CODEWITHSUNDEEP
pythonpandasnumpy
Dinosaurius
Dinosaurius

Reputation: 8638

Filter out nan rows in a specific column

df =

Col1 Col2 Col3
1    nan  4
2    5    4
3    3    nan

Given the dataframe df, I want to obtain a new dataframe df2 that does not contain nan in the column Col2. This is the expected result: df2 =

Col1 Col2 Col3
2    5    4
3    3    nan

I know that it's possible to use pandas.isnull and dropna, however how to specify only particular column to which filtering should be applied?

Upvotes: 35

Views: 67581

Answers (5)

Grant Shannon
Grant Shannon

Reputation: 5075

The simple implementation below follows on from the above - but shows filtering out nan rows in a specific column - in place - and for large data frames count rows with nan by column name (before and after)

import pandas as pd
import numpy as np

df = pd.DataFrame([[1,np.nan,'A100'],[4,5,'A213'],[7,8,np.nan],[10,np.nan,'GA23']])
df.columns = ['areaCode','Distance','accountCode']

dataframe

areaCode    Distance    accountCode
1           NaN         A100
4           5.0         A213
7           8.0         NaN
10          NaN         GA23

Before: count rows with nan (for each column):

df.isnull().sum()

count by column:

areaCode       0
Distance       2
accountCode    1
dtype: int64

remove unwanted rows in-place:

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

After: count rows with nan (for each column):

df.isnull().sum()

count by column:

areaCode       0
Distance       0
accountCode    1
dtype: int64

dataframe:

areaCode    Distance    accountCode
4           5.0         A213
7           8.0         NaN

Upvotes: 0

Alex Schwab
Alex Schwab

Reputation: 83

If you want to count and graph the number of nan's before dropping your column(s)

import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt

cols = df.columns
nans = [df[col].isna().sum() for col in cols] 

sns.set(font_scale=1.1)
ax = sns.barplot(cols, nans, palette='hls', log=False)
ax.set(xlabel='Feature', ylabel='Number of NaNs', title='Number of NaNs per feature')
for p, uniq in zip(ax.patches, nans):
    height = p.get_height()
    ax.text(p.get_x()+p.get_width()/2.,
            height + 10,
            uniq,
            ha="center") 
ax.set_xticklabels(ax.get_xticklabels(),rotation=90)
plt.show()

Upvotes: 0

MaxU - stand with Ukraine
MaxU - stand with Ukraine

Reputation: 210972

you can use DataFrame.dropna() method:

In [202]: df.dropna(subset=['Col2'])
Out[202]:
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

or (in this case) less idiomatic Series.notnull():

In [204]: df.loc[df.Col2.notnull()]
Out[204]:
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

or using DataFrame.query() method:

In [205]: df.query("Col2 == Col2")
Out[205]:
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

numexpr solution:

In [241]: import numexpr as ne

In [242]: col = df.Col2

In [243]: df[ne.evaluate("col == col")]
Out[243]:
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

Upvotes: 57

piRSquared
piRSquared

Reputation: 294516

Using numpy's isnan to mask and construct a new dataframe

m = ~np.isnan(df.Col2.values)
pd.DataFrame(df.values[m], df.index[m], df.columns)

   Col1  Col2  Col3
1   2.0   5.0   4.0
2   3.0   3.0   NaN

Timing
Bigger Data 

np.random.seed([3,1415])
df = pd.DataFrame(np.random.choice([np.nan, 1], size=(10000, 10))).add_prefix('Col')

%%timeit
m = ~np.isnan(df.Col2.values)
pd.DataFrame(df.values[m], df.index[m], df.columns)
1000 loops, best of 3: 326 µs per loop

%timeit df.query("Col2 == Col2")
1000 loops, best of 3: 1.48 ms per loop

%timeit df.loc[df.Col2.notnull()]
1000 loops, best of 3: 417 µs per loop

%timeit df[~df['Col2'].isnull()]
1000 loops, best of 3: 385 µs per loop

%timeit df.dropna(subset=['Col2'])
1000 loops, best of 3: 913 µs per loop

Upvotes: 6

jezrael
jezrael

Reputation: 863531

Use dropna:

df = df.dropna(subset=['Col2'])
print (df)
  Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

Another solution - boolean indexing with notnull:

df = df[df['Col2'].notnull()]
print (df)
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

What is same as:

df = df[~df['Col2'].isnull()]
print (df)
   Col1  Col2  Col3
1     2   5.0   4.0
2     3   3.0   NaN

Upvotes: 15

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