na

Deduplicating/collapsing records in an R dataframe

由 ☆樱花仙子☆ 提交于 2019-12-04 12:58:52
I have a dataset that is comprised of various individuals, where each individual has a unique id. Each individual can appear multiple times in the dataset, but it's my understanding that besides differing in one or two variables (there are about 80 for each individual) the values should be the same for each entry for the same user id in the dataset. I want to try to collapse the data if I can. My main obstacle is certain null values that I need to back populate. I'm looking for a function that can accomplish deduplication looking something like this: # Build sample dataset df1 = data.frame(id

R fill in NA with previous row value with condition

由 自闭症网瘾萝莉.ら 提交于 2019-12-04 12:30:14
问题 I need to fill in NA rows with the previous row value, but only until a criteria is not changed. As a simple example for days of week, meals and prices: Day = c("Mon", "Tues", "Wed", "Thus", "Fri", "Sat","Sun","Mon", "Tues", "Wed", "Thus", "Fri", "Sat","Sun") Meal = c("B","B","B","B","B","D","D","D","D","L","L", "L","L","L") Price = c(NA, 20, NA,NA,NA,NA,NA,15,NA,NA,10,10,NA,10) df = data.frame(Meal,Day ,Price ) df Meal Day Price 1 B Mon NA 2 B Tues 20 3 B Wed NA 4 B Thus NA 5 B Fri NA 6 D

data.table do not compute NA groups in by

由 回眸只為那壹抹淺笑 提交于 2019-12-04 11:54:12
问题 This question has a partial answer here but the question is too specific and I'm not able to apply it to my own problem. I would like to skip a potentially heavy computation of the NA group when using by . library(data.table) DT = data.table(X = sample(10), Y = sample(10), g1 = sample(letters[1:2], 10, TRUE), g2 = sample(letters[1:2], 10, TRUE)) set(DT, 1L, 3L, NA) set(DT, 1L, 4L, NA) set(DT, 6L, 3L, NA) set(DT, 6L, 4L, NA) DT[, mean(X*Y), by = .(g1,g2)] Here we can see there are up to 5

na.locf fill NAs up to maxgap even if gap > maxgap, with groups

由 99封情书 提交于 2019-12-04 11:30:45
问题 I've seen a solution to this, but can't get it to work for groups (Fill NA in a time series only to a limited number), and thought there must be a neater way to do this also? Say I have the following dt: dt <- data.table(ID = c(rep("A", 10), rep("B", 10)), Price = c(seq(1, 10, 1), seq(11, 20, 1))) dt[c(1:2, 5:10), 2] <- NA dt[c(11:13, 15:19) ,2] <- NA dt ID Price 1: A NA 2: A NA 3: A 3 4: A 4 5: A NA 6: A NA 7: A NA 8: A NA 9: A NA 10: A NA 11: B NA 12: B NA 13: B NA 14: B 14 15: B NA 16: B

Interpolate multiple NA values with R

由 断了今生、忘了曾经 提交于 2019-12-04 11:06:58
I want to interpolate multiple NA values in a matrix called, tester. This is a part of tester with only 1 column of NA values, in the whole 744x6 matrix other columns have multiple as well: ZONEID TIMESTAMP U10 V10 U100 V100 1 20121022 12:00 -1.324032e+00 -2.017107e+00 -3.278166e+00 -5.880225574 1 20121022 13:00 -1.295168e+00 NA -3.130429e+00 -6.414975148 1 20121022 14:00 -1.285004e+00 NA -3.068829e+00 -7.101699541 1 20121022 15:00 -9.605904e-01 NA -2.332645e+00 -7.478168285 1 20121022 16:00 -6.268261e-01 -3.057278e+00 -1.440209e+00 -8.026791079 I have installed the zoo package and used the

R data.table NA type consistency

由 痴心易碎 提交于 2019-12-04 05:30:50
问题 dt = data.table(x = c(1,1,2,2,2,2,3,3,3,3)) dt[, y := if(.N > 2) .N else NA, by = x] # fail dt[, y := if(.N > 2) .N else NA_integer_, by = x] # good This first grouping fails because NA has a type and it's not integer. Is there a way to tell data table to ignore that and try to make all NAs to whatever type that keeps consistency? I can manually set NA_integer here, but if I have lots of columns of different types, it's hard to set all NA type correct. BTW, what NA type should I use for Date

The difference of na.rm and na.omit in R

由 心不动则不痛 提交于 2019-12-04 05:27:20
I've just started with R and I've executed these statements: library(datasets) head(airquality) s <- split(airquality,airquality$Month) sapply(s, function(x) {colMeans(x[,c("Ozone", "Solar.R", "Wind")], na.rm = TRUE)}) lapply(s, function(x) {colMeans(na.omit(x[,c("Ozone", "Solar.R", "Wind")])) }) For the sapply , it returns the following: 5 6 7 8 9 Ozone 23.61538 29.44444 59.115385 59.961538 31.44828 Solar.R 181.29630 190.16667 216.483871 171.857143 167.43333 Wind 11.62258 10.26667 8.941935 8.793548 10.18000 And for lapply , it returns the following: $`5` Ozone Solar.R Wind 24.12500 182.04167

How to avoid NA columns in dcast() output?

由 折月煮酒 提交于 2019-12-04 04:27:27
问题 How can I avoid NA columns in dcast() output from the reshape2 package? In this dummy example the dcast() output will include an NA column: require(reshape2) data(iris) iris[ , "Species2"] <- iris[ , "Species"] iris[ 2:7, "Species2"] <- NA (x <- dcast(iris, Species ~ Species2, value.var = "Sepal.Width", fun.aggregate = length)) ## Species setosa versicolor virginica NA ##1 setosa 44 0 0 6 ##2 versicolor 0 50 0 0 ##3 virginica 0 0 50 0 For a somewhat similar usecase, table() does have an

How to select rows by group with the minimum value and containing NAs in R

由 女生的网名这么多〃 提交于 2019-12-04 04:19:45
问题 Here is an example: set.seed(123) data<-data.frame(X=rep(letters[1:3], each=4),Y=sample(1:12,12),Z=sample(1:100, 12)) data[data==3]<-NA What I am to realize is to select the unique row of X with minimum Y by ignoring NA s: a 4 68 b 1 4 c 2 64 What's the best way to do that? 回答1: Using the data.table package, this is trivial: library(data.table) d <- data.table(data) d[, min(Y, na.rm=TRUE), by=X] You can also use plyr and its ddply function: library(plyr) ddply(data, .(X), summarise, min(Y, na

Replacing NA's in each column of matrix with the median of that column

由 只谈情不闲聊 提交于 2019-12-04 03:56:52
问题 I am trying to replace the NA's in each column of a matrix with the median of of that column, however when I try to use lapply or sapply I get an error; the code works when I use a for-loop and when I change one column at a time, what am I doing wrong? Example: set.seed(1928) mat <- matrix(rnorm(100*110), ncol = 110) mat[sample(1:length(mat), 700, replace = FALSE)] <- NA mat1 <- mat2 <- mat mat1 <- lapply(mat1, function(n) { mat1[is.na(mat1[,n]),n] <- median(mat1[,n], na.rm = TRUE) } ) for (n
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