How to remove delay before Lottie animation starts playing - lottie

I have this Lottie animation:
https://lottiefiles.com/share/hmpf0dsc#_=_
But there is a delay before it starts. The delay feels around 800 - 1 second. How can I remove this in the JSON?
Here is the start of the JSON (not complete because too large for SO):
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Stroke","hd":false},{"ty":"tr","p":{"a":0,"k":[0,0],"ix":2},"a":{"a":0,"k":[0,0],"ix":1},"s":{"a":0,"k":[100,100],"ix":3},"r":{"a":0,"k":0,"ix":6},

I tried this following your link above. The animation starts playing straight away for me.
When refreshing the page, also plays straightaway.
When I say straight away, I mean after the resources are downloaded. Even on page refresh, there's 540kb of data that downloaded again, after about 4Mb has been cached.
I have a very fast Internet connection though. If I try this via my phone's connection, it takes 6 or 7 seconds to download and start playing.
Is it possible that the delay you are experiencing is the time it's taking to download that 4Mb on a slower connection?
Of that 540kb, 416Kb is vendor.????.js. Perhaps allowing that to be cached might help?

The start of the animation is a white heart, which doesn't show up on the background. Around frame 25 onwards you see the main part of the animation.
I'm not sure if you can easily remove this from the JSON, though as a work around you're able to play a segment of an animation as follows:
window.onload = function() {
const player = document.querySelector("lottie-player");
player.load("https://assets5.lottiefiles.com/packages/lf20_hmpf0dsc.json");
var lw = player.getLottie();
lw.loop = true;
lw.playSegments([30, 116], false);
}
<script src="https://unpkg.com/#lottiefiles/lottie-player#latest/dist/lottie-player.js" type="module"></script>
<lottie-player loop controls mode="normal" style="width: 400px">
</lottie-player>
Update with a simpler method:
In your lottie JSON, set IP to 30 and this should start the animation at frame 30 where it's visible.

Related

function which has as its input a vector where each element is a positive integer and returns each element of the input vector which is a cube number

I'm trying to create a function which has as its input a vector where each element is a positive integer and returns each element of the input vector which is a cube number.
This is my code so far:
positivecube<- function(x){
b<-round(x^(1/3), digits=6)
b%%1==0 &&
b!=1
return(x[b])
}
I'm struggling to remove the 1s from the final vector and to also convert them back to the original cube number. Any help would be great, thank you
You could return the subset of x where the absolute difference between the cube root and the cube root rounded to the nearest integer is very small:
positivecube <- function(x) x[abs(round(x^(1/3)) - x^(1/3)) < 1e-8]
positivecube(1:100)
#> [1] 1 8 27 64

R: select region on matrix

Assume X is a 5x5 matrix (which represents pixel luminosity from a picture):
I would like to select the element that fit within a given simple geometrical shape (e.g. square, circle, ovale) superposed to this matrix. In this example I would like to select the elements from the matrix that fall (even slightly) within the ovale.
Ultimately those elements position would be return into a vector logical vector elementInOvale on which I could perform a simple operation such as X[elementInOvale] <- 0
I have the feeling that this is a common problem which has already been solved, I just don't know how to formulate or where to find information about it, and to do with R.

