Groupby class and count missing values in features

Question

I have a problem and I cannot find any solution in the web or documentation, even if I think that it is very trivial.

What do I want to do?

I have a dataframe like this

CLASS FEATURE1 FEATURE2 FEATURE3
  X      A       NaN      NaN
  X     NaN       A       NaN
  B      A        A        A

I want to group by the label(CLASS) and display the number of NaN-Values that are counted in every feature so that it looks like this. The purpose of this is to get a general idea how missing values are distributed over the different classes.

CLASS FEATURE1 FEATURE2 FEATURE3
  X      1        1        2
  B      0        0        0

I know how to recieve the amount of nonnull-Values – df.groupby['CLASS'].count()

Is there something similar for the NaN-Values?

I tried to subtract the count() from the size() but it returned an unformatted output filled with the value NaN

Asked By: codlix

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Source

Answer 1

Compute a mask with isna, then group and find the sum:

df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum().reset_index()

  CLASS  FEATURE1  FEATURE2  FEATURE3
0     X       1.0       1.0       2.0
1     B       0.0       0.0       0.0

Another option is to subtract the size from the count using rsub along the 0^th axis for index aligned subtraction:

df.groupby('CLASS').count().rsub(df.groupby('CLASS').size(), axis=0)

Or,

g = df.groupby('CLASS')
g.count().rsub(g.size(), axis=0)

       FEATURE1  FEATURE2  FEATURE3
CLASS                              
B             0         0         0
X             1         1         2

There are quite a few good answers, so here are some timeits for your perusal:

df_ = df
df = pd.concat([df_] * 10000)

%timeit df.drop('CLASS', 1).isna().groupby(df.CLASS, sort=False).sum()
%timeit df.set_index('CLASS').isna().sum(level=0)    
%%timeit
g = df.groupby('CLASS')
g.count().rsub(g.size(), axis=0)

11.8 ms ± 108 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
9.47 ms ± 379 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
6.54 ms ± 81.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Actual performance depends on your data and setup, so your mileage may vary.

Answered By: cs95

Answer 2

Update due to Future warning:

FutureWarning: Using the level
keyword in DataFrame and Series aggregations is deprecated and will be
removed in a future version. Use groupby instead. df.sum(level=1)
should use df.groupby(level=1).sum().
df.set_index(‘CLASS’).isna().sum(level=0)

df.set_index('CLASS').isna().groupby(level=0).sum()

You can use set_index and sum:

# Will be deprecated soon.. do not use. You should use above statement instead.
df.set_index('CLASS').isna().sum(level=0)

Output:

       FEATURE1  FEATURE2  FEATURE3
CLASS                              
X           1.0       1.0       2.0
B           0.0       0.0       0.0

Answered By: Scott Boston

Answer 3

Using the diff between count and size

g=df.groupby('CLASS')

-g.count().sub(g.size(),0)

          FEATURE1  FEATURE2  FEATURE3
CLASS                              
B             0         0         0
X             1         1         2

And we can transform this question to the more generic question how to count how many NaN in dataframe with for loop

pd.DataFrame({x: y.isna().sum()for x , y in g }).T.drop('CLASS',1)
Out[468]: 
   FEATURE1  FEATURE2  FEATURE3
B         0         0         0
X         1         1         2

Answered By: BENY

Answer 4

A solution without using groupby, could be to use pivot_table and a custom aggregation function.
This solution may communicate the intent more clearly (at least to me) than the other solutions.

Generate the dataframe:

import pandas as pd
import numpy as np
df = pd.DataFrame({"CLASS":["X","X","B"],
                    "FEATURE1":["A", np.nan, "A",],
                    "FEATURE2":[np.nan,"A",  "A",],
                    "FEATURE3":[np.nan,np.nan,  "A",]
                  }
                 )

Generate the pivot table:

df.pivot_table(index="CLASS", 
               values=["FEATURE1","FEATURE2","FEATURE3"], 
               aggfunc= lambda x: x.isna().sum())
Out [2]:
       FEATURE1  FEATURE2  FEATURE3
CLASS                              
B             0         0         0
X             1         1         2

Performance:

Moreover, when looking at performance, this solution seems to be the fastest.

Answered By: Adrien Pacifico

Answer 5

Another solution (mostly for fun):

df.assign(
    **{col: df[col].isna() for col in df.columns if col not in "CLASS"},
).groupby("CLASS").sum()

Answered By: Adrien Pacifico

Groupby class and count missing values in features

Question:

Answers:

Performance: