python,algorithm,numpy,data-structures,cumsum

Yes, the formula you give, array.cumsum(0).cumsum(1).cumsum(2), will work. What the formula does is compute a few partial sums so that the sum of these sums is the volume sum. That is, every element needs to be summed exactly once, or, in other words, no element can be skipped and no...

Basically, this solution combines both your solutions, it iterates over segments of size 1000 and uses your cumsum solution to process 1000 elements at once. threshold=100; a=0; MAX=9999999; %segment size S=1000; for segment_start=1:S:MAX %process segment from _start to _end segment_end=min(segment_start+S-1,MAX); %which contains _count elements segment_count=segment_end-segment_start+1; %now use the cumsum solution...

This is the nature of using a fixed-precision representation with values that it cannot represent exactly. Just like 1/3 can't be represented exactly with a fixed number of decimal places, 0.1 can't be represented exactly with a fixed number of binary places. So just like 3 x (1/3) can't possibly...

I will rephrase what the question is asking to benefit those who are reading. The OP wishes to segment a matrix into chunks by splitting up the matrix into a bunch of columns. A cumsum is applied to each row individually for each column and are then concatenated together to...

library(data.table) data <- data.table(group1=c('A','A','A','B','B'),sum=c(1,2,4,3,7)) data[,list(cumsum = cumsum(sum)),by=list(group1)] ...

Without knowing more about the data, I cannot explain why the singular values appear the way they do here. However, generally in mathematics, larger singular values imply more "importance" to that data. I'm not sure why we are looking at the normalized cumulative sum; however, from these results we can...

Adding the extra row is probably the trickiest part. This seems to produce the output you are after do.call("rbind", unname(lapply(split(data, data$group), function(x) { x<-x[order(x$year),] cx <- c(which(x$year==1930),0)[1]+1 cs <- cumsum(c(0, x$val)) cbind(rbind(transform(x[1,], val=NA, year=min(x$year)-1), x), sum_rel=cs-cs[cx]) }))) # group year val sum_rel # 1 a 1927 NA -110 # 2...

Given your desired result (where you don't mind NA becoming 0), I guess the easiest thing is to first remove the NA values using is.na and then carry on as before. data[ is.na(data) ] <- 0 data.frame(t(apply(data,1,cumsum))) ...

If you don't want to use numpy or itertools, here is the code >>> a=[[1,2,4,6], ... [4,7,9,9], ... [1,9,10,20]] >>> a_r = a[::-1] # reverse original list >>> for i in range(len(a_r)-1): ... for j in range(len(a_r[0])): ... a_r[i][j] = a_r[i][j] - a_r[i+1][j] ... >>> a=a_r[::-1] # reverse result >>>...

Just use diff. Assuming your dataset is called "mydf" and you want to do this for all columns, try: mydf[] <- lapply(mydf, function(x) diff(c(0, x))) mydf # X1 X2 # 1 1 5 # 2 2 4 # 3 3 3 # 4 4 2 # 5 5 1 Since...

matlab,math,random,probability,cumsum

Just a side note before we start: Have you looked at the randsample function? This also allows you to draw values from a distribution where each value is encountered with a certain probability. BTW, you don't need the eval statement there. It's superfluous. Just reference the variable instead (signal5). Now...

You can temporarily replace NAs by zeroes: trial.matrix <- matrix(c(seq(3,13,by=2),rep(NA,3),10,12,14),ncol=2) xi <- apply(trial.matrix,2,function(cl) cl[which(!is.na(cl))[1]]) z2 <- diff(trial.matrix) # temporarily replace NAs in the second column by zeroes: nas <- which(is.na(z2[,2])) z2[nas,2] <- 0 new.df <- diffinv(z2,xi = t(xi)) # return NAs new.df[nas,2] <- NA # [,1] [,2] # [1,] 3...

python,pandas,timedelta,cumsum

Here's an example of how you might do this by iterating over each row in the dataframe. I created new data for the example for simplicity: df = pd.DataFrame({'TimeDelta': np.random.normal( 900, 60, size=100)}) print df.head() TimeDelta 0 971.021295 1 734.359861 2 867.000397 3 992.166539 4 853.281131 So let's do an...

One approach with diff and cumsum - %// Setup few arrays: cx = cumsum(x) %// Continuous Cumsumed version reset_mask = reset==1 %// We want to create a logical array version of %// reset for use as logical indexing next up %// Setup ID array of same size as input array...

You can do this by applying a rolling_sum after we groupby the Type. For example: >>> df["sum4"] = df.groupby("Type")["Value"].apply(lambda x: pd.rolling_sum(x,4)) >>> df Type Value sum4 0 A 1 NaN 1 A 2 NaN 2 A 1 NaN 3 A 1 5 4 A 2 6 5 A 2 6...

If you are interested in the total sum for frequency purposes, and not the intermediate running sum, the following may help. Given the following data: A <- '18 17 15 10 5 0 10 10 0 10 15 5 5 5 25 15 13 0 0 0 25 18 15...

You could probably get there using match quite easily. Assuming your data is: set.seed(1) df <- data.frame(variable=rbinom(10,prob=0.5,size=3)) Something like this seems to work out <- data.frame(a,b=b[match(names(a),names(b))],c=c[match(names(a),names(c))]) replace(out,is.na(out),0) # a b c #0 1 0 1 #1 4 2 2 #2 7 4 3 #3 10 5 5 ...