Pandas: How do I assign values based on multiple conditions for existing columns?

I would like to create a new column with a numerical value based on the following conditions:

a. if gender is male & pet1=pet2, points = 5

b. if gender is female & (pet1 is 'cat' or pet1='dog'), points = 5

c. all other combinations, points = 0

    gender    pet1      pet2
0   male      dog       dog
1   male      cat       cat
2   male      dog       cat
3   female    cat       squirrel
4   female    dog       dog
5   female    squirrel  cat
6   squirrel  dog       cat

I would like the end result to be as follows:

    gender    pet1      pet2      points
0   male      dog       dog       5
1   male      cat       cat       5
2   male      dog       cat       0
3   female    cat       squirrel  5
4   female    dog       dog       5
5   female    squirrel  cat       0
6   squirrel  dog       cat       0

How do I accomplish this?

Solution 1:

You can do this using np.where, the conditions use bitwise & and | for and and or with parentheses around the multiple conditions due to operator precedence. So where the condition is true 5 is returned and 0 otherwise:

In [29]:
df['points'] = np.where( ( (df['gender'] == 'male') & (df['pet1'] == df['pet2'] ) ) | ( (df['gender'] == 'female') & (df['pet1'].isin(['cat','dog'] ) ) ), 5, 0)
df

Out[29]:
     gender      pet1      pet2  points
0      male       dog       dog       5
1      male       cat       cat       5
2      male       dog       cat       0
3    female       cat  squirrel       5
4    female       dog       dog       5
5    female  squirrel       cat       0
6  squirrel       dog       cat       0

Solution 2:

numpy.select

This is a perfect case for np.select where we can create a column based on multiple conditions and it's a readable method when there are more conditions:

conditions = [
    df['gender'].eq('male') & df['pet1'].eq(df['pet2']),
    df['gender'].eq('female') & df['pet1'].isin(['cat', 'dog'])
]

choices = [5,5]

df['points'] = np.select(conditions, choices, default=0)

print(df)
     gender      pet1      pet2  points
0      male       dog       dog       5
1      male       cat       cat       5
2      male       dog       cat       0
3    female       cat  squirrel       5
4    female       dog       dog       5
5    female  squirrel       cat       0
6  squirrel       dog       cat       0

Solution 3:

using apply.

def f(x):
  if x['gender'] == 'male' and x['pet1'] == x['pet2']: return 5
  elif x['gender'] == 'female' and (x['pet1'] == 'cat' or x['pet1'] == 'dog'): return 5
  else: return 0

data['points'] = data.apply(f, axis=1)

Solution 4:

The apply method described by @RuggeroTurra takes a lot longer for 500k rows. I ended up using something like

df['result'] = ((df.a == 0) & (df.b != 1)).astype(int) * 2 + \
               ((df.a != 0) & (df.b != 1)).astype(int) * 3 + \
               ((df.a == 0) & (df.b == 1)).astype(int) * 4 + \
               ((df.a != 0) & (df.b == 1)).astype(int) * 5 

where the apply method took 25 seconds and this method above took about 18ms.

Solution 5:

You can also use the apply function. For example:

def myfunc(gender, pet1, pet2):
    if gender=='male' and pet1==pet2:
        myvalue=5
    elif gender=='female' and (pet1=='cat' or pet1=='dog'):
        myvalue=5
    else:
        myvalue=0
    return myvalue

And then using the apply function by setting axis=1

df['points'] = df.apply(lambda x: myfunc(x['gender'], x['pet1'], x['pet2']), axis=1)

We get:

     gender      pet1      pet2  points
0      male       dog       dog       5
1      male       cat       cat       5
2      male       dog       cat       0
3    female       cat  squirrel       5
4    female       dog       dog       5
5    female  squirrel       cat       0
6  squirrel       dog       cat       0