I have repeated-measures data.
I need to create a loop that will incrementally count each observation, within a participant, and label it.
I am new to writing loops. My logic was to say, for each item in the list of unique ids, count each row in that, and apply some function to that row.
Could someone point our what I am doing wrong?
data$Ob <- 0
for (i in unique(data$id)) {
count <- 1
for (u in data[data$id == i,]) {
data[data$id ==u,]$Ob <- count
count <- count + 1
print(count)
}
}
Thanks!
Justin
You can also use ave:
set.seed(1)
data <- data.frame(id = sample(4, 10, TRUE))
data$Ob = ave(data$id, data$id, FUN=seq_along)
data
id Ob
1 2 1
2 2 2
3 3 1
4 4 1
5 1 1
6 4 2
7 4 3
8 3 2
9 3 3
10 1 2
# Generate some dummy data
data <- data.frame(Ob=0, id=sample(4,20,TRUE))
# Go through every id value
for(i in unique(data$id)){
# Label observations
data$Ob[data$id == i] = 1:sum(data$id == i)
}
Be aware though that for loops are notoriously slow in R. In this simple case they work fine, but should you have millions and millions of rows in your data frame you'd better do something purely vectorized.
But you don't need a loop...
data <- data.frame (id = sample (4, 10, TRUE))
## id
## 1 3
## 2 4
## 3 1
## 4 3
## 5 3
## 6 4
## 7 2
## 8 1
## 9 1
## 10 4
data$Ob [order (data$id)] <- sequence (table (data$id))
## id Ob
## 1 3 1
## 2 4 1
## 3 1 1
## 4 3 2
## 5 3 3
## 6 4 2
## 7 2 1
## 8 1 2
## 9 1 3
## 10 4 3
(works also with character or factor IDs)
(isn't R just cool!?)
Related
This is my data frame:
time<-rep(c(1:5),4)
sim1<-rep(c(paste("sim",1)),5)
sim2<-rep(c(paste("sim",2)),5)
sim3<-rep(c(paste("sim",3)),5)
sim4<-rep(c(paste("sim",4)),5)
sim<-c(sim1,sim2,sim3,sim4)
id<-as.vector(replicate(4,sample(1:5)))
df<-data.frame(time,sim,id)
df$simnu<-as.numeric(df$sim)
Which should look something like this:
time sim id simnu
1 1 sim 1 1 1
2 2 sim 1 3 1
3 3 sim 1 2 1
4 4 sim 1 4 1
5 5 sim 1 5 1
6 1 sim 2 1 2
7 2 sim 2 5 2
8 3 sim 2 4 2
9 4 sim 2 2 2
10 5 sim 2 3 2
11 1 sim 3 2 3
12 2 sim 3 3 3
13 3 sim 3 4 3
14 4 sim 3 1 3
15 5 sim 3 5 3
16 1 sim 4 3 4
17 2 sim 4 5 4
18 3 sim 4 2 4
19 4 sim 4 1 4
20 5 sim 4 4 4
I have created this loop that subsets the data by simulation and then calculates the output I want:
surveillance<-5
n<-1
simsub<-df[which(df$simnu==1),names(df)%in%c("time","sim","id")]
while (n<=surveillance){
print (n)
rndid<-df[sample(nrow(simsub),1),]
print(rndid)
if(n<rndid$time){
n<-n+1
} else {
tinf<-sum(length(df[which(simsub$time<=n),1]))
prev<-tinf/length(simsub[,1])
print(paste(prev,"prevalence"))
break
}
}
My question is how do I run this loop for each simulation and return the values of this as a vector?
