如何删除在特定列中的值为NaN的Pandas DataFrame行

I have this DataFrame and want only the records whose EPS column is not NaN:

>>> df
                 STK_ID  EPS  cash
STK_ID RPT_Date                   
601166 20111231  601166  NaN   NaN
600036 20111231  600036  NaN    12
600016 20111231  600016  4.3   NaN
601009 20111231  601009  NaN   NaN
601939 20111231  601939  2.5   NaN
000001 20111231  000001  NaN   NaN

…i.e. something like df.drop(....) to get this resulting dataframe:

                  STK_ID  EPS  cash
STK_ID RPT_Date                   
600016 20111231  600016  4.3   NaN
601939 20111231  601939  2.5   NaN

How do I do that?

Don’t drop, just take the rows where EPS is not NA:

df = df[df['EPS'].notna()]

This question is already resolved, but…

…also consider the solution suggested by Wouter in his original comment. The ability to handle missing data, including dropna(), is built into pandas explicitly. Aside from potentially improved performance over doing it manually, these functions also come with a variety of options which may be useful.

In [24]: df = pd.DataFrame(np.random.randn(10,3))

In [25]: df.iloc[::2,0] = np.nan; df.iloc[::4,1] = np.nan; df.iloc[::3,2] = np.nan;

In [26]: df
Out[26]:
          0         1         2
0       NaN       NaN       NaN
1  2.677677 -1.466923 -0.750366
2       NaN  0.798002 -0.906038
3  0.672201  0.964789       NaN
4       NaN       NaN  0.050742
5 -1.250970  0.030561 -2.678622
6       NaN  1.036043       NaN
7  0.049896 -0.308003  0.823295
8       NaN       NaN  0.637482
9 -0.310130  0.078891       NaN

In [27]: df.dropna()     #drop all rows that have any NaN values
Out[27]:
          0         1         2
1  2.677677 -1.466923 -0.750366
5 -1.250970  0.030561 -2.678622
7  0.049896 -0.308003  0.823295

In [28]: df.dropna(how='all')     #drop only if ALL columns are NaN
Out[28]:
          0         1         2
1  2.677677 -1.466923 -0.750366
2       NaN  0.798002 -0.906038
3  0.672201  0.964789       NaN
4       NaN       NaN  0.050742
5 -1.250970  0.030561 -2.678622
6       NaN  1.036043       NaN
7  0.049896 -0.308003  0.823295
8       NaN       NaN  0.637482
9 -0.310130  0.078891       NaN

In [29]: df.dropna(thresh=2)   #Drop row if it does not have at least two values that are **not** NaN
Out[29]:
          0         1         2
1  2.677677 -1.466923 -0.750366
2       NaN  0.798002 -0.906038
3  0.672201  0.964789       NaN
5 -1.250970  0.030561 -2.678622
7  0.049896 -0.308003  0.823295
9 -0.310130  0.078891       NaN

In [30]: df.dropna(subset=[1])   #Drop only if NaN in specific column (as asked in the question)
Out[30]:
          0         1         2
1  2.677677 -1.466923 -0.750366
2       NaN  0.798002 -0.906038
3  0.672201  0.964789       NaN
5 -1.250970  0.030561 -2.678622
6       NaN  1.036043       NaN
7  0.049896 -0.308003  0.823295
9 -0.310130  0.078891       NaN

There are also other options (See docs at http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.dropna.html), including dropping columns instead of rows.

Pretty handy!

I know this has already been answered, but just for the sake of a purely pandas solution to this specific question as opposed to the general description from Aman (which was wonderful) and in case anyone else happens upon this:

import pandas as pd
df = df[pd.notnull(df['EPS'])]

You can use this:

df.dropna(subset=['EPS'], how='all', inplace=True)

Simplest of all solutions:

filtered_df = df[df['EPS'].notnull()]

The above solution is way better than using np.isfinite()

You could use dataframe method notnull or inverse of isnull, or numpy.isnan:

In [332]: df[df.EPS.notnull()]
Out[332]:
   STK_ID  RPT_Date  STK_ID.1  EPS  cash
2  600016  20111231    600016  4.3   NaN
4  601939  20111231    601939  2.5   NaN


In [334]: df[~df.EPS.isnull()]
Out[334]:
   STK_ID  RPT_Date  STK_ID.1  EPS  cash
2  600016  20111231    600016  4.3   NaN
4  601939  20111231    601939  2.5   NaN


In [347]: df[~np.isnan(df.EPS)]
Out[347]:
   STK_ID  RPT_Date  STK_ID.1  EPS  cash
2  600016  20111231    600016  4.3   NaN
4  601939  20111231    601939  2.5   NaN

Simple and easy way

df.dropna(subset=['EPS'],inplace=True)

source: https://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.dropna.html

yet another solution which uses the fact that np.nan != np.nan:

In [149]: df.query("EPS == EPS")
Out[149]:
                 STK_ID  EPS  cash
STK_ID RPT_Date
600016 20111231  600016  4.3   NaN
601939 20111231  601939  2.5   NaN

Another version:

df[~df['EPS'].isna()]

In datasets having large number of columns its even better to see how many columns contain null values and how many don’t.

print("No. of columns containing null values")
print(len(df.columns[df.isna().any()]))

print("No. of columns not containing null values")
print(len(df.columns[df.notna().all()]))

print("Total no. of columns in the dataframe")
print(len(df.columns))

For example in my dataframe it contained 82 columns, of which 19 contained at least one null value.

Further you can also automatically remove cols and rows depending on which has more null values
Here is the code which does this intelligently:

df = df.drop(df.columns[df.isna().sum()>len(df.columns)],axis = 1)
df = df.dropna(axis = 0).reset_index(drop=True)

Note: Above code removes all of your null values. If you want null values, process them before.

It may be added at that ‘&’ can be used to add additional conditions e.g.

df = df[(df.EPS > 2.0) & (df.EPS <4.0)]

Notice that when evaluating the statements, pandas needs parenthesis.

For some reason none of the previously submitted answers worked for me. This basic solution did:

df = df[df.EPS >= 0]

Though of course that will drop rows with negative numbers, too. So if you want those it’s probably smart to add this after, too.

df = df[df.EPS <= 0]

One of the solution can be

df = df[df.isnull().sum(axis=1) <= Cutoff Value]

Another way can be

df= df.dropna(thresh=(df.shape[1] - Cutoff_value))

I hope these are useful.