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• Using scipy.interpolate.splrep function

  - by Koustav Ghosal

  I am trying to fit a cubic spline to a given set of points. My points are not ordered. I CANNOT sort or reorder the points, since I need that information. But since the function scipy.interpolate.splrep works only on non-duplicate and monotonically increasing points I have defined a function that maps the x-coordinates to a monotonically increasing space. My old points are: xpoints=[4913.0, 4912.0, 4914.0, 4913.0, 4913.0, 4913.0, 4914.0, 4915.0, 4918.0, 4921.0, 4925.0, 4932.0, 4938.0, 4945.0, 4950.0, 4954.0, 4955.0, 4957.0, 4956.0, 4953.0, 4949.0, 4943.0, 4933.0, 4921.0, 4911.0, 4898.0, 4886.0, 4874.0, 4865.0, 4858.0, 4853.0, 4849.0, 4848.0, 4849.0, 4851.0, 4858.0, 4864.0, 4869.0, 4877.0, 4884.0, 4893.0, 4903.0, 4913.0, 4923.0, 4935.0, 4947.0, 4959.0, 4970.0, 4981.0, 4991.0, 5000.0, 5005.0, 5010.0, 5015.0, 5019.0, 5020.0, 5021.0, 5023.0, 5025.0, 5027.0, 5027.0, 5028.0, 5028.0, 5030.0, 5031.0, 5033.0, 5035.0, 5037.0, 5040.0, 5043.0] ypoints=[10557.0, 10563.0, 10567.0, 10571.0, 10575.0, 10577.0, 10578.0, 10581.0, 10582.0, 10582.0, 10582.0, 10581.0, 10578.0, 10576.0, 10572.0, 10567.0, 10560.0, 10550.0, 10541.0, 10531.0, 10520.0, 10511.0, 10503.0, 10496.0, 10490.0, 10487.0, 10488.0, 10488.0, 10490.0, 10495.0, 10504.0, 10513.0, 10523.0, 10533.0, 10542.0, 10550.0, 10556.0, 10559.0, 10560.0, 10559.0, 10555.0, 10550.0, 10543.0, 10533.0, 10522.0, 10514.0, 10505.0, 10496.0, 10490.0, 10486.0, 10482.0, 10481.0, 10482.0, 10486.0, 10491.0, 10497.0, 10506.0, 10516.0, 10524.0, 10534.0, 10544.0, 10552.0, 10558.0, 10564.0, 10569.0, 10573.0, 10576.0, 10578.0, 10581.0, 10582.0] Plots: The code for the mapping function and interpolation is: xnew=[] ynew=ypoints for c3,i in enumerate(xpoints): if np.isfinite(np.log(i*pow(2,c3))): xnew.append(np.log(i*pow(2,c3))) else: if c==0: xnew.append(np.random.random_sample()) else: xnew.append(xnew[c3-1]+np.random.random_sample()) xnew=np.asarray(xnew) ynew=np.asarray(ynew) constant1=10.0 nknots=len(xnew)/constant1 idx_knots = (np.arange(1,len(xnew)-1,(len(xnew)-2)/np.double(nknots))).astype('int') knots = [xnew[i] for i in idx_knots] knots = np.asarray(knots) int_range=np.linspace(min(xnew),max(xnew),len(xnew)) tck = interpolate.splrep(xnew,ynew,k=3,task=-1,t=knots) y1= interpolate.splev(int_range,tck,der=0) The code is throwing an error at the function interpolate.splrep() for some set of points like the above one. The error is: File "/home/neeraj/Desktop/koustav/res/BOS5/fit_spline3.py", line 58, in save_spline_f tck = interpolate.splrep(xnew,ynew,k=3,task=-1,t=knots) File "/usr/lib/python2.7/dist-packages/scipy/interpolate/fitpack.py", line 465, in splrep raise _iermessier(_iermess[ier][0]) ValueError: Error on input data But for other set of points it works fine. For example for the following set of points. xpoints=[1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1630.0, 1630.0, 1630.0, 1631.0, 1631.0, 1631.0, 1631.0, 1630.0, 1629.0, 1629.0, 1629.0, 1628.0, 1627.0, 1627.0, 1625.0, 1624.0, 1624.0, 1623.0, 1620.0, 1618.0, 1617.0, 1616.0, 1615.0, 1614.0, 1614.0, 1612.0, 1612.0, 1612.0, 1611.0, 1610.0, 1609.0, 1608.0, 1607.0, 1607.0, 1603.0, 1602.0, 1602.0, 1601.0, 1601.0, 1600.0, 1599.0, 1598.0] ypoints=[10570.0, 10572.0, 10572.0, 10573.0, 10572.0, 10572.0, 10571.0, 10570.0, 10569.0, 10565.0, 10564.0, 10563.0, 10562.0, 10560.0, 10558.0, 10556.0, 10554.0, 10551.0, 10548.0, 10547.0, 10544.0, 10542.0, 10541.0, 10538.0, 10534.0, 10532.0, 10531.0, 10528.0, 10525.0, 10522.0, 10519.0, 10517.0, 10516.0, 10512.0, 10509.0, 10509.0, 10507.0, 10504.0, 10502.0, 10500.0, 10501.0, 10499.0, 10498.0, 10496.0, 10491.0, 10492.0, 10488.0, 10488.0, 10488.0, 10486.0, 10486.0, 10485.0, 10485.0, 10486.0, 10483.0, 10483.0, 10482.0, 10480.0] Plots: Can anybody suggest what's happening ?? Thanks in advance......

