Groupby class and count missing values in features

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

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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

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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.

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

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

Another solution (mostly for fun):

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