For example I want to split a dataframe with 1,079,882 rows starting from 1,048,575 and follow by the next 1,048,575

I tried to do something like this

x<-split(df, rep(1:1048575, 1048575))
list2env(x,envir = .GlobalEnv)

but it does not work. Based on what I want to do I should get 2 dataframes. If my dataframe had 2,097,160 I should have three, First two with 1,048,575 rows and the third with 5 rows.

CodePudding user response：

I'll demonstrate on something a little smaller than 1e6 rows: mtcars.

Let's say you want the frame split into no more than 10 rows each. Since it has 32 rows, this means we should have three frames with 10 rows each and one frame with 2 rows.

We'll use the %/% integer (floor) division operator. It's important to note that we need the starting to start at 0 instead of R's default of 1, so we will subtract 1.

(seq_len(nrow(mtcars)) - 1) %/% 10
#  [1] 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3

From here, just split on that:

out <- split(mtcars, (seq_len(nrow(mtcars)) - 1) %/% 10)
sapply(out, nrow)
#  0  1  2  3 
# 10 10 10  2 

out
# $`0`
#                    mpg cyl  disp  hp drat    wt  qsec vs am gear carb
# Mazda RX4         21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4
# Mazda RX4 Wag     21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4
# Datsun 710        22.8   4 108.0  93 3.85 2.320 18.61  1  1    4    1
# Hornet 4 Drive    21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1
# Hornet Sportabout 18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2
# Valiant           18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1
# Duster 360        14.3   8 360.0 245 3.21 3.570 15.84  0  0    3    4
# Merc 240D         24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2
# Merc 230          22.8   4 140.8  95 3.92 3.150 22.90  1  0    4    2
# Merc 280          19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4
# $`1`
#                      mpg cyl  disp  hp drat    wt  qsec vs am gear carb
# Merc 280C           17.8   6 167.6 123 3.92 3.440 18.90  1  0    4    4
# Merc 450SE          16.4   8 275.8 180 3.07 4.070 17.40  0  0    3    3
# Merc 450SL          17.3   8 275.8 180 3.07 3.730 17.60  0  0    3    3
# Merc 450SLC         15.2   8 275.8 180 3.07 3.780 18.00  0  0    3    3
# Cadillac Fleetwood  10.4   8 472.0 205 2.93 5.250 17.98  0  0    3    4
# Lincoln Continental 10.4   8 460.0 215 3.00 5.424 17.82  0  0    3    4
# Chrysler Imperial   14.7   8 440.0 230 3.23 5.345 17.42  0  0    3    4
# Fiat 128            32.4   4  78.7  66 4.08 2.200 19.47  1  1    4    1
# Honda Civic         30.4   4  75.7  52 4.93 1.615 18.52  1  1    4    2
# Toyota Corolla      33.9   4  71.1  65 4.22 1.835 19.90  1  1    4    1
# $`2`
#                   mpg cyl  disp  hp drat    wt  qsec vs am gear carb
# Toyota Corona    21.5   4 120.1  97 3.70 2.465 20.01  1  0    3    1
# Dodge Challenger 15.5   8 318.0 150 2.76 3.520 16.87  0  0    3    2
# AMC Javelin      15.2   8 304.0 150 3.15 3.435 17.30  0  0    3    2
# Camaro Z28       13.3   8 350.0 245 3.73 3.840 15.41  0  0    3    4
# Pontiac Firebird 19.2   8 400.0 175 3.08 3.845 17.05  0  0    3    2
# Fiat X1-9        27.3   4  79.0  66 4.08 1.935 18.90  1  1    4    1
# Porsche 914-2    26.0   4 120.3  91 4.43 2.140 16.70  0  1    5    2
# Lotus Europa     30.4   4  95.1 113 3.77 1.513 16.90  1  1    5    2
# Ford Pantera L   15.8   8 351.0 264 4.22 3.170 14.50  0  1    5    4
# Ferrari Dino     19.7   6 145.0 175 3.62 2.770 15.50  0  1    5    6
# $`3`
#                mpg cyl disp  hp drat   wt qsec vs am gear carb
# Maserati Bora 15.0   8  301 335 3.54 3.57 14.6  0  1    5    8
# Volvo 142E    21.4   4  121 109 4.11 2.78 18.6  1  1    4    2

There are several advantages of this technique:

• it is memory-efficient in that the seq_len(.) only counts as high as your frame has rows; it does not try to do rep(highnumber, highnumber), which will usually exceed R's memory when you have 1e6 or so rows;
• it directly controls which row goes into which group
• if you prefer your names to be 1-based, then add one, as in (seq_len(.)-1) %/% 10 1; if you want the names to be something else, names(out) <- c(...) works, too

Another point: it is often much better in a sense to keep this in a list instead of transferring it to the global environment: once you become more comfortable with lapply and friends, it keeps your environment uncluttered, allows you to repeat one task for all frames in one motion (instead of copy/paste for each named variable), and is a very "canonical" approach to dealing with data (in R). See https://stackoverflow.com/a/24376207/3358227.

CodePudding user response：

You could do it this way:

1. Define the chunk of rows
2. define the lengths of rows
3. define the amount of chunks in your dataframe
4. split the dataframe by the defined amount

chunk <- 1048575
n <- nrow(df)
r  <- rep(1:ceiling(n/chunk),each=chunk)[1:n]
d <- split(df,r)
d