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• Interpolate air drag for my game?

  - by Valentin Krummenacher

  So I have a little game which works with small steps, however those steps vary in time, so for example I sometimes have 10 Steps/second and then I have 20 Steps/second. This changes automatically depending on how many steps the user's computer can take. To avoid inaccurate positioning of the game's player object I use y=v0*dt+g*dt^2/2 to determine my objects y-position, where dt is the time since the last step, v0 is the velocity of my object in the beginning of my step and g is the gravity. To calculate the velocity in the end of a step I use v=v0+g*dt what also gives me correct results, independent of whether I use 2 steps with a dt of for example 20ms or one step with a dt of 40ms. Now I would like to introduce air drag. For simplicity's sake I use a=k*v^2 where a is the air drag's acceleration (I am aware that it would usually result in a force, but since I assume 1kg for my object's mass the force is the same as the resulting acceleration), k is a constant (in this case I'm using 0.001) and v is the speed. Now in an infinitely small time interval a is k multiplied by the velocity in this small time interval powered by 2. The problem is that v in the next time interval would depend on the drag of the last which again depends on the v of the last interval and so on... In other words: If I use a=k*v^2 I get different results for my position/velocity when I use 2 steps of 20ms than when I use one step of 40ms. I used to have this problem for my position too, but adding +g*dt^2/2 to the formula for my position fixed the problem since it takes into account that the position depends on the velocity which changes slightly in every infinitely small time interval. Does something like that exist for air drag too? And no, I dont mean anything like Adding air drag to a golf ball trajectory equation or similar, for that kind of method only gives correct results when all my steps are the same. (I hope you can understand my intermediate english, it's not my main language so I would like to say sorry for all the silly mistakes I might have made in my question)

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• How do I interpolate between points without going beyond them?

  - by user1774893

  I have data of variable lengths (reaching movements recorded in 2D) and want to create a function that will resample this data to a uniform length (500 samples). However, I want matlab to only resample between the maximum and minimum values given, without adding any additional distance. For instance, if I resample the matrix [1:1:10], the resampled matrix should have a minimum value of 1 and a maximum value of 10. So far I've tried the following: x = [1:1:10]; interp(x, 500 / length(x)); This, however, gives values above my maximum specified value of 10. Is there any way I can get matlab to resample/interpolate solely between two points, without extending beyond them?

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• How to interpolate in MatLab

  - by G Sam

  I have a 1x1 Matrix of points which specifies speed of a drive with respect to time. This speed changes throughout the operation; which means that the difference between two points is changing. To give you an example: M = [1; 2; 3; 5; 7; 9; 11; 15; 19]. (Only that this is a 892x1 matrix) I want to make this matrix twice as long (so changing the relative speed per timestep), while retaining the way the speeds change. Eg: M' = [1; 1.5; 2; 2.5; 3; 4; 5; 6; 7; 8; 9; 10; 11; 13; 15; 17; 19]. Is there an easy way to do this in MatLab? So far I have tried upsampling (which fills the time step with zeros); interp (which fills it with low-pass interpolation. Thanks!

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• How to interpolate rotations?

