Pandas every nth row

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

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, :]

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.

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.