CODEWITHSUNDEEP
rxtszooquantmod
saradi
saradi

Reputation: 141

Convert daily to weekly/monthly data with R

I have daily prices series over a wide range of products; I want to convert to a new dataframe with weekly or monthly data.

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I first used xts in order to apply the to.weekly function...which works only for OHLC format. I am sure there may exist a function similar to to.weekly but for dataframe where the format is not OHLC.

There a different posts already related to this as the following: Does rollapply() allow an array of results from call to function? or Averaging daily data into weekly data

I eventually used:

length(bra)

[1] 2416

test<-bra[seq(1,2416,7),]

Would there be a more efficient approach? Thanks.

Upvotes: 7

Views: 37596

Answers (3)

KDJ
KDJ

Reputation: 83

I'm fairly new to R but stumbled on this when I had a similar problem. I needed to convert xts data that isn't OHLC. to.monthly states that it can handle univariate series but it also says in the details that it only supports returning OHLC. I think it might work by just setting OHLC=FALSE. Alternatively, the source of to.period uses the following function which worked for me even to convert several series (all with same time index)

data.monthly <- data[endpoints(data, on="months", k=1), ]

Short and clean and even copies the column names.

Upvotes: 7

Azam Yahya
Azam Yahya

Reputation: 696

Using tidyquant can help you achieve this without converting series into zoo or xts.

library(tidyquant)

library(zoo)

tt <- seq(Sys.Date(), by='day', length=365)  
vals <- data.frame(A=runif(365), B=rnorm(365), C=1:365)
z <- data.frame(vals, tt)

Now, use tidyquant library

z <- z  %>% tq_transmute(mutate_fun = apply.monthly, FUN = mean, na.rm = TRUE)

Upvotes: 1

Oscar Perpi&#241;&#225;n
Oscar Perpi&#241;&#225;n

Reputation: 4511

Let's try with this data:

library(zoo)
tt <- seq(Sys.Date(), by='day', length=365)
vals <- data.frame(A=runif(365), B=rnorm(365), C=1:365)
z <- zoo(vals, tt)

Now I define a function which extracts the year and the number of the week (drop %Y if you don't need to distinguish between years):

week <- function(x)format(x, '%Y.%W')

You can use this function to aggregate the zoo object with mean (for example):

aggregate(z, by=week, FUN=mean)

which produces this result:

                A           B  C
2013.18 0.3455357  0.34129269  3
2013.19 0.4506297  0.57665133  9
2013.20 0.3950585  0.46197173 16
2013.21 0.5990886 -0.02689994 23
2013.22 0.5115043  0.18726564 30
2013.23 0.5327597  0.16250339 37

Upvotes: 15

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