This question already has answers here:
Calculating cumulative sum for each row
(6 answers)
Sum of previous rows in a column R
(1 answer)
Closed 3 years ago.
I have a vector of alternating TRUE and FALSE values:
dat <- c(T,F,F,T,F,F,F,T,F,T,F,F,F,F)
I'd like to number each instance of TRUE with a unique sequential number and to assign each FALSE value the number associated with the TRUE value preceding it.
therefore, my desired output using the example dat above (which has 4 TRUE values):
1 1 1 2 2 2 2 3 3 4 4 4 4 4
What I tried:
I've tried the following (which works), but I know there must be a simpler solution!!
whichT <- which(dat==T)
whichF <- which(dat==F)
l1 <- lapply(1:length(whichT),
FUN = function(x)
which(whichF > whichT[x] & whichF < whichT[(x+1)])
)
l1[[length(l1)]] <- which(whichF > whichT[length(whichT)])
replaceFs <- unlist(
lapply(1:length(whichT),
function(x) l1[[x]] <- rep(x,length(l1[[x]]))
)
)
replaceTs <- 1:length(whichT)
dat2 <- dat
dat2[whichT] <- replaceTs
dat2[whichF] <- replaceFs
dat2
[1] 1 1 1 2 2 2 2 3 3 4 4 4 4 4
I need a simpler and quicker solution b/c my real data set is 181k rows long!
Base R solutions preferred, but any solution works
cumsum(dat) will do what you want. When used in mathematical functions TRUE gets converted to 1 and FALSE to 0 so taking the cumulative sum will add 1 every time you see a TRUE and add nothing when there is a FALSE which is what you want.
dat <- c(T,F,F,T,F,F,F,T,F,T,F,F,F,F)
cumsum(dat)
# [1] 1 1 1 2 2 2 2 3 3 4 4 4 4 4
Instead of doing the indexing, it can be easily done with cumsum from base R. Here, TRUE/FALSE gets coerced to 1/0 and when we do the cumulative sum, whereever there is 1, it gets increment by 1
cumsum(dat)
#[1] 1 1 1 2 2 2 2 3 3 4 4 4 4 4
cumsum() is the most straightforward way, however, you can also do:
Reduce("+", dat, accumulate = TRUE)
[1] 1 1 1 2 2 2 2 3 3 4 4 4 4 4
Related
Assume i have the following dataset:
dt<-data.frame(X=sample(5),Y=sample(5))
now, i need to compare these two features and select the one which is smaller.
X Y
1 4 3
2 5 2
3 2 4
4 3 5
5 1 1
Then the expected answer would be
3
2
2
3
1
I know
min(dt[1,])
could be helpful but it only gives me 1
Use pmin, which is the vectorized version of min:
pmin(dt$X,dt$Y)
Like thus:
> dt<-data.frame(X=sample(5),Y=sample(5))
> dt
X Y
1 3 2
2 4 3
3 1 5
4 2 4
5 5 1
> pmin(dt$X,dt$Y)
[1] 2 3 1 2 1
high <- apply(dt[,c("X","Y")], 1, max)
is another implementation
integer(0) or length 0 element happens when one of X or Y is of length(0)
For min or max, a length-one vector. For pmin or pmax, a vector of length the longest of the input vectors, or length zero if one of the inputs had zero length.
(from documentation)
max(which(1:3 == 5),10) works but pmax(which(1:3 == 5),10) gives integer(0)
This question already has answers here:
Create a variable capturing the most frequent occurence by group
(3 answers)
How to find the statistical mode?
(35 answers)
Closed 5 years ago.
I am trying to find the most frequent value by group. In the following example dataframe:
df<-data.frame(a=c(1,1,1,1,2,2,2,3,3),b=c(2,2,1,2,3,3,1,1,2))
> df
a b
1 1 2
2 1 2
3 1 1
4 1 2
5 2 3
6 2 3
7 2 1
8 3 1
9 3 2
I would like to add a column 'c' which has the most occurring value in 'b' when its values are grouped by 'a'. I would like the following output:
> df
a b c
1 1 2 2
2 1 2 2
3 1 1 2
4 1 2 2
5 2 3 3
6 2 3 3
7 2 1 3
8 3 1 1
9 3 2 1
I tried using table and tapply but didn't get it right. Is there a fast way to do that?
Thanks!
Building on Davids comments your solution is the following:
Mode <- function(x) {
ux <- unique(x)
ux[which.max(tabulate(match(x, ux)))]
}
library(dplyr)
df %>% group_by(a) %>% mutate(c=Mode(b))
Notice though that for the tie when df$a is 3 then the mode for b is 1.
We could get the 'Mode' of 'b' grouped by 'a' using ave
Mode <- function(x) {
ux <- unique(x)
ux[which.max(tabulate(match(x, ux)))]
}
df$c <- with(df, ave(b, a, FUN=Mode))
df$c
#[1] 2 2 2 2 3 3 3 1 1
Or using data.table
library(data.table)
setDT(df)[, c:= Mode(b), by=a][]
Here is a base R method that uses table to calculate a cross tab, max.col to find the mode per group, and rep together with rle to fill in the mode across groups.
