(base R) How to apply `diff()` to a column and append it as a new data.frame column?

Question

I have a data frame df1 like this:

time	Diamond.Hands	returns	volume	close
2021-02-16 10:00:00	0.4583333	0.0056710775	10059	53.20
2021-02-16 11:00:00	0.2352941	-0.0037586920	8664	53.01
2021-02-16 12:00:00	0.4400000	-0.0037586920	10059	52.40

# Log return
prices <- df1$close
log_returns <- diff(log(prices), lag=1)
df1$logreturns <- log_returns

returns the error:

Fehler in `$<-.data.frame`(`*tmp*`, logreturns, value = c(0.000187952260679136,  :
  Ersetzung hat 2219 Zeilen, Daten haben 2220

Do you have any ideas how to fix that?

Zheyuan Li · Accepted Answer · 2022-08-02T09:51:30.910

When you do

y <- diff(x, lag = m, differences = k)

the resulting vector y has m * k fewer elements than x. If you want to have both x and y as data.frame/matrix columns, you need to pad m * k number of leading NAs to y.

In your case, m = 1 and k = 1, so you need to pad one NA:

df1$logreturns <- c(NA, log_returns)

More concisely, we can pack your 3 lines of code into 1:

df1$logreturns <- c(NA, diff(log(df1$close)))

Remark:

If you want to know how to do mutate() + diff() in dplyr, then maybe something like:

df1 %>% mutate(logreturns = c(NA, diff(log(close))))

Here is another possibly related Q & A: Error when using "diff" function inside of dplyr mutate.

(base R) How to apply `diff()` to a column and append it as a new data.frame column?

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