Split Data Frame into Rows of Fixed Size

Split Data Frame into Rows of Fixed Size

I don't understand why a plyr solution is needed. split works perfectly well and even hadley himself didn't suggest a plyr/reshape2 solution when he looked at the earlier question:

split(dfrm, (0:nrow(dfrm) %/% 300)  # modulo division

Does produce a warning but since you were expecting a non-evenly divisible result you should ignore it.

How do I split a data frame into groups of a fixed size?

This will give you a list of DataFrames:

lst = [df.iloc[i:i+group_size] for i in range(0,len(df)-group_size+1,group_size)]

It just uses built-in indexing, so it should be pretty fast. The fidgeting with the stop index takes care of discarding the last frame if it's too small - you can also break it down with

lst = [df.iloc[i:i+group_size] for i in range(0,len(df),group_size)]
if len(lst[-1]) < group_size:
   lst.pop()

Having trouble splitting a dataframe into fixed chunks (per row)

According to np.array_split documentation, the second argument indices_or_sections specifies chunks boundaries rather than chunks sizes. I.e., if we pass an array with a first axis of length N and a list fracs with K elements, the resulting chunks will correspond to indexes [0, fracs[0]), [fracs[0], fracs[1]), ..., [fracs[K-1], N). So, if two consecutive elements of fracs are equal, this will result in a chunk of size 0.

The minimal modification of your code to achieve the expected result is to call np.cumsum on the resulting split_frac variable:

def dataframe_splitting(df:pd.DataFrame, fracs:list):
    split_frac = []
    print('Size of the dataframe:', df.shape)
    print('fracs:', fracs)
    for i in fracs:
        x = int(i*len(df))
        split_frac.append(x)
    chunks = np.array_split(df, np.cumsum(split_frac))  # note the cumsum here
    for x in chunks:
        print(x.shape)
    return chunks

Split a large pandas dataframe

Use np.array_split:

Docstring:
Split an array into multiple sub-arrays.

Please refer to the ``split`` documentation.  The only difference
between these functions is that ``array_split`` allows
`indices_or_sections` to be an integer that does *not* equally
divide the axis.

In [1]: import pandas as pd

In [2]: df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar',
   ...:                           'foo', 'bar', 'foo', 'foo'],
   ...:                    'B' : ['one', 'one', 'two', 'three',
   ...:                           'two', 'two', 'one', 'three'],
   ...:                    'C' : randn(8), 'D' : randn(8)})

In [3]: print df
     A      B         C         D
0  foo    one -0.174067 -0.608579
1  bar    one -0.860386 -1.210518
2  foo    two  0.614102  1.689837
3  bar  three -0.284792 -1.071160
4  foo    two  0.843610  0.803712
5  bar    two -1.514722  0.870861
6  foo    one  0.131529 -0.968151
7  foo  three -1.002946 -0.257468

In [4]: import numpy as np
In [5]: np.array_split(df, 3)
Out[5]: 
[     A    B         C         D
0  foo  one -0.174067 -0.608579
1  bar  one -0.860386 -1.210518
2  foo  two  0.614102  1.689837,
      A      B         C         D
3  bar  three -0.284792 -1.071160
4  foo    two  0.843610  0.803712
5  bar    two -1.514722  0.870861,
      A      B         C         D
6  foo    one  0.131529 -0.968151
7  foo  three -1.002946 -0.257468]

How to split a data frame into sub data frames of unequal size in a single dataframe

To get the desired output as mentioned in the question, i did the following process

step-1:

split the data frame in to lists using the following command

> split(DF[, -5], DF[, 5])

step-2:

then the following code is used to add a string to list's names

> colnames <- c("ABC", "AA", "BB", "CC", "DD")

> for (i in seq_along(DF)){
   colnames(DF[[i]]) <- paste0(names(DF[i]), colnames) 
  }

step-3:

then the unequal sized lists can be combined in a single data frame using the qpcR library as follows:

> library(qpcR)  ## required for combining multiple dataframes from a lists

> sample2 <- do.call(qpcR:::cbind.na, DF) # combines multiple dataframes in a list by column vise irrespective of row sizes

How to split Data Frame into Rows of Fixed size and separate them with dplyr

Based on the discussions you can use filter straight away and there's no need to arrange first:

even <- df %>% filter(No %% 2 == 0)
odd  <- df %>% filter(No %% 2 == 1)

Split up a dataframe by number of rows

Make your own grouping variable.

d <- split(my_data_frame,rep(1:400,each=1000))

You should also consider the ddply function from the plyr package, or the group_by() function from dplyr.

edited for brevity, after Hadley's comments.

If you don't know how many rows are in the data frame, or if the data frame might be an unequal length of your desired chunk size, you can do

chunk <- 1000
n <- nrow(my_data_frame)
r  <- rep(1:ceiling(n/chunk),each=chunk)[1:n]
d <- split(my_data_frame,r)

You could also use

r <- ggplot2::cut_width(1:n,chunk,boundary=0)

For future readers, methods based on the dplyr and data.table packages will probably be (much) faster for doing group-wise operations on data frames, e.g. something like

(my_data_frame 
   %>% mutate(index=rep(1:ngrps,each=full_number)[seq(.data)])
   %>% group_by(index)
   %>% [mutate, summarise, do()] ...
)

There are also many answers here

Split dataframe by row number follow by every specific row number count to create multiple data frames

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.

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