# calculate a cross tab, frequencies by group
myTab <- table(df$a, df$b)
# repeat the mode for each group, as calculated by colnames(myTab)[max.col(myTab)]
# repeating by the number of times the group ID is observed
df$c <- rep(colnames(myTab)[max.col(myTab)], rle(df$a)$length)
df
a b c
1 1 2 2
2 1 2 2
3 1 1 2
4 1 2 2
5 2 3 3
6 2 3 3
7 2 1 3
8 3 1 2
9 3 2 2
Note that this assumes the data has been sorted by group. Also, the default of max.col is to break ties (mulitple modes) at random. If you want the first or last value to be the mode, you can set this using the ties.method argument.
I have a few large data sets with many variables. There is a "key" variable that is the ID for the research participant. In these data sets, there are some IDs that are duplicated. I have written code to extract all data for duplicated IDs, but I would like a way to check if the remainder of the variables for those IDs are equal or not. Below is a simplistic example:
ID X Y Z
1 2 3 4
1 2 3 5
2 5 5 4
2 5 5 4
3 1 2 3
3 2 2 3
3 1 2 3
In this example, I would like to be able to identify that the rows for ID 1 and ID 3 are NOT all equal. Is there any way to do this in R?
You can use duplicated for this:
d <- read.table(text='ID X Y Z
1 2 3 4
1 2 3 5
2 5 5 4
2 5 5 4
3 1 2 3
3 2 2 3
3 1 2 3
4 1 1 1', header=TRUE)
tapply(duplicated(d), d[, 1], function(x) all(x[-1]))
## 1 2 3 4
## FALSE TRUE FALSE TRUE
Duplicated returns a vector indicating, for each row of a dataframe, whether it has been encountered earlier in the dataframe. We use tapply over this logical vector, splitting it in to groups based on ID and applying a function to each of these groups. The function we apply is all(x[-1]), i.e. we ask whether all rows for the group, other than the initial row, are duplicated?
Note that I added a group with a single record to ensure that the solution works in these cases as well.
Alternatively, you can reduce the dataframe to unique records with unique, and then split by ID and check whether each split has only a single row:
sapply(split(unique(d), unique(d)[, 1]), nrow) == 1
## 1 2 3 4
## FALSE TRUE FALSE TRUE
(If it's a big dataframe it's worth calculating unique(d) in advance rather than calling it twice.)
In R, in a vector, i.e. a 1-dim matrix, I would like to change components with value 3 to with value 1, and components with value 4 with value 2. How shall I do that? Thanks!
The idiomatic r way is to use [<-, in the form
x[index] <- result
If you are dealing with integers / factors or character variables, then == will work reliably for the indexing,
x <- rep(1:5,3)
x[x==3] <- 1
x[x==4] <- 2
x
## [1] 1 2 1 2 5 1 2 1 2 5 1 2 1 2 5
The car has a useful function recode (which is a wrapper for [<-), that will let you combine all the recoding in a single call
eg
library(car)
x <- rep(1:5,3)
xr <- recode(x, '3=1; 4=2')
x
## [1] 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
xr
## [1] 1 2 1 2 5 1 2 1 2 5 1 2 1 2 5
Thanks to #joran for mentioning mapvalues from the plyr package, another wrapper for [<-
x <- rep(1:5,3)
mapvalues(x, from = c(3,1), to = c(1,2))
plyr::revalue is a wrapper for mapvalues specifically factor or character variables.
This problem seems trivial but I'm at my wits end after hours of reading.
I need to generate a vector of the same length as the input vector that lists for each value of the input vector the total count for that value. So, by way of example, I would want to generate the last column of this dataframe:
> df
customer.id transaction.count total.transactions
1 1 1 4
2 1 2 4
3 1 3 4
4 1 4 4
5 2 1 2
6 2 2 2
7 3 1 3
8 3 2 3
9 3 3 3
10 4 1 1
I realise this could be done two ways, either by using run lengths of the first column, or grouping the second column using the first and applying a maximum.
I've tried both tapply:
> tapply(df$transaction.count, df$customer.id, max)
And rle:
> rle(df$customer.id)
But both return a vector of shorter length than the original:
[1] 4 2 3 1
Any help gratefully accepted!
You can do it without creating transaction counter with:
df$total.transactions <- with( df,
ave( transaction.count , customer.id , FUN=length) )
You can use rle with rep to get what you want:
x <- rep(1:4, 4:1)
> x
[1] 1 1 1 1 2 2 2 3 3 4
rep(rle(x)$lengths, rle(x)$lengths)
> rep(rle(x)$lengths, rle(x)$lengths)
[1] 4 4 4 4 3 3 3 2 2 1
For performance purposes, you could store the rle object separately so it is only called once.
Or as Karsten suggested with ddply from plyr:
require(plyr)
#Expects data.frame
dat <- data.frame(x = rep(1:4, 4:1))
ddply(dat, "x", transform, total = length(x))
You are probably looking for split-apply-combine approach; have a look at ddply in the plyr package or the split function in base R.