  - by uhuu

  I have two vectors describing rotations; a start rotation A and a target rotation B. How would I best go about interpolating A by a factor F to approach B? Using a simple lerp on the vectors fails to work when more than one dimension needs to be interpolated (i.e. produces undesirable rotations). Maybe building quaternions from the rotation vectors and using slerp is the way to go. But how, then, could I extract a vector describing the new rotation from the resulting quaternion? Thanks in advance.

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• numpy arange with multiple intervals

  - by Heiko Westermann

  Hi, i have an numpy array which represents multiple x-intervals of a function: In [137]: x_foo Out[137]: array([211, 212, 213, 214, 215, 216, 217, 218, 940, 941, 942, 943, 944, 945, 946, 947, 948, 949, 950]) as you can see, in x_foo are two intervals: one from 211 to 218, and one from 940 to 950. these are intervals, which i want to interpolate with scipy. for this, i need to adjust the spacing, e.g "211.0 211.1 211.2 ..." which you would normaly do with: arange( x_foo[0], x_foo[-1], 0.1 ) in the case of multiple intervals, this is not possible. so heres my question: is there a numpy-thonic way to do this in array-style? or do i need to write a function which loops over the whole array and split if the difference is 1? thanks!

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• How to interpolate hue values in HSV colour space?

  - by nick

  Hi, I'm trying to interpolate between two colours in HSV colour space to produce a smooth colour gradient. I'm using a linear interpolation, eg: h = (1 - p) * h1 + p * h2 s = (1 - p) * s1 + p * s2 v = (1 - p) * v1 + p * v2 (where p is the percentage, and h1, h2, s1, s2, v1, v2 are the hue, saturation and value components of the two colours) This produces a good result for s and v but not for h. As the hue component is an angle, the calculation needs to work out the shortest distance between h1 and h2 and then do the interpolation in the right direction (either clockwise or anti-clockwise). What formula or algorithm should I use? EDIT: By following Jack's suggestions I modified my JavaScript gradient function and it works well. For anyone interested, here's what I ended up with: // create gradient from yellow to red to black with 100 steps var gradient = hsbGradient(100, [{h:0.14, s:0.5, b:1}, {h:0, s:1, b:1}, {h:0, s:1, b:0}]); function hsbGradient(steps, colours) { var parts = colours.length - 1; var gradient = new Array(steps); var gradientIndex = 0; var partSteps = Math.floor(steps / parts); var remainder = steps - (partSteps * parts); for (var col = 0; col < parts; col++) { // get colours var c1 = colours[col], c2 = colours[col + 1]; // determine clockwise and counter-clockwise distance between hues var distCCW = (c1.h >= c2.h) ? c1.h - c2.h : 1 + c1.h - c2.h; distCW = (c1.h >= c2.h) ? 1 + c2.h - c1.h : c2.h - c1.h; // ensure we get the right number of steps by adding remainder to final part if (col == parts - 1) partSteps += remainder; // make gradient for this part for (var step = 0; step < partSteps; step ++) { var p = step / partSteps; // interpolate h var h = (distCW <= distCCW) ? c1.h + (distCW * p) : c1.h - (distCCW * p); if (h < 0) h = 1 + h; if (h > 1) h = h - 1; // interpolate s, b var s = (1 - p) * c1.s + p * c2.s; var b = (1 - p) * c1.b + p * c2.b; // add to gradient array gradient[gradientIndex] = {h:h, s:s, b:b}; gradientIndex ++; } } return gradient; }

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• How to make scipy.interpolate give a an extrapolated result beyond the input range?

  - by Salim Fadhley

  I'm trying to port a program which uses a hand-rolled interpolator (developed by a mathematitian colleage) over to use the interpolators provided by scipy. I'd like to use or wrap the scipy interpolator so that it has as close as possible behavior to the old interpolator. A key difference between the two functions is that in our original interpolator - if the input value is above or below the input range, our original interpolator will extrapolate the result. If you try this with the scipy interpolator it raises a ValueError. Consider this program as an example: import numpy as np from scipy import interpolate x = np.arange(0,10) y = np.exp(-x/3.0) f = interpolate.interp1d(x, y) print f(9) print f(11) # Causes ValueError, because it's greater than max(x) Is there a sensible way to make it so that instead of crashing, the final line will simply do a linear extrapolate, continuing the gradients defined by the first and last two pouints to infinity. Note, that in the real software I'm not actually using the exp function - that's here for illustration only!

