Obtain Date Column from Xts Object

Obtain date column from xts object

getSymbols does not return a data.frame by default; it returns an xts object. xts objects do not have row names. They have an index attribute that you can access with the index function.

Add Date Column to XTS Object

For turning a xts object into a dataframe with the date column you can use the following code. You use index to get the date index of the xts object and coredata for all the data contained in the xts object.

# my_xts is based on data from OP
df1 <- data.frame(Date = index(my_xts), coredata(my_xts) )

# show resulting structure
str(df1)

'data.frame':   12 obs. of  2 variables:
$ Date        : Date, format: "2016-01-01" "2016-02-01" "2016-03-01" "2016-04-01" ...
$ AAPL.Returns: num  -0.07524 -0.00668 0.12721 -0.13992 0.06529 ...

# outcome
df1
         Date AAPL.Returns
1  2016-01-01 -0.075242313
2  2016-02-01 -0.006677563
3  2016-03-01  0.127210629
4  2016-04-01 -0.139921096
5  2016-05-01  0.065286997
6  2016-06-01 -0.042659753
7  2016-07-01  0.090062774
8  2016-08-01  0.018136446
9  2016-09-01  0.065504290
10 2016-10-01  0.004334348
11 2016-11-01 -0.026598591
12 2016-12-01  0.047955150

How can I extract the timestamp of an xts object in R?

Use index(xtsobject)

data(sample_matrix)
sample.xts <- as.xts(sample_matrix, descr='my new xts object')
index(sample.xts)

> index(sample.xts)
  [1] "2007-01-02 EST" "2007-01-03 EST" "2007-01-04 EST" "2007-01-05 EST" "2007-01-06 EST" "2007-01-07 EST"
  [7] "2007-01-08 EST" "2007-01-09 EST" "2007-01-10 EST" "2007-01-11 EST" "2007-01-12 EST" "2007-01-13 EST"
 [13] "2007-01-14 EST" "2007-01-15 EST" "2007-01-16 EST" "2007-01-17 EST" "2007-01-18 EST" "2007-01-19 EST"
 [19] "2007-01-20 EST" "2007-01-21 EST" "2007-01-22 EST" "2007-01-23 EST" "2007-01-24 EST" "2007-01-25 EST"

Extracting Dates from xts object based on vaule

You could try

index(A[diff(A)<0])
#[1] "2014-12-31" "2015-01-04"

index(A[diff(A)==1])
#[1] "2014-12-29" "2015-01-02"

i am getting double date and time column after using xts? and getting error 'x' must be a time-series object

The data part should not include the times.

library(xts)

time <- c("21.11.2021 22:45", "21.11.2021 23:25")
time_p <- as.POSIXct(time, format='%d.%m.%Y %H:%M')
value1 <- c(1, 9)
value2 <- c(1, 21)

xts(cbind(value1, value2), time_p)
##                     value1 value2
## 2021-11-21 22:45:00      1      1
## 2021-11-21 23:25:00      9     21

or

DF <-  data.frame(time_p, value1, value2)
z <- read.zoo(DF)
as.xts(z)
# same

Extract week and day from an xts object into a column

You could use .indexwday to get the weekday of the index. It returns the wday vector from a POSIXlt representation of the index, so see the Details section of ?POSIXlt for a description.

require(quantmod)
getSymbols("SPY")
SPY$weekday <- .indexwday(SPY)

Using date and seconds from midnight columns to convert a data.frame to an xts object (R)

We can convert the 'DATE', 'TIME' column to Datetime class and convert the dataset to xts by specifying the order.by

library(xts)     
library(lubridate)
xts(df1[-(1:2)], order.by = as.POSIXct(paste(df1$DATE, 
             hms::hms(seconds_to_period(df1$TIME)))))
#                      Col1   Col2
#1993-01-04 09:35:38 10.250 10.000
#1994-01-05 09:35:41 10.250 10.111
#1997-03-16 09:35:46 10.250 10.222
#2001-08-28 09:36:05 10.251 10.444
#2006-04-20 09:40:07 10.251 10.555
#2017-11-10 09:36:01 10.251 10.333

NOTE: The index of xts needs a Datetime class object and not a formatted character class vector

data

df1 <- structure(list(DATE = c("1993-01-04", "1994-01-05", "1997-03-16", 
"2017-11-10", "2001-08-28", "2006-04-20"), TIME = c(34538L, 34541L, 
34546L, 34561L, 34565L, 34807L), Col1 = c(10.25, 10.25, 10.25, 
10.251, 10.251, 10.251), Col2 = c(10, 10.111, 10.222, 10.333, 
10.444, 10.555)), class = "data.frame", row.names = c("1", "2", 
"3", "4", "5", "6"))

How to divide the columns of an xts object by the columns of another xts object by its column name in R

The trick is to get the data in the same order by sorting the column names and use this to select the columns in order. This only works if all the column names are the same in both xts objects.

Note: I have renamed the df1 and df2 objects to df1_xts and df2_xts to show that we are dealing with xts objects and not data.frames to avoid any confusion.

# select the column names of both xts objects, via sort 
# and use matrix calculations to do the rest.
df1_xts[,sort(names(df1_xts))] / df2_xts[,sort(names(df2_xts))]

           Argentina Brasil Chile Colombia Peru
2017-01-01         4     16     1        8    1
2018-01-01         2      2     2        2    9
2019-01-01         4      4    14        1    2
2020-01-01         3      2    12        3    2
2021-01-01         2      1     6        2    2

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