How to unwrap rows to columns in r

Question

Is there an efficient way to unwrap rows in a r data frame to columns? This is an re-occurring problem I run into, when my data I get from an SQL script should be unwrapped into several columns. For example for timeseries forecasting where I don't have an rnn and I instead aim to use standard neural nets. Instead of having recurring nets I plan to flatten the data beforehand so that the net receives the row for t-1, t-2, t-3 etc. See my fancy paint-job below

Basically, for each row I want to concatenate n previous rows on the right, where n depends on how many previous times steps we want to use to predict values in the current row.

Mostly I'm looking for a smart and efficient way of doing it, preferably with existing libraries/functions in r. I can program it in several languages, but aim to find an r solution. I've done this in Java before (which was quite fast) and in r (which took ~forever=1 hour).

Currently I have a bit over 3000 rows and 10 columns. If for example I want to use 15 previous time steps we get 10+15*10 columns. Whether 15 is a good choice I don't know, and therefore I need to be able to quickly test e.g. n= 5, 10, 15, 20, 25, 50.

Edit

To be honest, I'm a beginner when it comes to r, therefore I ask for help instead of yet again coding my custom function for this.

dput gives:

structure(list(Date = structure(c(10L, 9L, 8L, 7L, 6L, 5L, 4L, 
3L, 2L, 1L), .Label = c("6/10/2016", "6/13/2016", "6/14/2016", 
"6/15/2016", "6/16/2016", "6/17/2016", "6/20/2016", "6/21/2016", 
"6/22/2016", "6/23/2016"), class = "factor"), Bid = c(5.04, 4.97, 
4.96, 4.93, 4.84, 5.09, 5.05, 4.96, 5.08, 5), Ask = c(5.04, 4.97, 
4.96, 4.94, 4.84, 5.09, 5.06, 4.97, 5.08, 5.01), Opening.price = c(4.98, 
4.97, 4.95, 4.94, 4.92, 5.01, 5.01, 5.01, 4.95, 5.05), High.price = c(5.07, 
4.98, 4.97, 4.99, 4.93, 5.14, 5.06, 5.1, 5.13, 5.09), Low.price = c(4.94, 
4.91, 4.89, 4.92, 4.81, 5.01, 4.94, 4.94, 4.89, 4.97), Closing.price = c(5.04, 
4.97, 4.95, 4.94, 4.86, 5.08, 5.05, 4.94, 5.06, 4.98), Average.price = c(5.02, 
4.96, 4.94, 4.94, 4.87, 5.08, 5.01, 5, 5.01, 5.01), Total.volume = c(18997216L, 
17969939L, 21430529L, 20725035L, 66884495L, 32994371L, 24600829L, 
24439514L, 26540825L, 24756699L), Turnover = c(95382241.29, 89106913.2, 
105823382.96, 102379207.58, 325592595.95, 167697936.93, 123243137.11, 
122189815.88, 133063486.77, 124080799.95), Trades = c(9220L, 
9317L, 10075L, 10230L, 16446L, 13544L, 11888L, 10923L, 11981L, 
9696L)), .Names = c("Date", "Bid", "Ask", "Opening.price", "High.price", 
"Low.price", "Closing.price", "Average.price", "Total.volume", 
"Turnover", "Trades"), class = "data.frame", row.names = c(NA, 
-10L))

The result when n = 2 (append 2 previous timesteps on the right):