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• What's the most effective way to interpolate between two colors? (pseudocode and bitwise ops expecte

  - by navand

  Making a Blackberry app, want a Gradient class. What's the most effective way (as in, speed and battery life) to interpolate two colors? Please be specific. // Java, of course int c1 = 0xFFAA0055 // color 1, ARGB int c2 = 0xFF00CCFF // color 2, ARGB float st = 0 // the current step in the interpolation, between 0 and 1 /* Help from here on. Should I separate each channel of each color, convert them to decimal and interpolate? Is there a simpler way? interpolatedChannel = red1+((red2-red1)*st) interpolatedChannel = interpolatedChannel.toString(16) ^ Is this the right thing to do? If speed and effectiveness is important in a mobile app, should I use bitwise operations? Help me! */

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• How do I interpolate air drag with a variable time step?

  - by Valentin Krummenacher

  So I have a little game which works with small steps, however those steps vary in time, so for example I sometimes have 10 Steps/second and then I have 20 Steps/second. This changes automatically depending on how many steps the user's computer can take. To avoid inaccurate positioning of the game's player object I use y=v0*dt+g*dt^2/2 to determine my objects y-position, where dt is the time since the last step, v0 is the velocity of my object in the beginning of my step and g is the gravity. To calculate the velocity in the end of a step I use v=v0+g*dt what also gives me correct results, independent of whether I use 2 steps with a dt of for example 20ms or one step with a dt of 40ms. Now I would like to introduce air drag. For simplicity's sake I use a=k*v^2 where a is the air drag's acceleration (I am aware that it would usually result in a force, but since I assume 1kg for my object's mass the force is the same as the resulting acceleration), k is a constant (in this case I'm using 0.001) and v is the speed. Now in an infinitely small time interval a is k multiplied by the velocity in this small time interval powered by 2. The problem is that v in the next time interval would depend on the drag of the last which again depends on the v of the last interval and so on... In other words: If I use a=k*v^2 I get different results for my position/velocity when I use 2 steps of 20ms than when I use one step of 40ms. I used to have this problem for my position too, but adding +g*dt^2/2 to the formula for my position fixed the problem since it takes into account that the position depends on the velocity which changes slightly in every infinitely small time interval. Does something like that exist for air drag too? And no, I dont mean anything like Adding air drag to a golf ball trajectory equation or similar, for that kind of method only gives correct results when all my steps are the same. (I hope you can understand my intermediate english, it's not my main language so I would like to say sorry for all the silly mistakes I might have made in my question)

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• Fixed timestep and interpolation question

  - by Eric

  I'm following Glenn Fiedlers excellent Fix Your Timestep! tutorial to step my 2D game. The problem I'm facing is in the interpolation phase in the end. My game has a Tween-function which lets me tween properties of my game entites. Properties such as scale, shear, position, color, rotation etc. Im curious of how I'd interpolate these values, since there's a lot of them. My first thought is to keep a previous value of every property (colorPrev, scalePrev etc.), and interpolate between those. Is this the correct method? To interpolate my characters I use their velocity; renderPostion = position + (velocity * interpolation), but I cannot apply that to color for example. So what is the desired method to interpolate various properties or a entity? Is there any rule of thumb to use?

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• Is a string formatter that pulls variables from its calling scope bad practice?

  - by Eric

  I have some code that does an awful lot of string formatting, Often, I end up with code along the lines of: "...".format(x=x, y=y, z=z, foo=foo, ...) Where I'm trying to interpolate a large number of variables into a large string. Is there a good reason not to write a function like this that uses the inspect module to find variables to interpolate? import inspect def interpolate(s): return s.format(**inspect.currentframe().f_back.f_locals) def generateTheString(x): y = foo(x) z = x + y # more calculations go here return interpolate("{x}, {y}, {z}")

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• What is the simplest way to interpolate and lookup in an x,y table in excel?

  - by dassouki

  I would like to do a lookup and interpolation based on x, y data for the following table. I'd like the equation to be as simple as possible to reduce the amount of possible errors. The full table is about 50 rows x 30 columns. I have about 20 of those tables. Here is an extract from one: A B C D 1 0.1 0.2 0.3 2 2.4 450 300 50 3 2.3 500 375 52 4 2.1 550 475 55 5 1.8 600 600 60 For example, the equation should find the value for x = 2.27 and y = 0.15

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• Can I interpolate two HEX color values without converting them to RGB?

