UPDATE: Wow, my original answer was a total misunderstanding of your question, so I've rewritten from scratch. With the new edits, it's clear your question is about how images are resampled for the screen while rendering. These images are texturemaps, in WebGL, and the process of getting them to the...

python,csv,pandas,resampling,merging-data

You can concat the two DataFrames, interpolate, then reindex on the DataFrame you want. I assume we have a certain number of DataFrames, where the Date is a DateTimeIndex in all of them. I will use two in this example, since you used two in the question, but the code...

python,pandas,dataframes,resampling

I think what you want is BMS (business month start): .resample('BMS', how='first') An alternative would be to groupby month and take the first with a plain ol' groupby (and e.g. use nth to get the first entry in each group): .groupby(pd.TimeGrouper('M')).nth(0) ...

You missed 1: in front of length(x) (i.e. from 1 to length of x by 1). Your code should be like this: x<- 1:1000 samp <- matrix(NA, ncol = 1, nrow = 1000) for(i in 1:length(x)){ samp[i,] <- sum(sample(x,5,replace = TRUE))} Which works great: > str(samp) int [1:1000, 1] 2715...

Ok, just found a solution. I'm new to R and to math ;) v <- seq(1,10) This resizes the vector by interpolating the values: v2 <- approx(v, n=33)$y Next, we normalize to keep the sum: norm <- sum(v)/sum(v2) v3 <- v2 * norm In this example, v3 will be [1]...

r,data.table,reshape2,resampling

Update: Fixed in commit 1253 of v1.9.3. From NEWS: dcast.data.table provides better error message when fun.aggregate is specified but it returns length != 1. Closes git #693. Thanks to Trevor Alexander for reporting here on SO. I agree that the error message should be more helpful in understanding the issue...

Heres a little script to do it without numpy. Maintains shape even if length required is larger than the length of the array. from math import floor def sample(input, count): output = [] sample_size = float(len(input)) / count for i in range(count): output.append(input[int(floor(i * sample_size))]) return output ...

arrays,perl,random,sampling,resampling

This problem can be reframed into pulling 10,000 random numbers between 0 and 1 billion, where no number is within 100 of another. Brute Force - 5 secs Because you're only pulling 10,000 numbers, and probably don't need to do it very often, I suggest approaching this type of problem...

adobe,photoshop,image-resizing,resampling

Save for web didn't resample your image. The quality and the definition are exactly the same than the orginal. BUT, "Save for the web" change a CMYN, 300 dpi image in a RGB,72 dpi image without resample anything.

I think you need to have a DatetimeIndex (rather than a MultiIndex): In [11]: df1 = df.reset_index('status') In [12]: df1 Out[12]: status TUFNWGTP TELFS t070101 t070102 t070103 t070104 TUDIARYDATE 2003-01-03 emp 8155462.672158 2 0 0 0 0 2003-01-04 emp 1735322.527819 1 0 0 0 0 2003-01-04 emp 3830527.482672 2 60...

c#,signal-processing,naudio,resampling

NAudio includes several different resamplers - one that uses ACM (WaveFormatConversionStream), one that uses Media Foundation (MediaFoundationResampler) and one written entirely in managed code (WdlResamplingSampleProvider). I discuss each of these in this post. For your case, you want to do "input driven" resampling, where you know how many input samples...

R doesn't have a =+ increment operator. x =+ 1 is equivalent to x = (+1) x = 1 So in your code, you set my.vect[screen] equal to 1, and you do it 100 times. Each time, you set 50 random indices to 1. You're explanation with replace = F...

It's generally unwise to use subset within functions, but even more unwise to index with a vector that is of a different length than the number of rows of the data-object. The boot function passes a series of index vectors of row-names that have been sampled with replacement from the...

image,matlab,image-processing,interpolation,resampling

I don't know any matching image processing function, so I would do it manually: %defines the image section you want I={[2.5:1:5],[3.5:1:5.5]} %spans a grid [P{1},P{2}]=ndgrid(I{:}) %interpolates using the grid IMG2=interpn(IMG,P{:}) This code is for 2D-images (grayscale), for coloured images: %defines the image section you want I={[2.5:1:5],[3.5:1:5.5]} %We dont want to...

matlab,resampling,downsampling

Well silly of me to forgot to scale properly in interp len=length(y); newlen=length(bands); scale=ceil(len/newlen) x1=x(1:scale:end) y1=y(1:scale:end); h=line(x1,y1,... 'LineWidth',1.2,... 'Color',[0 0 0],... 'Parent',hax); yyy=interp1(x1,y1,bands,'pchip',nan) h=line(1:20,yyy,... 'LineWidth',1.2,... 'Color',[0 1 1],... 'Parent',hax); ...

naudio,ms-media-foundation,resampling

To resample on the fly, just pass the reader directly into MediaFoundationResampler. You will now have an ISampleProvider so you won't be able to use WaveChannel32, but really that is an obsolete class now, and you should be able to do anything you need with other ISampleProvider classes from NAudio.

r,matrix,resampling,statistics-bootstrap

Without using a package, you could do it like this: # your data set.seed(1234) x <- matrix( round(rnorm(200, 5)), ncol=10) # reset seed for this sampling exercise; define sample size and # iterations set.seed(1) samp_size <- 5 iter <- 15 # here are 15 blocks of 5 numbers, which will...