ranking

Say I have 5 items A, B, C, D, E in a questionnaire and got respondents to rank them. The data looks like this, > df rank1 rank2 rank3 rank4 rank5 1 A B C D E 2 A C B D E 3 C A B E D 4 B A C D E 5 A B D C E How do I count the frequency of each rank by item so the output looks like this, item rank1 rank2 rank3 rank4 rank5 1 A 3 2 0 0 0 2 B 1 2 2 0 0 3 C 1 1 2 1 0 4 D 0 0 1 3 1 5 E 0 0 0 1 4 We can use table after converting to factor using base R lvls <- sort(unique(unlist(df))) sapply(df, function(x) table(factor(x, levels =lvls))) # rank1 rank2 rank3 rank4 rank5 #A 3 2 0 0 0 #B 1 2 2 0 0 #C 1

I want to rank multiple lists according to their elements how often they appear in each list. Example: list1 = 1,2,3,4 list2 = 4,5,6,7 list3 = 4,1,8,9 result = 4,1,2,3,4,5,6,7,8 (4 is counted three times, 1 two times and the rest once) I've tried the following but i need something more intelligent and something i can do with any ammount of lists. l = [] l.append([ 1, 2, 3, 4, 5]) l.append([ 1, 9, 3, 4, 5]) l.append([ 1, 10, 8, 4, 5]) l.append([ 1, 12, 13, 7, 5]) l.append([ 1, 14, 13, 13, 6]) x1 = set(l[0]) & set(l[1]) & set(l[2]) & set(l[3]) x2 = set(l[0]) & set(l[1]) & set(l[2]) & set(l[4])

I have two tables players and scores . I want to generate a report that looks something like this: player first score points foo 2010-05-20 19 bar 2010-04-15 29 baz 2010-02-04 13 Right now, my query looks something like this: select p.name player, min(s.date) first_score, s.points points from players p join scores s on s.player_id = p.id group by p.name, s.points I need the s.points that is associated with the row that min(s.date) returns. Is that happening with this query? That is, how can I be certain I'm getting the correct s.points value for the joined row? Side note: I imagine this is

I want to compute the Median of y in sub groups of this simple xy_table : x | y --groups--> gid | x | y --medians--> gid | x | y ------- ------------- ------------- 0.1 | 4 0.0 | 0.1 | 4 0.0 | 0.1 | 4 0.2 | 3 0.0 | 0.2 | 3 | | 0.7 | 5 1.0 | 0.7 | 5 1.0 | 0.7 | 5 1.5 | 1 2.0 | 1.5 | 1 | | 1.9 | 6 2.0 | 1.9 | 6 | | 2.1 | 5 2.0 | 2.1 | 5 2.0 | 2.1 | 5 2.7 | 1 3.0 | 2.7 | 1 3.0 | 2.7 | 1 In this example every x is unique and the table is already sorted by x . I now want to GROUP BY round(x) and get the tuple that holds the median of y in each group. I can already compute the median for the whole

I want to rank and unrank combinations with an Element distance constraint. Selected elements cannot repeated. For example: n := 10 elements choose from k := 5 elements being choosen d := 3 max distance between 2 choosen elements 1,3,5,8,9 matches the constraint 1,5,6,7,8 dont matches the constraint How can ranking the combination with given distance constraint, where 1,2,3,4,5 is smaller than 1,2,3,4,6 ? Is there a way do the ranking without compute the combinations with smaller Rank? You can do this by first creating and populating a two-dimensional array, which I will call NVT for "number

This question already has answers here : Closed 2 years ago . How to get ranks with no gaps when there are ties among values? (8 answers) I'm trying to order this dataframe by population and date, so I'm using the order() and rank() functions: > df <- data.frame(idgeoville = c(5, 8, 4, 3, 4, 5, 8, 8), date = c(rep(1950, 4), rep(2000, 4)), population = c(500, 450, 350, 350, 650, 500, 500, 450)) > df idgeoville date population 1 5 1950 500 2 8 1950 450 3 4 1950 350 4 3 1950 350 5 4 2000 650 6 5 2000 500 7 8 2000 500 8 8 2000 450 With ties.method = "first" I have no problem, finally I'm producing