Filtering pandas dataframe with multiple Boolean columns

So, the easiest way to do this:

students = [ ('jack1', 'Apples1' , 341) ,
             ('Riti1', 'Mangos1'  , 311) ,
             ('Aadi1', 'Grapes1' , 301) ,
             ('Sonia1', 'Apples1', 321) ,
             ('Lucy1', 'Mangos1'  , 331) ,
             ('Mike1', 'Apples1' , 351),
              ('Mik', 'Apples1' , np.nan)
              ]
#Create a DataFrame object
df = pd.DataFrame(students, columns = ['Name1' , 'Product1', 'Sale1']) 
print(df)


    Name1 Product1  Sale1
0   jack1  Apples1    341
1   Riti1  Mangos1    311
2   Aadi1  Grapes1    301
3  Sonia1  Apples1    321
4   Lucy1  Mangos1    331
5   Mike1  Apples1    351
6     Mik  Apples1    NaN

# Select rows in above DataFrame for which ‘Product’ column contains the value ‘Apples’,
subset = df[df['Product1'] == 'Apples1']
print(subset)

 Name1 Product1  Sale1
0   jack1  Apples1    341
3  Sonia1  Apples1    321
5   Mike1  Apples1    351
6     Mik  Apples1    NA

# Select rows in above DataFrame for which ‘Product’ column contains the value ‘Apples’, AND notnull value in Sale

subsetx= df[(df['Product1'] == "Apples1")  & (df['Sale1'].notnull())]
print(subsetx)
    Name1   Product1    Sale1
0   jack1   Apples1      341
3   Sonia1  Apples1      321
5   Mike1   Apples1      351

# Select rows in above DataFrame for which ‘Product’ column contains the value ‘Apples’, AND Sale = 351

subsetx= df[(df['Product1'] == "Apples1")  & (df['Sale1'] == 351)]
print(subsetx)

   Name1 Product1  Sale1
5  Mike1  Apples1    351

# Another example
subsetData = df[df['Product1'].isin(['Mangos1', 'Grapes1']) ]
print(subsetData)

Name1 Product1  Sale1
1  Riti1  Mangos1    311
2  Aadi1  Grapes1    301
4  Lucy1  Mangos1    331

Here is the source of this code: https://thispointer.com/python-pandas-select-rows-in-dataframe-by-conditions-on-multiple-columns/
I added minor changes to it.

Hooray! More options!

np.where

df[np.where(df.C | df.D, True, False)]

             A   B      C      D
0     John Doe  45   True  False
2  Alan Holmes  55  False   True
3   Eric Lamar  29   True   True  

---

pd.Series.where on df.index

df.loc[df.index.where(df.C | df.D).dropna()]

               A   B      C      D
0.0     John Doe  45   True  False
2.0  Alan Holmes  55  False   True
3.0   Eric Lamar  29   True   True

---

df.select_dtypes

df[df.select_dtypes([bool]).any(1)]   

             A   B      C      D
0     John Doe  45   True  False
2  Alan Holmes  55  False   True
3   Eric Lamar  29   True   True

---

Abusing np.select

df.iloc[np.select([df.C | df.D], [df.index])].drop_duplicates()

             A   B      C      D
0     John Doe  45   True  False
2  Alan Holmes  55  False   True
3   Eric Lamar  29   True   True

In [82]: d
Out[82]:
             A   B      C      D
0     John Doe  45   True  False
1   Jane Smith  32  False  False
2  Alan Holmes  55  False   True
3   Eric Lamar  29   True   True

Solution 1:

In [83]: d.loc[d.C | d.D]
Out[83]:
             A   B      C      D
0     John Doe  45   True  False
2  Alan Holmes  55  False   True
3   Eric Lamar  29   True   True

Solution 2:

In [94]: d[d[['C','D']].any(1)]
Out[94]:
             A   B      C      D
0     John Doe  45   True  False
2  Alan Holmes  55  False   True
3   Eric Lamar  29   True   True

Solution 3:

In [95]: d.query("C or D")
Out[95]:
             A   B      C      D
0     John Doe  45   True  False
2  Alan Holmes  55  False   True
3   Eric Lamar  29   True   True

PS If you change your solution to:

df[(df['C']==True) | (df['D']==True)]

it'll work too

Pandas docs - boolean indexing

---

why we should NOT use "PEP complaint" df["col_name"] is True instead of df["col_name"] == True?

In [11]: df = pd.DataFrame({"col":[True, True, True]})

In [12]: df
Out[12]:
    col
0  True
1  True
2  True

In [13]: df["col"] is True
Out[13]: False               # <----- oops, that's not exactly what we wanted

Or

d[d.eval('C or D')]

Out[1065]:
             A   B      C      D
0     John Doe  45   True  False
2  Alan Holmes  55  False   True
3   Eric Lamar  29   True   True