find peaks and valleys for each column of a dataframe - r

Question

I have a dataframe df, I would like to find peaks and valleys for each column and then replace the points where peaks and valleys are present with the value 1.
Here I made an example by applying it to only one column.
Is it possible to do this for all the columns in the dataframe?
df <- data.frame(a = sample(1:10,10),
b = sample(1:10,10),
c = sample(1:10,10),
d = sample(1:10,10),
e = sample(1:10,10))
vallys<- findValleys(df$b, thresh =0)
peaks <- findPeaks(df$b, thresh = 0)
df$b <- rep(0, nrow(df))
df$b <- replace(df$b, peaks, values=1)
df$b <- replace(df$b, vallys, values=1)
Thank you

Answer 1

The easiest thing is to put your code into a function.
library(quantmod)
replace_peaks_valleys <- function(x) {
valleys <- findValleys(x, thresh = 0)
peaks <- findPeaks(x, thresh = 0)
new_col <- rep(0, length(x))
new_col <- replace(new_col, peaks, values = 1)
new_col <- replace(new_col, valleys, values = 1)
return(new_col)
}
Then you can choose whether to do it in base R, dplyr or data.table.
base R
As you want to assign back to your original data frame, in base R you can do (note the square brackets or it will return a list):
df[] <- lapply(df, replace_peaks_valleys)
head(df)
# a b c d e
# 1 0 0 0 0 0
# 2 0 0 0 0 0
# 3 1 1 1 1 1
# 4 1 0 1 1 0
# 5 1 1 0 1 0
# 6 0 1 1 1 1
dplyr
Alternatively, with dplyr you can just do:
library(dplyr)
df |>
mutate(
across(
a:e, replace_peaks_valleys
)
)
# a b c d e
# 1 0 0 0 0 0
# 2 0 0 0 0 0
# 3 1 1 1 1 1
# 4 1 0 1 1 0
# <etc>
data.table
You can also do this with data.table:
library(data.table)
dt <- setDT(df)
dt[, lapply(.SD, replace_peaks_valleys)]
# a b c d e
# 1: 0 0 0 0 0
# 2: 0 0 0 0 0
# 3: 1 0 1 1 1
# 4: 1 1 0 0 0
# <etc>
N.B. I used set.seed(1) before I ran your code - if you do this as well you should exactly the same output.
Function definition
I just copied and pasted your code and made it into a function. You could change it so you assign 0 or 1 to the existing vector, rather than creating a new vector every time:
replace_peaks_valleys2 <- function(x) {
valleys <- findValleys(x, thresh = 0)
peaks <- findPeaks(x, thresh = 0)
x[] <- 0
x[c(peaks,valleys)] <- 1
return(x)
}