  - by navand

  I'm trying to make a Gradient Class for a Blackberry app. At first I thought about converting the HEX values to RGB and then interpolating them before converting the result back into HEX, but since I will be doing this for every pixel line of an area, and the calculations will be made by a mobile, I thought that maybe there's a more efficient way of doing it. Maybe involving those pesky bitwise operators which I know nothing of... or something. So, is there a way of interpolating without converting to RGB and back? If so, is it faster than the original way? In any case, can you help me make the most efficient color interpolation? Thank you in advance!

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• How can I manually interpolate string escapes in a Perl string?

  - by Ryan Thompson

  In perl suppose I have a string like 'hello\tworld\n', and what I want is: 'hello world ' That is, "hello", then a literal tab character, then "world", then a literal newline. Or equivalently, "hello\tworld\n" (note the double quotes). In other words, is there a function for taking a string with escape sequences and returning an equivalent string with all the escape sequences interpolated?

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• How can I interpolate the result of a function in a Perl here doc?

  - by Zerobu

  I was wondering if it was possible to call a function inside a perl here doc sub Function { } print<<HERE; Function() HERE

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• How can I interpolate literal \t and \n in Perl strings?

  - by Michael

  Say I have an environment variable myvar: myvar=\tapple\n When the following command will print out this variable perl -e 'print "$ENV{myvar}"' I will literally have \tapple\n, however, I want those control chars to be evaluated and not escaped. How would I achieve it? In the real world $ENV residing in substitution, but I hope the answer will cover that.

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• Mac 10.6 Universal Binary scipy: cephes/specfun "_aswfa_" symbol not found

  - by Markus

  Hi folks, I can't get scipy to function in 32 bit mode when compiled as a i386/x86_64 universal binary, and executed on my 64 bit 10.6.2 MacPro1,1. My python setup With the help of this answer, I built a 32/64 bit intel universal binary of python 2.6.4 with the intention of using the arch command to select between the architectures. (I managed to make some universal binaries of a few libraries I wanted using lipo.) That all works. I then installed scipy according to the instructions on hyperjeff's article, only with more up-to-date numpy (1.4.0) and skipping the bit about moving numpy aside briefly during the installation of scipy. Now, everything except scipy seems to be working as far as I can tell, and I can indeed select between 32 and 64 bit mode using arch -i386 python and arch -x86_64 python. The error Scipy complains in 32 bit mode: $ arch -x86_64 python -c "import scipy.interpolate; print 'success'" success $ arch -i386 python -c "import scipy.interpolate; print 'success'" Traceback (most recent call last): File "<string>", line 1, in <module> File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scipy/interpolate/__init__.py", line 7, in <module> from interpolate import * File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scipy/interpolate/interpolate.py", line 13, in <module> import scipy.special as spec File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scipy/special/__init__.py", line 8, in <module> from basic import * File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scipy/special/basic.py", line 8, in <module> from _cephes import * ImportError: dlopen(/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scipy/special/_cephes.so, 2): Symbol not found: _aswfa_ Referenced from: /Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scipy/special/_cephes.so Expected in: flat namespace in /Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scipy/special/_cephes.so Attempt at tracking down the problem It looks like scipy.interpolate imports something called _cephes, which looks for a symbol called _aswfa_ but can't find it in 32 bit mode. Browsing through scipy's source, I find an ASWFA subroutine in specfun.f. The only scipy product file with a similar name is specfun.so, but both that and _cephes.so appear to be universal binaries: $ cd /Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scipy/special/ $ file _cephes.so specfun.so _cephes.so: Mach-O universal binary with 2 architectures _cephes.so (for architecture i386): Mach-O bundle i386 _cephes.so (for architecture x86_64): Mach-O 64-bit bundle x86_64 specfun.so: Mach-O universal binary with 2 architectures specfun.so (for architecture i386): Mach-O bundle i386 specfun.so (for architecture x86_64): Mach-O 64-bit bundle x86_64 Ho hum. I'm stuck. Things I may try but haven't figured out how yet include compiling specfun.so myself manually, somehow. I would imagine that scipy isn't broken for all 32 bit machines, so I guess something is wrong with the way I've installed it, but I can't figure out what. I don't really expect a full answer given my fairly unique (?) setup, but if anyone has any clues that might point me in the right direction, they'd be greatly appreciated. (edit) More details to address questions: I'm using gfortran (GNU Fortran from GCC 4.2.1 Apple Inc. build 5646). Python 2.6.4 was installed more-or-less like so: cd /tmp curl -O http://www.python.org/ftp/python/2.6.4/Python-2.6.4.tar.bz2 tar xf Python-2.6.4.tar.bz2 cd Python-2.6.4 # Now replace buggy pythonw.c file with one that supports the "arch" command: curl http://bugs.python.org/file14949/pythonw.c | sed s/2.7/2.6/ > Mac/Tools/pythonw.c ./configure --enable-framework=/Library/Frameworks --enable-universalsdk=/ --with-universal-archs=intel make -j4 sudo make frameworkinstall Scipy 0.7.1 was installed pretty much as described as here, but it boils down to a simple sudo python setup.py install. It would indeed appear that the symbol is undefined in the i386 architecture if you look at the _cephes library with nm, as suggested by David Cournapeau: $ nm -arch x86_64 /Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scipy/special/_cephes.so | grep _aswfa_ 00000000000d4950 T _aswfa_ 000000000011e4b0 d _oblate_aswfa_data 000000000011e510 d _oblate_aswfa_nocv_data (snip) $ nm -arch i386 /Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/scipy/special/_cephes.so | grep _aswfa_ U _aswfa_ 0002e96c d _oblate_aswfa_data 0002e99c d _oblate_aswfa_nocv_data (snip) however, I can't yet explain its absence.

