Using conditional to generate new column in pandas dataframe

Question

I have a pandas dataframe that looks like this:

   portion  used
0        1   1.0
1        2   0.3
2        3   0.0
3        4   0.8

I'd like to create a new column based on the used column, so that the df looks like this:

   portion  used    alert
0        1   1.0     Full
1        2   0.3  Partial
2        3   0.0    Empty
3        4   0.8  Partial

• Create a new alert column based on
• If used is 1.0, alert should be Full.
• If used is 0.0, alert should be Empty.
• Otherwise, alert should be Partial.

What's the best way to do that?

Answer 1

Use np.select() for >2 conditions

Given >2 conditions like OP's example, np.select() is much cleaner than nesting multiple levels of np.where() (and is just as fast).

• Either define the conditions/choices as two lists (paired element-wise) with an optional default value ("else" case):

  conditions = [
      df.used.eq(0),
      df.used.eq(1),
  ]
  choices = [
      'Empty',
      'Full',
  ]
  df['alert'] = np.select(conditions, choices, default='Partial')
  

• Or define the conditions/choices as a dictionary for maintainability (easier to keep them paired properly when making additions/revisions):

  conditions = {
      'Empty': df.used.eq(0),
      'Full': df.used.eq(1),
  }
  df['alert'] = np.select(conditions.values(), conditions.keys(), default='Partial')
  

---

np.select() is very fast

Timings with 5 conditions (full, high, medium, low, empty):

^{df = pd.DataFrame({'used': np.random.randint(10 + 1, size=10)}).div(10)}

Answer 2

You can define a function which returns your different states "Full", "Partial", "Empty", etc and then use df.apply to apply the function to each row. Note that you have to pass the keyword argument axis=1 to ensure that it applies the function to rows.

import pandas as pd

def alert(row):
  if row['used'] == 1.0:
    return 'Full'
  elif row['used'] == 0.0:
    return 'Empty'
  elif 0.0 < row['used'] < 1.0:
    return 'Partial'
  else:
    return 'Undefined'

df = pd.DataFrame(data={'portion':[1, 2, 3, 4], 'used':[1.0, 0.3, 0.0, 0.8]})

df['alert'] = df.apply(alert, axis=1)

#    portion  used    alert
# 0        1   1.0     Full
# 1        2   0.3  Partial
# 2        3   0.0    Empty
# 3        4   0.8  Partial

Answer 3

df['TaxStatus'] = np.where(df.Public == 1, True, np.where(df.Public == 2, False))

This would appear to work, except for the ValueError: either both or neither of x and y should be given

Answer 4

Can't comment so making a new answer: Improving on Ffisegydd's approach, you can use a dictionary and the dict.get() method to make the function to pass in to .apply() easier to manage:

import pandas as pd

def alert(c):
    mapping = {1.0: 'Full', 0.0: 'Empty'}
    return mapping.get(c['used'], 'Partial')

df = pd.DataFrame(data={'portion':[1, 2, 3, 4], 'used':[1.0, 0.3, 0.0, 0.8]})

df['alert'] = df.apply(alert, axis=1)

Depending on the use case, you might like to define the dict outside of the function definition as well.

Answer 5

Alternatively you could do:

import pandas as pd
import numpy as np
df = pd.DataFrame(data={'portion':np.arange(10000), 'used':np.random.rand(10000)})

%%timeit
df.loc[df['used'] == 1.0, 'alert'] = 'Full'
df.loc[df['used'] == 0.0, 'alert'] = 'Empty'
df.loc[(df['used'] >0.0) & (df['used'] < 1.0), 'alert'] = 'Partial'

Which gives the same output but runs about 100 times faster on 10000 rows:

100 loops, best of 3: 2.91 ms per loop

Then using apply:

%timeit df['alert'] = df.apply(alert, axis=1)

1 loops, best of 3: 287 ms per loop

I guess the choice depends on how big is your dataframe.

Answer 6

Use np.where, is usually fast

In [845]: df['alert'] = np.where(df.used == 1, 'Full', 
                                 np.where(df.used == 0, 'Empty', 'Partial'))

In [846]: df
Out[846]:
   portion  used    alert
0        1   1.0     Full
1        2   0.3  Partial
2        3   0.0    Empty
3        4   0.8  Partial

---

_Timings

In [848]: df.shape
Out[848]: (100000, 3)

In [849]: %timeit df['alert'] = np.where(df.used == 1, 'Full', np.where(df.used == 0, 'Empty', 'Partial'))
100 loops, best of 3: 6.17 ms per loop

In [850]: %%timeit
     ...: df.loc[df['used'] == 1.0, 'alert'] = 'Full'
     ...: df.loc[df['used'] == 0.0, 'alert'] = 'Empty'
     ...: df.loc[(df['used'] >0.0) & (df['used'] < 1.0), 'alert'] = 'Partial'
     ...:
10 loops, best of 3: 21.9 ms per loop

In [851]: %timeit df['alert'] = df.apply(alert, axis=1)
1 loop, best of 3: 2.79 s per loop