My suggestion for you is to take a look at the lapply function (resp. sapply and vapply), and avoid using while, to be honest it's a bit tricky to help without really knowing what is happening in your code, but in any case here's an example how you can use lapply, however since I don't know what your code should return I can't be sure that the output is correct
I added comments and questions with your original lines, hope this helps
# first define a function that takes one simnu and returns whatever you want it to return
my_calc_fun <- function(sim_nr){
## you can subset the DF without which, names, or %in%
# simres[[i]] <- my_df[which(my_df$simnu==i),names(my_df)%in%c("time","sim","id")]
sim_df <- my_df[my_df$simnu == sim_nr, c("time","sim","id")]
for(n in 1:surveillance){
## I'm not sure that is what you meant to do,
## you are sampling the full DF, but you want a sample
## from the subset i.e., simres[[i]]
# rndid<-my_df[sample(nrow(simres[[i]]),1),]
row_id <- sample(nrow(sim_df), 1)
rndid <- sim_df[row_id, ]
if(n >= rndid$time){
## what are you trying to sum here?
## because you are giving the function one number length(....)
## and just like above you are subsetting the full DF here
# tinf<-sum(length(my_df[which(simres[[i]]$time<=n),1]))
tinf <- length(sim_df[sim_df$time<=n, 1])
# is this the value you want to return for each simnu?
prev <- tinf/length(sim_df["time"])
break
}
}
return(c('simnu'=sim_nr, 'prev' = prev))
}
# apply this function on all values of simnu and save to list
result_all <- lapply(unique(my_df$simnu), my_calc_fun)
result_all
I have a simple dataframe with group IDs and elements of each group, like this:
x <- data.frame("ID" = c(1,1,1,2,2,2,3,3,3), "Values" = c(3,5,7,2,4,5,2,4,6))
Each ID may have a different number of elements. Now I want to find all IDs that have distinct elements with other IDs. In this example, ID1 and ID3 will be selected because they have distinct elements (3,5,7 vs 2,4,6). I also want to copy these unique IDs and their elements into a new dataframe, similar to the original.
How would I do that in R? My skills with R is quite limited.
Thank you very much!
Bests,
Seems like a good question for igraph cliques with one edge to another clique but I cant seem to wrap my head on how to use it.
Anyway, here is an option applying join to identify IDs with same Values and then anti-join to remove those IDs using data.table:
library(data.table)
DT <- as.data.table(x)
for (i in DT[, unique(ID)]) {
dupeID <- DT[DT[ID==i], on=.(Values), .(ID=unique(x.ID[x.ID!=i.ID]))]
DT <- DT[!dupeID , on=.(ID)]
}
output:
ID Values
1: 1 3
2: 1 5
3: 1 7
4: 3 2
5: 3 4
6: 3 6
x <- data.frame("ID" = c(1,1,1,2,2,2,3,3,3), "Values" = c(3,5,7,2,4,5,2,4,6))
gps = split(x, x$ID)
nGroups = length(gps)
k = 1
results = data.frame(ID = NULL, Values = NULL)
for(i in 1:(nGroups - 1)){
j = i + 1
while(j <= nGroups){
if(length(intersect(gps[[i]]$Values, gps[[j]]$Values)) == 0){
print(c(i,j))
results = rbind(results, gps[[i]], gps[[j]])
}
j = j + 1
}
}
results
> results
ID Values
1 1 3
2 1 5
3 1 7
7 3 2
8 3 4
9 3 6
You can try the following code, where the y is the list of data frames (including all data frames that have exclusive Value)
xs <- split(x,x$ID)
id <- names(xs)
y <- list()
ids <- seq_along(xs)
repeat {
if (length(ids)==0) break;
y[[length(y)+1]] <- xs[[ids[1]]]
p <- ids[[1]]
qs <- p
for (q in ids[-1]) {
if (length(intersect(xs[[p]]$Value,xs[[q]]$Value))==0) {
y[[length(y)]] <- rbind(y[[length(y)]],xs[[q]])
qs <- c(qs,q)
}
}
ids <- setdiff(ids,qs)
}
Example
x <- data.frame("ID" = c(1,1,1,2,2,2,3,3,3,4,4),
"Values" = c(3,5,7,2,4,5,2,4,6,1,3))
> x
ID Values
1 1 3
2 1 5
3 1 7
4 2 2
5 2 4
6 2 5
7 3 2
8 3 4
9 3 6
10 4 1
11 4 3
then you will get
> y
[[1]]
ID Values
1 1 3
2 1 5
3 1 7
7 3 2
8 3 4
9 3 6
[[2]]
ID Values
4 2 2
5 2 4
6 2 5
10 4 1
11 4 3
Given a data.frame:
df <- data.frame(grp1 = c(1,1,1,2,2,2,3,3,3,4,4,4),
grp2 = c(1,2,3,3,4,5,6,7,8,6,9,10))
#> df
# grp1 grp2
#1 1 1
#2 1 2
#3 1 3
#4 2 3
#5 2 4
#6 2 5
#7 3 6
#8 3 7
#9 3 8
#10 4 6
#11 4 9
#12 4 10
Both coluns are grouping variables, such that all 1's in column grp1 are known to be grouped together, and so on with all 2's, etc. Then the same goes for grp2. All 1's are known to be the same, all 2's the same.