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• Convert methods from Java-actionscript to ObjectiveC

  - by eco_bach

  Hi I'm tring to convert the following 3 methods from java-actionscript to Objective C. Part of my confusion I think is not knowing what Number types, primitives I should be using. ie in actionscript you have only Number, int, and uint. These are the 3 functions I am trying to convert public function normalize(value:Number, minimum:Number, maximum:Number):Number { return (value - minimum) / (maximum - minimum); } public function interpolate(normValue:Number, minimum:Number, maximum:Number):Number { return minimum + (maximum - minimum) * normValue; } public function map(value:Number, min1:Number, max1:Number, min2:Number, max2:Number):Number { return interpolate( normalize(value, min1, max1), min2, max2); } This is what I have so far -(float) normalize:(float*)value withMinimumValue:(float*)minimum withMaximumValue:(float*)maximum { return (value - minimum) / (maximum - minimum); } -(float) interpolate:(float*)normValue withMinimumValue:(float*)minimum withMaximumValue:(float*)maximum { return minimum + (maximum - minimum) * normValue; } -(float) map:(float*)value withMinimumValue1:(float*)min1 withMaximumValue1:(float*)max1 withMinimumValue2:(float*)min2 withMaximumValue2:(float*)max2 { return interpolate( normalize(value, min1, max1), min2, max2); }

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• Glitch-free cross-fades in HTML5

  - by Alexander Gladysh

  In my HTML5 canvas game, I need to cross-fade two sprites which have some glow around them. (Glow is backed into sprites.) Initially, the first sprite is visible. During the cross-fade the first sprite should vanish, and be replaced with the second one. How exactly the cross-fade is done — does not matter, as long as it is smooth and there are no visual glitches. I've tried two techniques: During the cross-fade I simultaneously interpolate alpha of the first sprite from 1.0 to 0.0, and alpha of the second sprite — from 0.0 to 1.0. With this technique I can see background in the middle of the cross-fade. That's because both sprites are semi-transparent most of the time. During the cross-fade I first interpolate alpha of the second sprite from 0.0 to 1.0 (first sprite alpha is at 1.0), and then interpolate alpha of the first sprite from 1.0 to 0.0. With this technique background is not seen, but the glow around sprites flashes during the cross-fide — when both sprites are near the full visibility. In non-HTML5 game I'd use shaders to do cross-fade separately in RGB and alpha channels. Is there a trick to do the cross-fade I need in HTML5 without visual glitches?

