I have written a function and am trying to iterate so that each value from a list becomes the input to the function once.
repeat100 <- function(B) {
se100 <- replicate(100, bootandse(B))
bandse <- data.frame("B" = B, "SE" = se100)
}
The output of this function is a dataframe with one column for the value of B that was inputted and another column for the SE. Like this, but with 100 rows.
B SE
1 2
1 4
1 3
I have an object B with several values. I am trying to iterate repeat100() over each value of B.
B <- c(1, 4, 6, 20, 30)
How can I get the resulting output to be a dataframe that has the output of each repeat100() for each value of B stacked like this?
B SE
1 2
1 3
1 2
4 5
4 3
4 2
Use lapply with rbind to combine the data in one dataframe.
result <- do.call(rbind, lapply(B, repeat100))
Or purrr::map_df which is shorter.
purrr::map_df(B, repeat100)
We can use rbindlist from data.table
library(data.table)
rbindlist(lapply(B, repeat100))
Related
does anyone know how to have a row in R that is calculated from another row automatically? i.e.
lets say in excel, i want to make a row C, which is made up of (B2/B1)
e.g. C1 = B2/B1
C2 = B3/B2
...
Cn = Cn+1/Cn
but in excel, we only need to do one calculation then drag it down. how do we do it in R?
In R you work with columns as vectors so the operations are vectorized. The calculations as described could be implemented by the following commands, given a data.frame df (i.e. a table) and the respective column names as mentioned:
df["C1"] <- df["B2"]/df["B1"]
df["C2"] <- df["B3"]/df["B2"]
In R you usually would name the columns according to the content they hold. With that, you refer to the columns by their name, although you can also address the first column as df[, 1], the first row as df[1, ] and so on.
EDIT 1:
There are multiple ways - and certainly some more elegant ways to get it done - but for understanding I kept it in simple base R:
Example dataset for demonstration:
df <- data.frame("B1" = c(1, 2, 3),
"B2" = c(2, 4, 6),
"B3" = c(4, 8, 12))
Column calculation:
for (i in 1:ncol(df)-1) {
col_name <- paste0("C", i)
df[col_name] <- df[, i+1]/df[, i]
}
Output:
B1 B2 B3 C1 C2
1 1 2 4 2 2
2 2 4 8 2 2
3 3 6 12 2 2
So you iterate through the available columns B1/B2/B3. Dynamically create a column name in every iteration, based on the number of the current iteration, and then calculate the respective column contents.
EDIT 2:
Rowwise, as you actually meant it apparently, works similarly:
a <- c(10,15,20, 1)
df <- data.frame(a)
for (i in 1:nrow(df)) {
df$b[i] <- df$a[i+1]/df$a[i]
}
Output:
a b
1 10 1.500000
2 15 1.333333
3 20 0.050000
4 1 NA
You can do this just using vectors, without a for loop.
a <- c(10,15,20, 1)
df <- data.frame(a)
df$b <- c(df$a[-1], 0) / df$a
print(df)
a b
1 10 1.500000
2 15 1.333333
3 20 0.050000
4 1 0.000000
Explanation:
In the example data, df$a is the vector 10 15 20 1.
df$a[-1] is the same vector with its first element removed, 15 20 1.
And using c() to add a new element to the end so that the vector has the same lenght as before:
c(df$a[-1],0) which is 15 20 1 0
What we want for column b is this vector divided by the original df$a.
So:
df$b <- c(df$a[-1], 0) / df$a
I have a dataframe with multiple rows. I want to call a function is using any two rows. For example, Let's say I have this data and this myFunc which accepts two args:
df <- data.frame(q1=c(1,2,5), q2=c(5,5,5), q3=c(5,2,5), q4=c(5,5,5), q5=c(2,3,1))
df
q1 q2 q3 q4 q5
1 1 5 5 5 2
2 2 5 2 5 3
3 5 5 5 5 1
myFunc<-function(a,b) sum((df[a,]==df[b,] & df[a,]==5)*1)
A want to apply myFunc for row 1 and 2, myFunc(1,2) and I expect 2, myFunc compute how many "5" are have in common under the same column, between row 1 and 2.
Since I have thousands of rows, and I want to match all pairs, I want do this without writing a for loop, maybe with the do call or apply function family.
I tried this:
a=c(1,2) # match the row 1 and 2
b=c(2,3) # match the row 2 and 3
my_list=list(a,b)
do.call("myFunc", my_list)
But I got 4, instead of 2 and 2, any ideas?
The question recently changed. My understanding of it is that the input should be a list of pairs of row numbers and the output should be the same length as that list such that each component of the output is the number of columns with both entries equal to 5 in both rows defined by the corresponding pair. Thus for df shown in the question the list L shown below would correspond to c(myFunc(1, 2), myFunc(2, 3)) where myFunc is as defined in the question.
