CODEWITHSUNDEEP
rlapply
user9357993
user9357993

Reputation: 41

Using lapply to standardise the ts() function

I am new to R and writing functions. I've spent hours trying to figure this out and searching Google, but can't seem to find anything. Hopefully you can help? I want to use lapply() to analyze the data below using the ts() function.

My code looks like this:

library(dplyr)

#group out different sites
 mylist <- data %>%
   group_by(Site)
 mylist

#Write ts() function
 alpha_function = function(x) {
    ts_alpha = ts(x$Temperature, frequency=12, start=c(0017, 7, 20))
     return(data.frame(ts_alpha))    
 }

 #Run list through lapply()
 results = lapply(mylist, alpha_function())

But I get this error: argument "x" is missing with no default.

I have a data set that looks like:

Site(factor)    Date(POSIXct)         Temperature(num)
1               0017-03-04            2.73
2               0017-03-04            3.73
3               0017-03-04            2.71
4               0017-03-04            2.22
5               0017-03-04            2.89
etc.

I have over 3,000 temperature readings at different dates for 5 different sites.

Thanks in advance!

Upvotes: 0

Views: 328

Answers (2)

Aur&#232;le
Aur&#232;le

Reputation: 12839

A recommended approach when working with dplyr and the tidyverse is to keep things in data frames:

library(tidyverse)
library(zoo)

dat %>% 
  nest(-Site) %>% 
  mutate(data = map(data, ~ zoo(.x$Temperature, .x$Date)))

# # A tibble: 5 x 2
#   Site  data     
#   <fct> <list>   
# 1 a     <S3: zoo>
# 2 b     <S3: zoo>
# 3 c     <S3: zoo>
# 4 d     <S3: zoo>
# 5 e     <S3: zoo>

Or if we must have ts rather than zoo objects, we can use as.ts(zoo(...)).

In case we still prefer regular lists, we can use base split() and lapply():

dat %>% 
  split(.$Site) %>% 
  lapply(function(.x) zoo(.x$Temperature, .x$Date))

# List of 5
#  $ a:‘zoo’ series from 2017-03-04 12:00:00 to 2017-05-06 00:30:00
#   Data: num [1:3000] 5.37 5.49 5.32 5.44 5.43 ...
#   Index:  POSIXct[1:3000], format: "2017-03-04 12:00:00" ...
#  $ b:‘zoo’ series from 2017-03-04 12:00:00 to 2017-05-06 00:30:00
#   Data: num [1:3000] 5.36 5.22 5.15 5.41 5.41 ...
#   Index:  POSIXct[1:3000], format: "2017-03-04 12:00:00" ...
#  $ c:‘zoo’ series from 2017-03-04 12:00:00 to 2017-05-06 00:30:00
#   Data: num [1:3000] 6.08 6.11 6.22 6.13 6.03 ...
#   Index:  POSIXct[1:3000], format: "2017-03-04 12:00:00" ...
#  $ d:‘zoo’ series from 2017-03-04 12:00:00 to 2017-05-06 00:30:00
#   Data: num [1:3000] 5.06 4.96 5.23 5.16 5.29 ...
#   Index:  POSIXct[1:3000], format: "2017-03-04 12:00:00" ...
#  $ e:‘zoo’ series from 2017-03-04 12:00:00 to 2017-05-06 00:30:00
#   Data: num [1:3000] 5.1 5.08 5.14 5.13 5.22 ...
#   Index:  POSIXct[1:3000], format: "2017-03-04 12:00:00" ...

(where dat is generated as follows:

n_sites <- 5
n_dates <- 3000

set.seed(123) ; dat <- tibble(
  Site = factor(rep(letters[1:n_sites], each = n_dates)), 
  Date = rep(seq.POSIXt(as.POSIXct("2017-03-04 12:00:00"), by = "30 min", length.out = n_dates), times = n_sites),
  Temperature = as.vector(replicate(n_sites, runif(1, 5, 6) + cumsum(rnorm(n_dates, 0, 0.1))))
)

Upvotes: 1

P Ackerman
P Ackerman

Reputation: 2406

I'm not exactly an R guy, but I would wager this line: results = lapply(mylist, alpha_function()) should be results = lapply(mylist, alpha_function).

What you have calls the alpha function when you are trying to supply it to lapply, when what you really (most likely) want to do is provide a reference to the function without calling it. (The error you are getting indicates that alpha_function needs an x parameter when being called like alpha_function()).

Upvotes: 2

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