Split Up a Dataframe by Number of Rows

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 pandas dataframe in two if it has more than 10 rows

This will return the split DataFrames if the condition is met, otherwise return the original and None (which you would then need to handle separately). Note that this assumes the splitting only has to happen one time per df and that the second part of the split (if it is longer than 10 rows (meaning that the original was longer than 20 rows)) is OK.

df_new1, df_new2 = df[:10, :], df[10:, :] if len(df) > 10 else df, None

Note you can also use df.head(10) and df.tail(len(df) - 10) to get the front and back according to your needs. You can also use various indexing approaches: you can just provide the first dimensions index if you want, such as df[:10] instead of df[:10, :] (though I like to code explicitly about the dimensions you are taking). You can can also use df.iloc and df.ix to index in similar ways.

Be careful about using df.loc however, since it is label-based and the input will never be interpreted as an integer position. .loc would only work "accidentally" in the case when you happen to have index labels that are integers starting at 0 with no gaps.

But you should also consider the various options that pandas provides for dumping the contents of the DataFrame into HTML and possibly also LaTeX to make better designed tables for the presentation (instead of just copying and pasting). Simply Googling how to convert the DataFrame to these formats turns up lots of tutorials and advice for exactly this application.

Split a data frame into six equal parts based on number of rows without knowing the number of rows - pandas

You can use np.array_split():

dfs = np.array_split(df, 6)

for index, df in enumerate(dfs):
    df.to_csv(f'df{index+1}.csv')

>>> print(dfs)

[   ID Job  Salary
 0   1   A     100
 1   2   B     200
 2   3   B      20,

    ID Job  Salary
 3   4   C     150
 4   5   A     500
 5   6   A     600,

    ID Job  Salary
 6   7   A     200
 7   8   B     150,

    ID Job  Salary
 8   9   C     110
 9  10   B     200,

     ID Job  Salary
 10  11   B     220
 11  12   A     150,

     ID Job  Salary
 12  13   C      20
 13  14   B      50]

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.

Pandas - splitting dataframe into equal rows and assign number in new column as case_id in increasing order from 1 and so on

You can try this:

import pandas as pd

n = 4 # number of rows in each chunk
data = {"id": [0,1,2,3,4,5,6,7],
        "col1": ["a", "b", "c", "d", "e", "f", "g", "h"],
        "col2": ["a", "b", "c", "d", "e", "f", "g", "h"]
        }
df = pd.DataFrame.from_dict(data)
length = len(df)
df["new_col_case_id"] = df["id"].apply(lambda x: int(x/n) + 1)
df = df.set_index("id") #optional
print(df)

output:

   col1 col2  new_col_case_id
id                           
0     a    a                1
1     b    b                1
2     c    c                1
3     d    d                1
4     e    e                2
5     f    f                2
6     g    g                2
7     h    h                2

Split pandas dataframe rows into multiple rows

IIUC, group the IDs by chunks of 2 using a list comprehension, then explode the two IDs/distance columns:

df['IDs'] = [[l[i:i+2] for i in range(0,len(l),2)] for l in df['IDs']]
df = df.explode(['IDs', 'distance'])

NB. this requires len(IDs) to be 2 times len(distance) for each row!

output:

                        IDs distance
2022-01-01 12:00:00  [A, B]        1
2022-01-01 12:00:01  [A, B]      1.1
2022-01-01 12:00:01  [A, C]      2.8
2022-01-01 12:00:02  [A, B]        1
2022-01-01 12:00:02  [A, D]        3
2022-01-01 12:00:02  [C, D]      0.5

Split up a dataframe by number of NAs in each row

Try this following useful comments. You can split() and use apply() to build a group:

#Code
new <- split(DF,apply(DF[,-1],1,function(x)sum(is.na(x))))

Output:

$`1`
  ID C1 C2 C3 C4 C5
1 aa 12 13 10 NA 12
4 jj 31 14 NA 41 11
5 ss NA 15 11 12 11

$`2`
  ID C1 C2 C3 C4 C5
2 ff 12 NA NA 23 13
3 ee 67 23 NA NA 21

A more practical way (Many thanks and credits to @RuiBarradas):

#Code2
new <- split(DF, rowSums(is.na(DF[-1])))

Same output.

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