I don’t know if Boost has more specific functions, but you can do it with the standard library. Given std::vector<double> v, this is the naive way: #include <numeric> double sum = std::accumulate(v.begin(), v.end(), 0.0); double mean = sum / v.size(); double sq_sum = std::inner_product(v.begin(), v.end(), v.begin(), 0.0); double stdev = std::sqrt(sq_sum / v.size() – mean … Read more
You can take the mean of Timedelta. So find the minimum value and subtract it from the series to get a series of Timedelta. Then take the mean and add it back to the minimum. dob = df_test.DOB m = dob.min() (m + (dob – m).mean()).to_pydatetime() datetime.datetime(2014, 7, 12, 0, 0) One-line df_test.DOB.pipe(lambda d: (lambda … Read more
A more general approach would employ unique to find the unique index values: [U, ix, iu] = unique(x(:, 1)); and then accumarray: [c, r] = meshgrid(1:size(x, 2), iu); y = accumarray([r(:), c(:)], x(:), [], @mean); Explanation The input values to process are actually the second parameter of accumarray. The first parameter of accumarray is a … Read more
Can you do a df.dtypes ? In the example below type is Int as it works fine. import pandas as pd ##group by 1 columns df = pd.DataFrame({‘ data’: [4610, 4611, 4612, 4613], ‘Category’: [2, 2, 7, 7]}) print df.groupby(‘Category’). mean() ##Mutiple columns to group by df1 = pd.DataFrame({‘ data’: [4610, 4611, 4612, 4613], ‘Category’: … Read more
With base, using aggregate > aggregate(WLN~GROUP+WORD, mean, data=df) GROUP WORD WLN 1 1 1 3.333333 2 1 2 2.333333 3 2 3 1.333333 4 2 4 1.000000 where df is @Metrics’ data. Another alternative is using summaryBy from doBy package > library(doBy) > summaryBy(WLN~GROUP+WORD, FUN=mean, data=df) GROUP WORD WLN.mean 1 1 1 3.333333 2 1 … Read more
I would use colMeans(matrix(a, 60)) .colMeans(a, 60, length(a) / 60) # more efficient (without reshaping to matrix) Enhancement on user adunaic‘s request This only works if there are 60×100 data points. If you have an incomplete 60 at the end then this errors. It would be good to have a general solution for others looking … Read more
I think you’re asking how to compute the mean of a variable in a data frame, given the name of the column. There are two typical approaches to doing this, one indexing with [[ and the other indexing with [: data(iris) mean(iris[[“Petal.Length”]]) # [1] 3.758 mean(iris[,”Petal.Length”]) # [1] 3.758 mean(iris[[“Sepal.Width”]]) # [1] 3.057333 mean(iris[,”Sepal.Width”]) # … Read more
Since you asked for a one-liner: rnorm2 <- function(n,mean,sd) { mean+sd*scale(rnorm(n)) } r <- rnorm2(100,4,1) mean(r) ## 4 sd(r) ## 1
So, you can simulate your situation like this; # Simulate some data: # Create 332 data frames set.seed(1) df.list<-replicate(332,data.frame(sulfate=rnorm(100),nitrate=rnorm(100)),simplify=FALSE) # Generate names like 001.csv and 010.csv file.names<-paste0(‘specdata/’,sprintf(‘%03d’,1:332),’.csv’) # Write them to disk invisible(mapply(write.csv,df.list,file.names)) And here is a function that would read those files: pollutantmean <- function(directory, pollutant, id = 1:332) { file.names <- list.files(directory) file.numbers … Read more
If you want to first take mean on the combination of [‘cluster’, ‘org’] and then take mean on cluster groups, you can use: In [59]: (df.groupby([‘cluster’, ‘org’], as_index=False).mean() .groupby(‘cluster’)[‘time’].mean()) Out[59]: cluster 1 15 2 54 3 6 Name: time, dtype: int64 If you want the mean of cluster groups only, then you can use: In … Read more