Thus, if we look at the 3rd and 4th row, based on column 1 we know that the first 3 rows can be grouped together and the second 3 rows can be grouped together. Then since rows 3 and 4 share the same grp2 value, we know that all 6 rows, in fact, can be grouped together.
Based off the same logic we can see that the last six rows can also be grouped together (since rows 7 and 10 share the same grp2).
Aside from writing a fairly involved set of for() loops, is there a more straight forward approach to this? I haven't been able to think one one yet.
The final output that I'm hoping to obtain would look something like:
# > df
# grp1 grp2 combinedGrp
# 1 1 1 1
# 2 1 2 1
# 3 1 3 1
# 4 2 3 1
# 5 2 4 1
# 6 2 5 1
# 7 3 6 2
# 8 3 7 2
# 9 3 8 2
# 10 4 6 2
# 11 4 9 2
# 12 4 10 2
Thank you for any direction on this topic!
I would define a graph and label nodes according to connected components:
gmap = unique(stack(df))
gmap$node = seq_len(nrow(gmap))
oldcols = unique(gmap$ind)
newcols = paste0("node_", oldcols)
df[ newcols ] = lapply(oldcols, function(i) with(gmap[gmap$ind == i, ],
node[ match(df[[i]], values) ]
))
library(igraph)
g = graph_from_edgelist(cbind(df$node_grp1, df$node_grp2), directed = FALSE)
gmap$group = components(g)$membership
df$group = gmap$group[ match(df$node_grp1, gmap$node) ]
grp1 grp2 node_grp1 node_grp2 group
1 1 1 1 5 1
2 1 2 1 6 1
3 1 3 1 7 1
4 2 3 2 7 1
5 2 4 2 8 1
6 2 5 2 9 1
7 3 6 3 10 2
8 3 7 3 11 2
9 3 8 3 12 2
10 4 6 4 10 2
11 4 9 4 13 2
12 4 10 4 14 2
Each unique element of grp1 or grp2 is a node and each row of df is an edge.
One way to do this is via a matrix that defines links between rows based on group membership.
This approach is related to #Frank's graph answer but uses an adjacency matrix rather than using edges to define the graph. An advantage of this approach is it can deal immediately with many > 2 grouping columns with the same code. (So long as you write the function that determines links flexibly.) A disadvantage is you need to make all pair-wise comparisons between rows to construct the matrix, so for very long vectors it could be slow. As is, #Frank's answer would work better for very long data, or if you only ever have two columns.
The steps are
compare rows based on groups and define these rows as linked (i.e., create a graph)
determine connected components of the graph defined by the links in 1.
You could do 2 a few ways. Below I show a brute force way where you 2a) collapse links, till reaching a stable link structure using matrix multiplication and 2b) convert the link structure to a factor using hclust and cutree. You could also use igraph::clusters on a graph created from the matrix.
1. construct an adjacency matrix (matrix of pairwise links) between rows
(i.e., if they in the same group, the matrix entry is 1, otherwise it's 0). First making a helper function that determines whether two rows are linked
linked_rows <- function(data){
## helper function
## returns a _function_ to compare two rows of data
## based on group membership.
## Use Vectorize so it works even on vectors of indices
Vectorize(function(i, j) {
## numeric: 1= i and j have overlapping group membership
common <- vapply(names(data), function(name)
data[i, name] == data[j, name],
FUN.VALUE=FALSE)
as.numeric(any(common))
})
}
which I use in outer to construct a matrix,
rows <- 1:nrow(df)
A <- outer(rows, rows, linked_rows(df))
2a. collapse 2-degree links to 1-degree links. That is, if rows are linked by an intermediate node but not directly linked, lump them in the same group by defining a link between them.