L <- list(1:2, 2:3)
myFunc2 <- function(x) myFunc(x[1], x[2])
sapply(L, myFunc2)
## [1] 2 2
Note that *1 in myFunc is unnecessary since sum will coerce a logical argument to numeric.
An alternative might be to specify the first row numbers as a vector and the second row numbers as another vector. In terms of L that would be a <- sapply(L, "[", 1); b <- sapply(L, "[", 2). Then use mapply.
a <- c(1, 2) # L[[1]][1], L[[2]][1]
b <- c(2, 3) # L[[1]][2], L[[2]][2]
mapply(myFunc, a, b)
## [1] 2 2
Try passing the rows instead of the row index
df <- data.frame(q1=c(1,2,5), q2=c(5,5,5), q3=c(5,2,5), q4=c(5,5,5), q5=c(2,3,1))
myFunc<-function(a,b) sum((a==b & a==5)*1)
myFunc(df[1,],df[2,])
This worked for me (returned 2)
Let's say i have the following list of df's (in reality i have many more dfs).
seq <- c("12345","67890")
li <- list()
for (i in 1:length(seq)){
li[[i]] <- list()
names(li)[i] <- seq[i]
li[[i]] <- data.frame(A = c(1,2,3),
B = c(2,4,6))
}
What i would like to do is calculate the mean within the same cell position between the lists, keeping the same amount of rows and columns as the original lists. How could i do this? I believe I can use the apply() function, but i am unsure how to do this.
The expected output (not surprising):
A B
1 1 2
2 2 4
3 3 6
In reality, the values within each list are not necessarily the same.
If there are no NAs, then we can Reduce to get the sum of observations for each element and divide by the length of the list
Reduce(`+`, li)/length(li)
# A B
#1 1 2
#2 2 4
#3 3 6
If there are NA values, then it may be better to use mean (which has na.rm argument). For this, we can convert it to array and then use apply
apply(array(unlist(li), dim = c(dim(li[[1]]), length(li))), c(1, 2), mean)
An equivalent option in tidyverse would be
library(tidyverse)
reduce(li, `+`)/length(li)
In R, I have a data table and a character vector with a subset of the data table's column names. I need to compute the z-scores (i.e. number of standard deviations from the mean) of each column with a specified name, and put the averages of the z-scores in a new column. I found a solution with explicit for-loops (posted below), but this must be a common enough task that some library function could be made to do the work more elegantly. Is there a better way?
Here's my solution:
#! /usr/bin/env RSCRIPT
library(data.table)
# Sample data table.
dt <- data.table(a=1:3, b=c(5, 6, 3), c=2:4)
# List of column names.
cols <- c('a', 'b')
# Convert columns to z-scores, and add each to a new list of vectors.
zscores <- list()
for (colIx in 1:length(cols)) {
zscores[[colIx]] <- scale(dt[,get(cols[colIx])], center=TRUE, scale=TRUE)
}
# Average corresponding entries of each vector of z-scores.
avg <- numeric(nrow(dt))
for (rowIx in 1:nrow(dt)) {
avg[rowIx] <- mean(sapply(1:length(cols),
function(colIx) {zscores[[colIx]][rowIx]}))
}
# Add new vector to the table, and print out the new table.
dt[,d:=avg]
print(dt)
This gives what you might expect.
a b c d
1: 1 5 2 -0.39089105
2: 2 6 3 0.43643578
3: 3 3 4 -0.04554473
scale can be applied to matrix(-like) object, you can get desired output by
> set(dt, NULL, 'd', rowMeans(scale(dt[, cols, with = F])))
> dt
a b c d
1: 1 5 2 -0.39089105
2: 2 6 3 0.43643578
3: 3 3 4 -0.04554473
I am working in R with a data frame d:
ID <- c("A","A","A","B","B")
eventcounter <- c(1,2,3,1,2)
numberofevents <- c(3,3,3,2,2)
d <- data.frame(ID, eventcounter, numberofevents)
> d
ID eventcounter numberofevents
1 A 1 3
2 A 2 3
3 A 3 3
4 B 1 2
5 B 2 2
where numberofevents is the highest value in the eventcounter for each ID.
Currently, I am trying to create an additional vector z <- c(6,6,6,3,3).
If the numberofevents == 3, it is supposed to calculate sum(1:3), equally to 3 + 2 + 1 = 6.
If the numberofevents == 2, it is supposed to calculate sum(1:2) equally to 2 + 1 = 3.
Working with a large set of data, I thought it might be convenient to create this additional vector
by using the sum function in R d$z<-sum(1:d$numberofevents), i.e.
sum(1:3) # for the rows 1-3
and
sum(1:2) # for the rows 4-5.
However, I always get this warning:
Numerical expression has x elements: only the first is used.
You can try ave
d$z <- with(d, ave(eventcounter, ID, FUN=sum))
Or using data.table
library(data.table)
setDT(d)[,z:=sum(eventcounter), ID][]
Try using apply sapply or lapply functions in R.
sapply(numberofevents, function(x) sum(1:x))
It works for me.