interpolation

Bicubic Interpolation on Scattered Data in Matlab

由 巧了我就是萌 提交于 2019-12-07 14:29:58
问题 I was wondering if there is a way to do bicubic interpolation on a scattered data set (2d)? I did some online searches and figured out that bicubic patches (not sure what these are) need to be fitted on scattered data. Not sure how to proceed from here. Any help will be appreciated! Note: I understand that Matlab recommends TriScatteredInterp and griddata to interpolate on scattered points. However, TriScatteredInterp is a triangulation interpolation and does not uses splines to interpolate

How to do interpolation on datetime and float

由 陌路散爱 提交于 2019-12-07 11:59:24
I am doing 1d interpolation using scipy on time-series. My x-axis data is in datetime format and y axis is in float like: 3/15/2012 16:00:00 32.94 3/16/2012 16:00:00 32.95 3/19/2012 16:00:00 32.61 Now during slope calculation slope = (y_hi-y_lo) / (x_hi-x_lo) i am getting the error TypeError: unsupported operand type(s) for /: 'float' and 'datetime.timedelta' which is an obvious error. Can someone point me toward the right direction, How to handle it ? Your issue is that you are trying to divide a float by a datetime.timedelta object which is, as you said, obviously throwing a TypeError. You

What is the algorithm used to interpolate in Matlab's imresize function?

由 怎甘沉沦 提交于 2019-12-07 11:57:13
问题 I am using the Matlab/Octave imresize() function which resamples a given 2D array. I want to understand how a particular interpolation algorithm used in imresize works. (I am using octave on windows) e.g. A = 1 2 3 4 is a 2D array. Then I use the command b=imresize(a,2,'linear'); basically upsampling row and columns by 2. The output is 1.0000 1.3333 1.6667 2.0000 1.6667 2.0000 2.3333 2.6667 2.3333 2.6667 3.0000 3.3333 3.0000 3.3333 3.6667 4.0000 I don't understand how this linear

d3.js 3D array interpolation

由 空扰寡人 提交于 2019-12-07 10:12:32
问题 Code is here: http://jsfiddle.net/S48QX/. I want to draw a image based on a 3D data set, for example: var data = [ {x:1.428, y:0.500, energy:0.458}, {x:1.428, y:1.191, energy:0.616}, {x:1.428, y:1.882, energy:0.795}, {x:1.428, y:2.573, energy:0.642}, {x:1.428, y:3.264, energy:0.536}, {x:1.428, y:3.955, energy:0.498}, {x:1.428, y:4.646, energy:0.494}, {x:1.428, y:5.337, energy:0.517}, ... } It's like scattered plot, but I need every pixel to be set, not just a bunch of color dots on the image.

what kind of algorithm for generating height-map from contour line?

由 佐手、 提交于 2019-12-07 09:07:26
问题 I'm looking for interpolating some contour lines to generating a 3D view. The contours are not stored in a picture, coordinates of each point of the contour are simply stored in a std::vector. for convex contours : , it seems (I didn't check by myself) that the height can be easily calculates (linear interpolation) by using the distance between the two closest points of the two closest contours. my contours are not necessarily convex : , so it's more tricky... actualy I don't have any idea

Using pandas reindex with floats: interpolation

由 混江龙づ霸主 提交于 2019-12-07 08:58:24
问题 Can you explain this bizarre behaviour? df=pd.DataFrame({'year':[1986,1987,1988],'bomb':arange(3)}).set_index('year') In [9]: df.reindex(arange(1986,1988.125,.125)) Out[9]: bomb 1986.000 0 1986.125 NaN 1986.250 NaN 1986.375 NaN 1986.500 NaN 1986.625 NaN 1986.750 NaN 1986.875 NaN 1987.000 1 1987.125 NaN 1987.250 NaN 1987.375 NaN 1987.500 NaN 1987.625 NaN 1987.750 NaN 1987.875 NaN 1988.000 2 In [10]: df.reindex(arange(1986,1988.1,.1)) Out[10]: bomb 1986.0 0 1986.1 NaN 1986.2 NaN 1986.3 NaN 1986

Correct usage of scipy.interpolate.RegularGridInterpolator

由 柔情痞子 提交于 2019-12-07 08:14:19
问题 I am a little confused by the documentation for scipy.interpolate.RegularGridInterpolator. Say for instance I have a function f: R^3 => R which is sampled on the vertices of the unit cube. I would like to interpolate so as to find values inside the cube. import numpy as np # Grid points / sample locations X = np.array([[0,0,0], [0,0,1], [0,1,0], [0,1,1], [1,0,0], [1,0,1], [1,1,0], [1,1,1.]]) # Function values at the grid points F = np.random.rand(8) Now, RegularGridInterpolator takes a points

interpolation of grouped data using data.table

由 允我心安 提交于 2019-12-07 07:39:11
问题 This is a continuation of a question that I had originally posted at http://r.789695.n4.nabble.com/subset-between-data-table-list-and-single-data-table-object-tp4673202.html . Matthew had suggested that I post my question here so I am doing that now. This is my input below: library(data.table) library(pracma) # for the interp1 function tempbigdata1 <- data.table(c(14.80, 14.81, 14.82), c(7900, 7920, 7930), c("02437100", "02437100", "02437100")) tempbigdata2 <- data.table(c(9.98, 9.99, 10.00),

Android Is it possible to use concurrent interpolators?

由 こ雲淡風輕ζ 提交于 2019-12-07 06:49:30
问题 I have a set of two animations, both animations run together using the overshoot interpolator <?xml version="1.0" encoding="utf-8"?> <set xmlns:android="http://schemas.android.com/apk/res/android" android:interpolator="@android:anim/overshoot_interpolator" > <translate android:duration="6000" android:fromXDelta="100%" android:toXDelta="0%" /> <scale android:duration="6000" android:fromXScale="1.0" android:toXScale="0.6" android:pivotX="0" android:fromYScale="1.0" android:toYScale="1.0"

How to get the last “step” in a step function using d3?

由 家住魔仙堡 提交于 2019-12-07 05:55:45
问题 I'm using a simple line graph in d3 with interpolation of "step-after". var line = d3.svg.line() .x(function(d) { return x(d.date); }) .y(function(d) { return y(d.close); }); // Note that with step-after, the last data point shows no "step". Why? line.interpolate('step-after'); (http://jsfiddle.net/hrabinowitz/zCmHw/1/) This shows a horizontal line after each data point except the last one . For small numbers of points, if you think of the x axis as time, this is unintuitive, because the last
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