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• Client Side Prediction for a Look Vector

  - by Mike Sawayda

  So I am making a first person networked shooter. I am working on client-side prediction where I am predicting player position and look vectors client-side based on input messages received from the server. Right now I am only worried about the look vectors though. I am receiving the correct look vector from the server about 20 times per second and I am checking that against the look vector that I have client side. I want to interpolate the clients look vector towards the correct one that is server side over a period of time. Therefore no matter how far you are away from the servers look vector you will interpolate to it over the same amount of time. Ex. if you were 10 degrees off it would take the same amount of time as if you were 2 degrees off to be correctly lined up with the server copy. My code looks something like this but the problem is that the amount that you are changing the clients copy gets infinitesimally small so you will actually never reach the servers copy. This is because I am always calculating the difference and only moving by a percentage of that every frame. Does anyone have any suggestions on how to interpolate towards the servers copy correctly? if(rotationDiffY > ClientSideAttributes::minRotation) { if(serverRotY > clientRotY) { playerObjects[i]->collisionObject->rotation.y += (rotationDiffY * deltaTime); } else { playerObjects[i]->collisionObject->rotation.y -= (rotationDiffY deltaTime); } }

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• Color picker does not give gradient appearance

  - by ykaratoprak

  i added below codes. But it generates to me 16 color. but i need 16 color between "red" and "khaki". i don't need gradient flow. My colors look like gradient flow. My colors must not closer to each other. Because i will use this codes return values in chart columns. they are too near each other. static class Program { [STAThread] static void Main() { Form form = new Form(); Color start = Color.Red, end = Color.Khaki; for (int i = 0; i < 16; i++) { int r = Interpolate(start.R, end.R, 15, i), g = Interpolate(start.G, end.G, 15, i), b = Interpolate(start.B, end.B, 15, i); Button button = new Button(); button.Dock = DockStyle.Top; button.BackColor = Color.FromArgb(r, g, b); form.Controls.Add(button); button.BringToFront(); } Application.Run(form); } static int Interpolate(int start, int end, int steps, int count) { float s = start, e = end, final = s + (((e - s) / steps) * count); return (int)final; } }

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• Scipy interpolation on a numpy array

  - by dassouki

  I have a lookup table that is defined the following way: TR_ua1 = np.array([ [3.6, 6.5, 9.1, 11.5, 13.8], [3.9, 7.3, 10.0, 13.1, 15.9], [4.5, 9.2, 12.2, 14.8, 18.2] ]) The header row elements are (hh) <1,2,3,4,5+ The header column (inc) elements are <10000, 20000, 20001+ The user will input a value ex (1.3, 25,000) or (0.2, 50,000). Scipy.interpolate() should interpolate to determine the correct value. Currently, the only way i can do this is with a bunch of if/elifs as exemplified below. I'm pretty sure there is a better, more efficient way of doing this Here's what i've got so far import numpy as np from scipy import interplate if (ua == 1): if (inc <= low_inc): #low_inc = 10,000 if (hh <= 1): return TR_ua1[0][0] elif (hh >= 1 & hh < 2): return interpolate( (1,2), (TR_ua1[0][1], TR_ua1[0][2]) )

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• Python Least-Squares Natural Splines

  - by Eldila

  I am trying to find a numerical package which will fit a natural which minimizes weighted least squares. There is a package in scipy which does what I want for unnatural splines. import numpy as np import matplotlib.pyplot as plt from scipy import interpolate import random x = np.arange(0,5,1.0/2) xs = np.arange(0,5,1.0/500) y = np.sin(x+1) for i in range(len(y)): y[i] += .2*random.random() - .1 knots = np.array([1,2,3,4]) tck = interpolate.splrep(x,y,s=1,k=3,t=knots,task=-1) ynew = interpolate.splev(xs,tck,der=0) plt.figure() plt.plot(xs,ynew,x,y,'x')

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• OpenGL Color Interpolation across vertices

  - by gutsblow

  Right now, I have more than 25 vertices that form a model. I want to interpolate color linearly between the first and last vertex. The Problem is when I write the following code glColor3f(1.0,0.0,0.0); vertex3f(1.0,1.0,1.0); vertex3f(0.9,1.0,1.0); . .`<more vertices>; glColor3f(0.0,0.0,1.0); vertex3f(0.0,0.0,0.0); All the vertices except that last one are red. Now I am wondering if there is a way to interpolate color across these vertices without me having to manually interpolate color natively (like how opengl does it automatically) at each vertex since, I will be having a lot more number of colors at various vertices. Any help would be extremely appreciated. Thank you!

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