Pandas: Remove NaN only at beginning and end of dataframe

Question:

I’ve got a pandas DataFrame that looks like this:

       sum
1948   NaN
1949   NaN
1950     5
1951     3
1952   NaN
1953     4
1954     8
1955   NaN

and I would like to cut off the NaNs at the beginning and at the end ONLY (i.e. only the values incl. NaN from 1950 to 1954 should remain).
I already tried .isnull() and dropna(), but somehow I couldn’t find a proper solution.
Can anyone help?

Asked By: user3017048

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Source

Answers:

Here is one way to do it.

import pandas as pd

# your data
# ==============================
df

      sum
1948  NaN
1949  NaN
1950    5
1951    3
1952  NaN
1953    4
1954    8
1955  NaN

# processing
# ===============================
idx = df.fillna(method='ffill').dropna().index
res_idx = df.loc[idx].fillna(method='bfill').dropna().index
df.loc[res_idx]

      sum
1950    5
1951    3
1952  NaN
1953    4
1954    8
Answered By: Jianxun Li

Use the built in first_valid_index and last_valid_index they are designed specifically for this and slice your df:

In [5]:

first_idx = df.first_valid_index()
last_idx = df.last_valid_index()
print(first_idx, last_idx)
df.loc[first_idx:last_idx]
1950 1954
Out[5]:
      sum
1950    5
1951    3
1952  NaN
1953    4
1954    8
Answered By: EdChum

Here is a an approach with Numpy:

import numpy as np

x    = np.logical_not(pd.isnull(df))
mask = np.logical_and(np.cumsum(x)!=0, np.cumsum(x[::-1])[::-1]!=0)

In [313]: df.loc[mask['sum'].tolist()]

Out[313]:
      sum
1950    5
1951    3
1952  NaN
1953    4
1954    8
Answered By: Colonel Beauvel

One-liner:

df.query('[email protected]().isna().any(axis=1)&[email protected]().isna().any(axis=1)')
Answered By: VovaM