Problem 378. Inhomogenous Depth Scale Interpolation

Created by @bmtran (Bryant Tran)

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I recently came across a dataset for physical ocean parameters such as temperature and salinity that provided data in the form of a 3-Dimensional array of depths and a 3-Dimensional array of the parameters where each point on the first two dimensions corresponds to a latitude and longitude, respectively, and the points in the third dimension corresponds to depths.Given a target depth, return the linearly interpolated values of the data at that depth without using the keywords <tt>for</tt> or <tt>while</tt> in your code.For example, if Z and V are arrays with size [ M, N, D ], the output matrix V2 would be of size [ M N ].Test Case Details:<ul><li>The depth array, Z, will always start at 0 and end at 1, which are also the bounds of the target depth d. This is so that you do not have to handle "edge" cases.</li><li>The depth array, Z, will always be sorted in the third dimension.</li><li>Both input arrays will be the same size.</li><li>The test cases are randomly generated and your solution is tested against a simple for-looping code.</li></ul>

I recently came across a dataset for physical ocean parameters such as temperature and salinity that provided data in the form of a 3-Dimensional array of depths and a 3-Dimensional array of the parameters where each point on the first two dimensions corresponds to a latitude and longitude, respectively, and the points in the third dimension corresponds to depths.

Given a target depth, return the linearly interpolated values of the data at that depth without using the keywords |for| or |while| in your code.

For example, if _Z_ and _V_ are arrays with size [ _M_, _N_, _D_ ], the output matrix _V2_ would be of size [ _M_ _N_ ].

Test Case Details:

* The depth array, _Z_, will always start at 0 and end at 1, which are also the bounds of the target depth _d_. This is so that you do not have to handle "edge" cases.
* The depth array, _Z_, will always be sorted in the third dimension.
* Both input arrays will be the same size.
* The test cases are randomly generated and your solution is tested against a simple for-looping code.

Solve

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46 Solutions
7 Solvers

Last Solution submitted on Jul 31, 2022

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1 Comment

Rafael S.T. Vieira on 10 Nov 2020

Please, bmtran, don't use isequal with floats, prefer to use abs(a-b) < tol. Unless I use the function interp1 exactly like you did, my solution cannot pass the test suite. Please try my commented code (and the difference between the solutions is at the order of 1e-15).

Solution Comments

Solution 48055

2 Comments

2 Comments

Alfonso Nieto-Castanon on 22 Feb 2012

perhaps you could specify a target numerical precision? (isequal on floats can be flaky)

Rafael S.T. Vieira on 10 Nov 2020

Alfonso's solution suffers from the same problem of precision apparently. Using isequal requires that one does the exact same operation that you did (to obain the exact same error*), and closes the door for alternative or smaller solutions.

* Floating-point numbers have some precision problems: the classic example is 0.1 which becomes an infinite number as a binary.

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