One iteration involves: i) matrix multiply to get the square of A, and
ii) set any non-zero entry in the squared matrix to 1 (as if it were a first degree, pairwise link)
## define as a function to use below
lump_links <- function(A) {
A <- A %*% A
A[A > 0] <- 1
A
}
repeat this till the links are stable
oldA <- 0
i <- 0
while (any(oldA != A)) {
oldA <- A
A <- lump_links(A)
}
2b. Use the stable link structure in A to define groups (connected components of the graph). You could do this a variety of ways.
One way, is to first define a distance object, then use hclust and cutree. If you think about it, we want to define linked (A[i,j] == 1) as distance 0. So the steps are a) define linked as distance 0 in a dist object, b) construct a tree from the dist object, c) cut the tree at zero height (i.e., zero distance):
df$combinedGrp <- cutree(hclust(as.dist(1 - A)), h = 0)
df
In practice you can encode steps 1 - 2 in a single function that uses the helper lump_links and linked_rows:
lump <- function(df) {
rows <- 1:nrow(df)
A <- outer(rows, rows, linked_rows(df))
oldA <- 0
while (any(oldA != A)) {
oldA <- A
A <- lump_links(A)
}
df$combinedGrp <- cutree(hclust(as.dist(1 - A)), h = 0)
df
}
This works for the original df and also for the structure in #rawr's answer
df <- data.frame(grp1 = c(1,1,1,2,2,2,3,3,3,4,4,4,5,5,6,7,8,9),
grp2 = c(1,2,3,3,4,5,6,7,8,6,9,10,11,3,12,3,6,12))
lump(df)
grp1 grp2 combinedGrp
1 1 1 1
2 1 2 1
3 1 3 1
4 2 3 1
5 2 4 1
6 2 5 1
7 3 6 2
8 3 7 2
9 3 8 2
10 4 6 2
11 4 9 2
12 4 10 2
13 5 11 1
14 5 3 1
15 6 12 3
16 7 3 1
17 8 6 2
18 9 12 3
PS
Here's a version using igraph, which makes the connection with #Frank's answer more clear:
lump2 <- function(df) {
rows <- 1:nrow(df)
A <- outer(rows, rows, linked_rows(df))
cluster_A <- igraph::clusters(igraph::graph.adjacency(A))
df$combinedGrp <- cluster_A$membership
df
}
Hope this solution helps you a bit:
Assumption: df is ordered on the basis of grp1.
## split dataset using values of grp1
split_df <- split.default(df$grp2,df$grp1)
parent <- vector('integer',length(split_df))
## find out which combinations have values of grp2 in common
for (i in seq(1,length(split_df)-1)){
for (j in seq(i+1,length(split_df))){
inter <- intersect(split_df[[i]],split_df[[j]])
if (length(inter) > 0){
parent[j] <- i
}
}
}
ans <- vector('list',length(split_df))
index <- which(parent == 0)
## index contains indices of elements that have no element common
for (i in seq_along(index)){
ans[[index[i]]] <- rep(i,length(split_df[[i]]))
}
rest_index <- seq(1,length(split_df))[-index]
for (i in rest_index){
val <- ans[[parent[i]]][1]
ans[[i]] <- rep(val,length(split_df[[i]]))
}
df$combinedGrp <- unlist(ans)
df
grp1 grp2 combinedGrp
1 1 1 1
2 1 2 1
3 1 3 1
4 2 3 1
5 2 4 1
6 2 5 1
7 3 6 2
8 3 7 2
9 3 8 2
10 4 6 2
11 4 9 2
12 4 10 2
Based on https://stackoverflow.com/a/35773701/2152245, I used a different implementation of igraph because I already had an adjacency matrix of sf polygons from st_intersects():
library(igraph)
library(sf)
# Use example data
nc <- st_read(system.file("shape/nc.shp", package="sf"))
nc <- nc[-sample(1:nrow(nc),nrow(nc)*.75),] #drop some polygons
# Find intersetions
b <- st_intersects(nc, sparse = F)
g <- graph.adjacency(b)
clu <- components(g)
gr <- groups(clu)
# Quick loop to assign the groups
for(i in 1:nrow(nc)){
for(j in 1:length(gr)){
if(i %in% gr[[j]]){
nc[i,'group'] <- j
}
}
}
# Make a new sfc object
nc_un <- group_by(nc, group) %>%
summarize(BIR74 = mean(BIR74), do_union = TRUE)
plot(nc_un['BIR74'])
I have the dataframe ds
CountyID ZipCode Value1 Value2 Value3 ... Value25
1 1 0 etc etc etc
2 1 3
3 1 0
4 1 1
5 2 2
6 3 3
7 4 7
8 4 2
9 5 1
10 6 0
and would like to aggregate based on ds$ZipCode and set ds$CountyID equal to the primary county based on the highest ds$Value1. For the above example, it would look like this:
CountyID ZipCode Value1 Value2 Value3 ... Value25
2 1 4 etc etc etc
5 2 2
6 3 3
7 4 9
9 5 1
10 6 0
All the ValueX columns are the sum of that column grouped by ZipCode.
