lapply

问题 In this case (more details could be found in this question: Count how many observations in the rest of the dat fits multiple conditions? (R)) This is a dataset called event, containing thousands of events (observations) and I selected several rows to show you the data structure. It contains the "STATEid", "date" of occurrence, and geographical coordinates in two variables "LON" "LAT". I am writing to calculate a new variable (column) for each row. This new variable should be: "Given any

问题 My goal is to make another column by summing the observation from the present day and all previous observations from the same ID by using the date (the data set is sorted in date and chr nr(ID). I will need the aggregation to start over when a new "id" is presented. there might be som NA's, they should be considered as null "Doseringer_pr_kg_dyr" is the observation. CHR_NR DATO_AFSLUT Doseringer_pr_kg_dyr brugstid 10358 2018-08-06 29416.67 31 10358 2018-09-06 104682.27 36 10358 2018-10-12

问题 In continuation of this issue comparison Mann-Whitney test between groups, I decided to create a new topic. Solution of Rui Barradas helped me calculate Mann-Whitney for group 1-2 and 1-3. lst <- split(mydat, mydat$group) lapply(lst[-1], function(DF) wilcox.test(DF$var, lst[[1]]$var, exact = FALSE)) So now i want get the descriptive statistics. I use library:psych describeBy(mydat$var,mydat$group) So i get the following output group: 1 vars n mean sd median trimmed mad min max range skew

问题 Let's work with this data sample timeseries<-structure(list(Data = structure(c(10L, 14L, 18L, 22L, 26L, 29L, 32L, 35L, 38L, 1L, 4L, 7L, 11L, 15L, 19L, 23L, 27L, 30L, 33L, 36L, 39L, 2L, 5L, 8L, 12L, 16L, 20L, 24L, 28L, 31L, 34L, 37L, 40L, 3L, 6L, 9L, 13L, 17L, 21L, 25L), .Label = c("01.01.2018", "01.01.2019", "01.01.2020", "01.02.2018", "01.02.2019", "01.02.2020", "01.03.2018", "01.03.2019", "01.03.2020", "01.04.2017", "01.04.2018", "01.04.2019", "01.04.2020", "01.05.2017", "01.05.2018", "01

问题 d1 <- data.frame(col_one = c(1,2,3),col_two = c(4, 5, 6)) d2 <- data.frame(col_one = c(1, 1, 1), col_two = c(6, 5, 4)) d3 <- data.frame(col_one = c(7, 1, 1), col_two = c(8, 5, 4)) my.list <- list(d1, d2,d3) for (i in 1:3) { table<- lapply(my.list, function(data, count) { sql <- #sqldf( paste0( "select *,count(col_one) from data where col_one = ", count," group by col_one" ) #) print(sql) }, count = i) } output: [1] "select *,count(col_one) from data where col_one = 1 group by col_one" [1]

问题 I have a data frame which I then split into three (or any number) of dataframes. What I’m trying to do is to automatically process each column in each dataframe and add lagged versions of existing variables. For example if there were three variables in each data.frame (V1, V2, V3) I would like to automatically (without hardcoding) add V1.lag, V2.lag and V3.lag. Here is what I have so far, but I’m stuck now. Any help would be highly apprecaited. dd<-data.frame(matrix(rnorm(216),72,3),c(rep("A"

问题 I can't combine the use of lapply and paste to combine two columns for multiple dataframes contained within a list. I've looked through multiple sources, but can't find the answer. This answer Apply paste over a list of vectors to get a list of strings is about combining rows within lists, not combining columns to obtain a vector. This answer explains how to select columns but not paste them together using lapply on a list of dataframes This page explains how to access columns in lists, but

问题 Using BASE R, I wonder how to answer the following question: Are there any value on X or Y (i.e., variables of interest names) that occurs only in one element in m (as a cluster) but not others? If yes, produce my desired output below. For example: Here we see X == 3 only occurs in element m[[3]] but not m[[1]] and m[[2]] . Here we also see Y == 99 only occur in m[[1]] but not others. Note: the following is a toy example, a functional answer is appreciated. AND X & Y may or may not be numeric

问题 I'm trying to apply a function to all similarly spelled data frames in my global environment in R. I want to apply this function to all these data frames, but I can't figure out how to do it without me specifying 1 by 1. I want to return the data frame to the global environment with the same spelling as it was before. mtcars_test = mtcars iris_test = iris #....etc......could be 2 of them or 88 of them...but they will all end in "_test" # figure out what data frames I am working with list_of

问题 I want to apply a function with multiple variables with different values to a list. I know how to do this with one changing variable sapply(c(1:10), function(x) x * 2) # [1] 2 4 6 8 10 12 14 16 18 20 but not with two. I show you first manually what I want (actually I use lapply() but sapply() is more synoptic in SO): # manual a <- sapply(c(1:10), function(x, y=2) x * y) b <- sapply(c(1:10), function(x, y=3) x * y) c <- sapply(c(1:10), function(x, y=4) x * y) c(a, b, c) # [1] 2 4 6 8 10 12 14