Transposing row vector X to get compatible dimensions warning in Scilab

I am learning to use Scilab, I tried plotting a function wich I know have a discontinuity at a certain value but the plot I got had a non expectable behavior so I tried to plot a very well known function "y=1/x".
I created the "x" vector
x=[-10:1:10];
Then created the "y" function
y=1/x;
And then used the plot command
plot(x,y)
I got the following warning
WARNING: Transposing row vector X to get compatible dimensions
And my plot is a straight line, I don't know what I'm doing wrong.
Take care: 1/x compute some thing like 1*pinv(x) not the array [1/x(1),1/x(2),...]. to obtain the previous result use 1 ./x for element wise disvision.
Well, try typing x and y into the consol, to see, how your variables look like:
-->x
x =
column 1 to 12
- 10. - 9. - 8. - 7. - 6. - 5. - 4. - 3. - 2. - 1. 0. 1.
column 13 to 21
2. 3. 4. 5. 6. 7. 8. 9. 10.
-->y
y =
- 0.0129870
- 0.0116883
- 0.0103896
- 0.0090909
- 0.0077922
- 0.0064935
- 0.0051948
- 0.0038961
- 0.0025974
- 0.0012987
0.
0.0012987
0.0025974
0.0038961
0.0051948
0.0064935
0.0077922
0.0090909
0.0103896
0.0116883
0.0129870
So x is a row array, and y is a column array, that's why scilab is complaining about incompatible dimensions. To remove the warning, you must transpose one of your arrays, to make them identical in dimensions. You can do it many places, e.g:
y=1/x';
or
plot(x',y);
or
plot(x,y');
Note: in Scilab 5.4.x and earlier versions there was no such warning, Scilab silently transposed one of the arrays.

Refraction vector not normalized

How can I get the not-normalized output refracted vector, with an also not-normalized incident vector?
I'm following that formulas, work with normalized input, but if I pass not-normalized doesn't. Tried to divide the dot product by input vector length but also nothing.
Wikipedia Snell's Law Vector form
If you divide the dot product by the incident vector length, then your thetas will be correct.
After that, if you multiply n by the incident vector length, then your vreflected and vrefracted vectors will be correct.

How do I normalize an image?

If I have a series of pixels, which range from say -500 to +1000, how would I normalize all the pixels on the same gradient so that they fall between a specific range, say 0 and 255?
Some pseudocode like this would scale values linearly from one range to another
oldmin=-500
oldmax=1000
oldrange=oldmax-oldmin;
newmin=0
newmax=255;
newrange=newmax-newmin;
foreach(oldvalue)
{
//where in the old scale is this value (0...1)
scale=(oldvalue-oldmin)/oldrange;
//place this scale in the new range
newvalue=(newrange*scale)+newmin
}
Your question isn't very clear so I'm going to assume that you're doing some kind of image processing and the results you get are values from -500 to 1000 and now you need to save the color to a file where every value needs to be between 0 and 255.
How you do this is really very dependent in the application, what is really the meaning of the results and what exactly you want to do. The two main options are:
clamp the values - anything under 0 you replace by 0 and anything above 255 you replace by 255. You'll want to do this, for instance, if your image processing is some kind of interpolation which really shouldn't reach these values
Linear normalization - linearly may your minimal value to 0 and your maximal value to 255. Of course you'll first need to find the minimum and maximum. You do:
v = (origv - min)/(max - min) * 255.0
What this does is first map the values to [0,1] and then stretch them back to [0,255].
A third option is to mix and match between these two options. Your application might demand that you treat negative values as unneeded values and clamp them to 0 and positive values to linearly map to [0,255].
First make it all positive. If the minimum is -500 then add 500 to all values. Then the minimum would be 0, and the maximum would be 1500.
Then it is just a rule of three and you have it:
[value in 0,255] = 255*(Pixel/1500)
Some pseudo code may help:
foreach( pixel_value in pixel_values): # between -500 and 1000
position = (pixel_value + 500) / 1500 # gives you a 0 to 1 decimal
new_value = int(postion * 255) # or instead of casting, you could round it off
That's python code by the way.
Create two variables, MinInputValue and MaxInputValue. Initialize MinInputValue to a very large positive number (higher than the largest pixel value you ever expect to see) and MaxInputValue to a very large negative number (lower than the lowest pixel value you ever expect to see).
Loop over every pixel in the image. For each pixel, if the pixel value PixelValue is lower than MinInputValue, set MinInputValue to PixelValue. If the pixel value is higher than MaxInputValue, set MaxInputValue to PixelValue.
Create a new variable, InputValueRange, and set it to MaxInputValue - MinInputValue.
Once this is done, loop over every pixel in the image again. For each pixel PixelValue, calculate the output pixel value as 255.0 * (PixelValue - MinInputValue) / InputValueRange. You can assign this new value back to the original PixelValue, or you can set the corresponding pixel in an output image of the same size.

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