Using R lag for Julia - julia

Question

I am attempting to create a lag +1 forward for a particular column in my data frame.
My data is like this:
julia> head(df)
6×9 DataFrames.DataFrame. Omitted printing of 1 columns
│ Row │ Date │ Open │ High │ Low │ Close │ Adj Close │ Volume │ Close_200sma │
├─────┼────────────┼─────────┼─────────┼─────────┼─────────┼───────────┼─────────┼──────────────┤
│ 1 │ 1993-02-02 │ 43.9687 │ 43.9687 │ 43.75 │ 43.9375 │ 27.6073 │ 1003200 │ NaN │
│ 2 │ 1993-02-03 │ 43.9687 │ 44.25 │ 43.9687 │ 44.25 │ 27.8036 │ 480500 │ NaN │
│ 3 │ 1993-02-04 │ 44.2187 │ 44.375 │ 44.125 │ 44.3437 │ 27.8625 │ 201300 │ NaN │
│ 4 │ 1993-02-05 │ 44.4062 │ 44.8437 │ 44.375 │ 44.8125 │ 28.1571 │ 529400 │ NaN │
│ 5 │ 1993-02-08 │ 44.9687 │ 45.0937 │ 44.4687 │ 45.0 │ 28.2749 │ 531500 │ NaN │
│ 6 │ 1993-02-09 │ 44.9687 │ 45.0625 │ 44.7187 │ 44.9687 │ 28.2552 │ 492100 │ NaN
│
So this is my attempt at lagging forward, in R I may rep NA, 1 and then append this to the front of the subsetted data. Here is my Julia:
# Lag data +1 forward
lag = df[1:nrow(df)-1,[:Long]] # shorten vector by 1 (remove last element)
v = Float64[]
v = vec(convert(Array, lag)) # convert df column to vector
z = fill(NaN, 1) # rep NaN, 1 time (add this to front) to push all forward +1
lags = Float64[]
lags= vec[z; [v]] # join both arrays z=NA first , make vector same nrow(df)
When I join the NaN and my array I have a length(lags) of 2.
The data is split in two:
julia> length(lags[2])
6255
I see the longer length when access the second portion.
If I join the other way, NaN at end, numbers first. I obtain correct length:
# try joining other way
lags_flip = [v; [z]]
julia> length(lags_flip)
6256
I can also add this back to my data frame: (Nan at bottom, i want at front)
# add back to data frame
df[:add] = lags_flip
1
1
1
1
1
1
1
1
[NaN]
My question is when joining the Nan and my data like this:
lags_flip = [v; [z]]
I obtain correct length, when I do it the other way:
Nan first:
lags= [z; [v]]
Then it doesnt appear correct.
How can I offset by data +1 forward, placing a Nan in front and adding back to my df? I feel im close but missing something

Answer 1

EDIT:
A a second thought - probably messing with length of column in a DataFrame is not the best thing to do and I assume you want a new column anyway. In this case this could be a basic approach:
df[:LagLong] = [missing; df[1:end-1,:Long]]
or if you want NaN (but probably you want missing as explained below):
df[:LagLong] = [NaN; df[1:end-1,:Long]]
PREVIOUS REPLY:
You can do it in place:
julia> x = [1.0,2.0,3.0]
3-element Array{Float64,1}:
1.0
2.0
3.0
julia> pop!(unshift!(x, NaN))
3.0
julia> x
3-element Array{Float64,1}:
NaN
1.0
2.0
Replace x in pop!(unshift!(x, NaN)) by an appropriate column selector like df[:Long].
Note, however, that NaN is not NA in R. In Julia NA is missing. And now there is a branch:
if your column allows missing values (it will show Union{Missing, [Something]} in showcols) then you do the same as above pop!(unshift!(df[:Long], missing)).
if it does not allow missings you have two options. First is to first call allowmissing!(df, :Long) to allow missings and go forward as described above. The other is similar to the approach you have proposed: df[:Long] = [missing; df[1:end-1, :Long]].