I've tried a bunch of different strategies over the last couple days, but none of them work. The best I've come up with is
#initialize the dataframe
ds_temp = data.frame()
#loop through each subset based on unique zipcodes
for (zip in unique(ds$ZipCode) {
sub <- subset(ds, ds$ZipCode == zip)
len <- length(sub)
maxIndex <- which.max(sub$Value1)
#do the aggregation
row <- aggregate(sub[3:27], FUN=sum, by=list(
CountyID = rep(sub$CountyID[maxIndex], len),
ZipCode = sub$ZipCode))
rbind(ds_temp, row)
}
ds <- ds_temp
I haven't been able to test this on the real data, but with dummy datasets (such as the one above), I keep getting the error "arguments must have the same length). I've messed around with rep() and fixed vectors (eg c(1,2,3,4)) but no matter what I do, the error persists. I also occasionally get an error to the effect of
cannot subset data of type 'closure'.
Any ideas? I've also tried messing around with data.frame(), ddply(), data.table(), dcast(), etc.
You can try this:
data.frame(aggregate(df[,3:27], by=list(df$ZipCode), sum),
CountyID = unlist(lapply(split(df, df$ZipCode),
function(x) x$CountyID[which.max(x$Value1)])))
Fully reproducible sample data:
df<-read.table(text="
CountyID ZipCode Value1
1 1 0
2 1 3
3 1 0
4 1 1
5 2 2
6 3 3
7 4 7
8 4 2
9 5 1
10 6 0", header=TRUE)
data.frame(aggregate(df[,3], by=list(df$ZipCode), sum),
CountyID = unlist(lapply(split(df, df$ZipCode),
function(x) x$CountyID[which.max(x$Value1)])))
# Group.1 x CountyID
#1 1 4 2
#2 2 2 5
#3 3 3 6
#4 4 9 7
#5 5 1 9
#6 6 0 10
In response to your comment on Frank's answer, you can preserve the column names by using the formula method in aggregate. Using Franks's data df, this would be
> cbind(aggregate(Value1 ~ ZipCode, df, sum),
CountyID = sapply(split(df, df$ZipCode), function(x) {
with(x, CountyID[Value1 == max(Value1)]) }))
# ZipCode Value1 CountyID
# 1 1 4 2
# 2 2 2 5
# 3 3 3 6
# 4 4 9 7
# 5 5 1 9
# 6 6 0 10
I'm trying to select the column with the highest value for each row in a data.frame. So for instance, the data is set up as such.
> df <- data.frame(one = c(0:6), two = c(6:0))
> df
one two
1 0 6
2 1 5
3 2 4
4 3 3
5 4 2
6 5 1
7 6 0
Then I'd like to set another column based on those rows. The data frame would look like this.
> df
one two rank
1 0 6 2
2 1 5 2
3 2 4 2
4 3 3 3
5 4 2 1
6 5 1 1
7 6 0 1
I imagine there is some sort of way that I can use plyr or sapply here but it's eluding me at the moment.
There might be a more efficient solution, but
ranks <- apply(df, 1, which.max)
ranks[which(df[, 1] == df[, 2])] <- 3
edit: properly spaced!