Pandas every nth row

Question

Dataframe.resample() works only with timeseries data. I cannot find a way of getting every nth row from non-timeseries data. What is the best method?

Asked By: mikael

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Source

Answer 1

I’d use iloc, which takes a row/column slice, both based on integer position and following normal python syntax. If you want every 5th row:

df.iloc[::5, :]

Answered By: chrisb

Answer 2

Though @chrisb’s accepted answer does answer the question, I would like to add to it the following.

A simple method I use to get the nth data or drop the nth row is the following:

df1 = df[df.index % 3 != 0]  # Excludes every 3rd row starting from 0
df2 = df[df.index % 3 == 0]  # Selects every 3rd raw starting from 0

This arithmetic based sampling has the ability to enable even more complex row-selections.

This assumes, of course, that you have an index column of ordered, consecutive, integers starting at 0.

Answered By: metastableB

Answer 3

I had a similar requirement, but I wanted the n’th item in a particular group. This is how I solved it.

groups = data.groupby(['group_key'])
selection = groups['index_col'].apply(lambda x: x % 3 == 0)
subset = data[selection]

Answered By: Steztric

Answer 4

There is an even simpler solution to the accepted answer that involves directly invoking df.__getitem__.

df = pd.DataFrame('x', index=range(5), columns=list('abc'))
df

   a  b  c
0  x  x  x
1  x  x  x
2  x  x  x
3  x  x  x
4  x  x  x

For example, to get every 2 rows, you can do

df[::2]

   a  b  c
0  x  x  x
2  x  x  x
4  x  x  x

There’s also GroupBy.first/GroupBy.head, you group on the index:

df.index // 2
# Int64Index([0, 0, 1, 1, 2], dtype='int64')

df.groupby(df.index // 2).first()
# Alternatively,
# df.groupby(df.index // 2).head(1)

   a  b  c
0  x  x  x
1  x  x  x
2  x  x  x

The index is floor-divved by the stride (2, in this case). If the index is non-numeric, instead do

# df.groupby(np.arange(len(df)) // 2).first()
df.groupby(pd.RangeIndex(len(df)) // 2).first()

   a  b  c
0  x  x  x
1  x  x  x
2  x  x  x

Answered By: cs95

Answer 5

A solution I came up with when using the index was not viable ( possibly the multi-Gig .csv was too large, or I missed some technique that would allow me to reindex without crashing ).
Walk through one row at a time and add the nth row to a new dataframe.

import pandas as pd
from csv import DictReader

def make_downsampled_df(filename, interval):    
    with open(filename, 'r') as read_obj:
        csv_dict_reader = DictReader(read_obj)
        column_names = csv_dict_reader.fieldnames
        df = pd.DataFrame(columns=column_names)
    
        for index, row in enumerate(csv_dict_reader):
            if index % interval == 0:
               print(str(row))
               df = df.append(row, ignore_index=True)

    return df

Answered By: Kyle

Answer 6

df.drop(labels=df[df.index % 3 != 0].index, axis=0) #  every 3rd row (mod 3)

Answered By: bitbang

Answer 7

Adding reset_index() to metastableB’s answer allows you to only need to assume that the rows are ordered and consecutive.

df1 = df[df.reset_index().index % 3 != 0]  # Excludes every 3rd row starting from 0
df2 = df[df.reset_index().index % 3 == 0]  # Selects every 3rd row starting from 0

df.reset_index().index will create an index that starts at 0 and increments by 1, allowing you to use the modulo easily.

Answered By: evilpegasus

Pandas every nth row

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