subset

I am subsetting a dataframe according to multiple criteria across several columns. I am choosing the rows in the dataframe that contain any one of several values defined in the vector "criteria" in any one of three different columns. I have some code that works, but wonder what other (more elegant?) ways there are to do this. Here is what I've done: criteria <-c(1:10) subset1 <-subset(data, data[, "Col1"] %in% criteria | data[, "Col2"] %in% criteria | data[, "Col3"] %in% criteria) Suggestions warmly welcomed. (I am an R beginner, so very simple explanations about what you are suggesting are

I have a pandas dataset that I want to downsize (remove all values under x). The mask is df[my_column] > 50 I would typically just use df = df[mask] , but want to avoid making a copy every time, particularly because it gets error prone when used in functions (as it only gets altered in the function scope). What is the best way to subset a dataset inplace? I was thinking of something along the lines of df.drop(df.loc[mask].index, inplace = True) Is there a better way to do this, or any situation where this won't work at all? You are missing the inplace parameter : df.drop(df[df.my_column < 50]

I have a vector of dates e.g. dates <- c('2013-01-01', '2013-04-02', '2013-06-10', '2013-09-30') And a dataframe which contains a date column e.g. df <- data.frame( 'date' = c('2013-01-04', '2013-01-22', '2013-10-01', '2013-10-10'), 'a' = c(1,2,3,4), 'b' = c('a', 'b', 'c', 'd') ) And I would would like to subset the dataframe so it only contains rows where the date is less than 5 days after any of the dates in the 'dates' vector. i.e. The initial dataframe looks like this date a b 2013-01-04 1 a 2013-01-22 2 b 2013-10-01 3 c 2013-10-10 4 d After the query I would only be left with the first

I'm sorry for deleting the original question, here it is: We have a bag or an array of n integers, we need to find the product of each of the (n-1) subsets. e.g: S = {1, 0, 3, 6} ps[1] = 0*3*6 = 0; ps[2] = 1*3*6 = 18; etc. After discussions, we need to take care of the three cases and they are illustrated in the following: 1. S is a set (contains one zero element) for i=1 to n if s[i]=0 sp[i] = s[1] * s[2] * ...* s[i-1] * s[i+1] *.....*s[n] else sp[i] = 0; 2. S is a bag (contains more than one zero element) for i=1 to n sp[i] = 0; 3. S is a set (contains no zero elements) product = 1 for i=1