This question already has answers here:
Calculating cumulative sum for each row
(6 answers)
Closed 7 years ago.
I've come across the aggregate() function and things like seq_along(), but I'm not sure how to solve this yet:
For the following:
x
1
5
10
20
I'd like to get the following output:
y
1
6
16
36
It seemed to me that doing something like x[1:seq_along(x)] would do the trick but it seems not because seq_along(x) is a sequence rather than a number.
As mentioned by #DavidArenburg, you can use the cumsum function:
x <- c(1, 5, 10, 20)
cumsum(x)
[1] 1 6 16 36
Related
This question already has answers here:
Assign a value, if a number is in between two numbers
(5 answers)
Nested ifelse statement
(10 answers)
Closed last year.
I have data like this
df:
A rate
10 ...
20
How to rate A[] with this rule with the simplest code in R
#if A<20 rate=1 if 20<A[]<30 rate=2
Thanks
There are several ways to do this, perhaps the simplest and most general being findInterval:
set.seed(1)
A <- sample(40, 5)
A
#> [1] 4 39 1 34 23
rate <- findInterval(A, c(-Inf, 20, 30, Inf))
rate
#> [1] 1 3 1 3 2
Created on 2022-02-17 by the reprex package (v2.0.1)
This question already has answers here:
Check to see if a value is within a range?
(7 answers)
Closed 5 years ago.
I'm starting to learn R, as it's needed for work. I have never done statistical work, so I'm a bit lost.
I'm looking to get the value of x between two numbers.
So, for example, the range is 3:7 I need to print 4,5,6
I have tried
x <- 3:7
x[x>3 && x<7]
and
x <- 3
v <- 7
cbind(x, findInterval(x, v))
Any advice/guidelines
An option is between from data.table
x[data.table::between(x, 3, 7, incbounds = FALSE)]
#[1] 4 5 6
This question already has answers here:
Finding ALL duplicate rows, including "elements with smaller subscripts"
(9 answers)
How can I remove all duplicates so that NONE are left in a data frame?
(3 answers)
Closed 5 years ago.
I would like to keep the non-duplicated values from a vector, but without retaining one element from duplicated values. unique() does not work for this. Neither would duplicated().
For example:
> test <- c(1,1,2,3,4,4,4,5,6,6,7,8,9,9)
> unique(test)
[1] 1 2 3 4 5 6 7 8 9
Whereas I would like the result to be: 2,3,5,7,8
Any ideas on how to approach this? Thank you!
We can use duplicated
test[!(duplicated(test)|duplicated(test, fromLast=TRUE))]
#[1] 2 3 5 7 8
You can use ave to count the length of sub-groups divided by unique values in test and retain only the ones whose length is 1 (the ones that have no duplicates)
test[ave(test, test, FUN = length) == 1]
#[1] 2 3 5 7 8
If test is comprised of characters, use seq_along as first argument of ave
test[ave(seq_along(test), test, FUN = length) == 1]
This question already has answers here:
Extract the maximum value within each group in a dataframe [duplicate]
(3 answers)
Closed 7 years ago.
I am searching for an efficient and fast way to do the following:
I have a data frame with, say, 2 variables, A and B, where the values for A can occur several times:
mat<-data.frame('VarA'=rep(seq(1,10),2),'VarB'=rnorm(20))
VarA VarB
1 0.95848233
2 -0.07477916
3 2.08189370
4 0.46523827
5 0.53500190
6 0.52605101
7 -0.69587974
8 -0.21772252
9 0.29429577
10 3.30514605
1 0.84938361
2 1.13650996
3 1.25143046
Now I want to get a vector giving me for every unique value of VarA
unique(mat$VarA)
the maximum of VarB conditional on VarA.
In the example here that would be
1 0.95848233
2 1.13650996
3 2.08189370
etc...
My data-frame is very big so I want to avoid the use of loops.
Try this:
library(dplyr)
mat %>% group_by(VarA) %>%
summarise(max=max(VarB))
Try to use data.table package.
library(data.table)
mat <- data.table(mat)
result <- mat[,max(VarB),VarA]
print(result)
Try this:
library(plyr)
ddply(mat, .(VarA), summarise, VarB=min(VarB))
This question already has answers here:
Calculate the mean of every 13 rows in data frame
(4 answers)
Closed 1 year ago.
I want to calculate meas of three consecutive variables a vector.
Ex:
Vec<-rep(1:10)
I would like the output to be like the screenshot below:
You can create the following function to calculate means by groups of 3 (or any other number):
f <- function(x, k=3)
{
for(i in seq(k,length(x),k))
x[(i/k)] <- mean(x[(i-k+1):i])
return(x[1:(length(x)/k)])
}
f(1:15)
[1] 2 5 8 11 14
We can create a grouping variable using gl and then get the mean with ave
ave(Vec, as.numeric(gl(length(Vec), 3, length(Vec))))