converted this value back to a normal Float64, you’d get the same number you started with. This is why I talked about writing the floating point literals yourself. If you write them out, you’re writing some decimal fraction — something divisible by a power of 10 — that is able to be rationalized in this manner. If you instead do 0.02*87, then you’re also multiply that trailing ...00004163336... in the full expansion of 0.2 87 times, and that adds up to the point that you’re more than a half-step away from the rational number you really wanted. Matlab, on the other hand, seems to have some “closeness” heuristics that are much more forgiving.

A really awesome side-effect of the high precision arithmetic backing this is that the intermediate values are also much more frequently what you intended. The classic example is the third value in 0.1:0.1:0.9 — it’s 0.3 in Julia, but 0.30000000000000004 in most other languages.

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#17

Actually, it’s 0.29999999999999998889776975… in Julia; it just prints as 0.3 because that is the shortest literal that parses to the same value. The fraction 3/10 is not representable in binary floating point.

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#18

Right, it’s 0.3 but not 0.3 if you see what I’m saying.

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