structure(list(Date = structure(c(8L, 7L, 6L, 5L, 4L, 3L, 2L, 
1L), .Label = c("6/14/2016", "6/15/2016", "6/16/2016", "6/17/2016", 
"6/20/2016", "6/21/2016", "6/22/2016", "6/23/2016"), class = "factor"), 
Bid = c(5.04, 4.97, 4.96, 4.93, 4.84, 5.09, 5.05, 4.96), 
Ask = c(5.04, 4.97, 4.96, 4.94, 4.84, 5.09, 5.06, 4.97), 
Opening.price = c(4.98, 4.97, 4.95, 4.94, 4.92, 5.01, 5.01, 
5.01), High.price = c(5.07, 4.98, 4.97, 4.99, 4.93, 5.14, 
5.06, 5.1), Low.price = c(4.94, 4.91, 4.89, 4.92, 4.81, 5.01, 
4.94, 4.94), Closing.price = c(5.04, 4.97, 4.95, 4.94, 4.86, 
5.08, 5.05, 4.94), Average.price = c(5.02, 4.96, 4.94, 4.94, 
4.87, 5.08, 5.01, 5), Total.volume = c(18997216L, 17969939L, 
21430529L, 20725035L, 66884495L, 32994371L, 24600829L, 24439514L
), Turnover = c(95382241.29, 89106913.2, 105823382.96, 102379207.58, 
325592595.95, 167697936.93, 123243137.11, 122189815.88), 
Trades = c(9220L, 9317L, 10075L, 10230L, 16446L, 13544L, 
11888L, 10923L), X1_Bid = c(4.97, 4.96, 4.93, 4.84, 5.09, 
5.05, 4.96, 5.08), X1_Ask = c(4.97, 4.96, 4.94, 4.84, 5.09, 
5.06, 4.97, 5.08), X1_Opening.price = c(4.97, 4.95, 4.94, 
4.92, 5.01, 5.01, 5.01, 4.95), X1_High.price = c(4.98, 4.97, 
4.99, 4.93, 5.14, 5.06, 5.1, 5.13), X1_Low.price = c(4.91, 
4.89, 4.92, 4.81, 5.01, 4.94, 4.94, 4.89), X1_Closing.price = c(4.97, 
4.95, 4.94, 4.86, 5.08, 5.05, 4.94, 5.06), X1_Average.price = c(4.96, 
4.94, 4.94, 4.87, 5.08, 5.01, 5, 5.01), X1_Total.volume = c(17969939L, 
21430529L, 20725035L, 66884495L, 32994371L, 24600829L, 24439514L, 
26540825L), X1_Turnover = c(89106913.2, 105823382.96, 102379207.58, 
325592595.95, 167697936.93, 123243137.11, 122189815.88, 133063486.77
), X1_Trades = c(9317L, 10075L, 10230L, 16446L, 13544L, 11888L, 
10923L, 11981L), X2_Bid = c(4.96, 4.93, 4.84, 5.09, 5.05, 
4.96, 5.08, 5), X2_Ask = c(4.96, 4.94, 4.84, 5.09, 5.06, 
4.97, 5.08, 5.01), X2_Opening.price = c(4.95, 4.94, 4.92, 
5.01, 5.01, 5.01, 4.95, 5.05), X2_High.price = c(4.97, 4.99, 
4.93, 5.14, 5.06, 5.1, 5.13, 5.09), X2_Low.price = c(4.89, 
4.92, 4.81, 5.01, 4.94, 4.94, 4.89, 4.97), X2_Closing.price = c(4.95, 
4.94, 4.86, 5.08, 5.05, 4.94, 5.06, 4.98), X2_Average.price = c(4.94, 
4.94, 4.87, 5.08, 5.01, 5, 5.01, 5.01), X2_Total.volume = c(21430529L, 
20725035L, 66884495L, 32994371L, 24600829L, 24439514L, 26540825L, 
24756699L), X2_Turnover = c(105823382.96, 102379207.58, 325592595.95, 
167697936.93, 123243137.11, 122189815.88, 133063486.77, 124080799.95
), X2_Trades = c(10075L, 10230L, 16446L, 13544L, 11888L, 
10923L, 11981L, 9696L)), .Names = c("Date", "Bid", "Ask", 
"Opening.price", "High.price", "Low.price", "Closing.price", 
"Average.price", "Total.volume", "Turnover", "Trades", "X1_Bid", 
"X1_Ask", "X1_Opening.price", "X1_High.price", "X1_Low.price", 
"X1_Closing.price", "X1_Average.price", "X1_Total.volume",  "X1_Turnover", 
"X1_Trades", "X2_Bid", "X2_Ask", "X2_Opening.price", "X2_High.price", 
"X2_Low.price", "X2_Closing.price", "X2_Average.price", "X2_Total.volume", 
"X2_Turnover", "X2_Trades"), class = "data.frame", row.names = c(NA, 
-8L))

edit 2

Here's somebody who's haven a question about rnns and how they are different from classic nets when the previous time points are input directly. rnn vs classic net Part 1 of the question explains how I want to get x(t-1), x(t-2)... on the right of the data I want to predict, for each row.

Answer 1

# I put your dataframe into x. 
# Then I extract just the numeric columns into x1,
# because operations on matrices tend to be faster 
# than operations on dataframes.
    x1<-as.matrix(x[,-1])
# Now I increase the size to a 3000 row matrix to match your comment
    x2<-x1[sample(1:10,3000,replace=TRUE),]
# record some sizes:
    N<-50; n<-nrow(x2); J<-1:(n-(N-1))
# Create the new numeric matrix:
    z<-x2[J,]; for (i in 1:(N-1)){z<-cbind(z,x2[i+J,])}
#This took an imperceptible length of time (< 1 second) on my MacBook Pro
    dim(z)
    [1] 2951  500
# Note that looping is inelegant at times but here it's just fine.
# appending columns to a matrix is fast because it goes 
# in the order in which matrices are stored.

# You could run your test cases with z[,1:50], z[,1:100] ... 
# to test the right number of past entries