Pandas conditional creation of a series/dataframe column

PythonPandasNumpyDataframe

Python Problem Overview


I have a dataframe along the lines of the below:

    Type       Set
1    A          Z
2    B          Z           
3    B          X
4    C          Y

I want to add another column to the dataframe (or generate a series) of the same length as the dataframe (equal number of records/rows) which sets a colour 'green' if Set == 'Z' and 'red' if Set equals anything else.

What's the best way to do this?

Python Solutions


Solution 1 - Python

If you only have two choices to select from:

df['color'] = np.where(df['Set']=='Z', 'green', 'red')

For example,

import pandas as pd
import numpy as np

df = pd.DataFrame({'Type':list('ABBC'), 'Set':list('ZZXY')})
df['color'] = np.where(df['Set']=='Z', 'green', 'red')
print(df)

yields

  Set Type  color
0   Z    A  green
1   Z    B  green
2   X    B    red
3   Y    C    red

If you have more than two conditions then use np.select. For example, if you want color to be

  • yellow when (df['Set'] == 'Z') & (df['Type'] == 'A')
  • otherwise blue when (df['Set'] == 'Z') & (df['Type'] == 'B')
  • otherwise purple when (df['Type'] == 'B')
  • otherwise black,

then use

df = pd.DataFrame({'Type':list('ABBC'), 'Set':list('ZZXY')})
conditions = [
    (df['Set'] == 'Z') & (df['Type'] == 'A'),
    (df['Set'] == 'Z') & (df['Type'] == 'B'),
    (df['Type'] == 'B')]
choices = ['yellow', 'blue', 'purple']
df['color'] = np.select(conditions, choices, default='black')
print(df)

which yields

  Set Type   color
0   Z    A  yellow
1   Z    B    blue
2   X    B  purple
3   Y    C   black

Solution 2 - Python

List comprehension is another way to create another column conditionally. If you are working with object dtypes in columns, like in your example, list comprehensions typically outperform most other methods.

Example list comprehension:

df['color'] = ['red' if x == 'Z' else 'green' for x in df['Set']]

%timeit tests:

import pandas as pd
import numpy as np

df = pd.DataFrame({'Type':list('ABBC'), 'Set':list('ZZXY')})
%timeit df['color'] = ['red' if x == 'Z' else 'green' for x in df['Set']]
%timeit df['color'] = np.where(df['Set']=='Z', 'green', 'red')
%timeit df['color'] = df.Set.map( lambda x: 'red' if x == 'Z' else 'green')

1000 loops, best of 3: 239 µs per loop
1000 loops, best of 3: 523 µs per loop
1000 loops, best of 3: 263 µs per loop

Solution 3 - Python

Another way in which this could be achieved is

df['color'] = df.Set.map( lambda x: 'red' if x == 'Z' else 'green')

Solution 4 - Python

The following is slower than the approaches timed here, but we can compute the extra column based on the contents of more than one column, and more than two values can be computed for the extra column.

Simple example using just the "Set" column:

def set_color(row):
    if row["Set"] == "Z":
        return "red"
    else:
        return "green"

df = df.assign(color=df.apply(set_color, axis=1))

print(df)

  Set Type  color
0   Z    A    red
1   Z    B    red
2   X    B  green
3   Y    C  green

Example with more colours and more columns taken into account:

def set_color(row):
    if row["Set"] == "Z":
        return "red"
    elif row["Type"] == "C":
        return "blue"
    else:
        return "green"

df = df.assign(color=df.apply(set_color, axis=1))

print(df)

  Set Type  color
0   Z    A    red
1   Z    B    red
2   X    B  green
3   Y    C   blue
Edit (21/06/2019): Using plydata

It is also possible to use plydata to do this kind of things (this seems even slower than using assign and apply, though).

from plydata import define, if_else

Simple if_else:

df = define(df, color=if_else('Set=="Z"', '"red"', '"green"'))

print(df)

  Set Type  color
0   Z    A    red
1   Z    B    red
2   X    B  green
3   Y    C  green

Nested if_else:

df = define(df, color=if_else(
    'Set=="Z"',
    '"red"',
    if_else('Type=="C"', '"green"', '"blue"')))

print(df)                            

  Set Type  color
0   Z    A    red
1   Z    B    red
2   X    B   blue
3   Y    C  green

Solution 5 - Python

Here's yet another way to skin this cat, using a dictionary to map new values onto the keys in the list:

def map_values(row, values_dict):
    return values_dict[row]

values_dict = {'A': 1, 'B': 2, 'C': 3, 'D': 4}

df = pd.DataFrame({'INDICATOR': ['A', 'B', 'C', 'D'], 'VALUE': [10, 9, 8, 7]})

df['NEW_VALUE'] = df['INDICATOR'].apply(map_values, args = (values_dict,))

What's it look like:

df
Out[2]: 
  INDICATOR  VALUE  NEW_VALUE
0         A     10          1
1         B      9          2
2         C      8          3
3         D      7          4

This approach can be very powerful when you have many ifelse-type statements to make (i.e. many unique values to replace).

And of course you could always do this:

df['NEW_VALUE'] = df['INDICATOR'].map(values_dict)

But that approach is more than three times as slow as the apply approach from above, on my machine.

And you could also do this, using dict.get:

df['NEW_VALUE'] = [values_dict.get(v, None) for v in df['INDICATOR']]

Solution 6 - Python

You can simply use the powerful .loc method and use one condition or several depending on your need (tested with pandas=1.0.5).

Code Summary:

df=pd.DataFrame(dict(Type='A B B C'.split(), Set='Z Z X Y'.split()))
df['Color'] = "red"
df.loc[(df['Set']=="Z"), 'Color'] = "green"

#practice!
df.loc[(df['Set']=="Z")&(df['Type']=="B")|(df['Type']=="C"), 'Color'] = "purple"

Explanation:

df=pd.DataFrame(dict(Type='A B B C'.split(), Set='Z Z X Y'.split()))

# df so far: 
  Type Set  
0    A   Z 
1    B   Z 
2    B   X 
3    C   Y

add a 'color' column and set all values to "red"

df['Color'] = "red"

Apply your single condition:

df.loc[(df['Set']=="Z"), 'Color'] = "green"


# df: 
  Type Set  Color
0    A   Z  green
1    B   Z  green
2    B   X    red
3    C   Y    red

or multiple conditions if you want:

df.loc[(df['Set']=="Z")&(df['Type']=="B")|(df['Type']=="C"), 'Color'] = "purple"

You can read on Pandas logical operators and conditional selection here: https://stackoverflow.com/questions/21415661/logical-operators-for-boolean-indexing-in-pandas

Solution 7 - Python

You can use pandas methods where and mask:

df['color'] = 'green'
df['color'] = df['color'].where(df['Set']=='Z', other='red')
# Replace values where the condition is False

or

df['color'] = 'red'
df['color'] = df['color'].mask(df['Set']=='Z', other='green')
# Replace values where the condition is True

Alternatively, you can use the method transform with a lambda function:

df['color'] = df['Set'].transform(lambda x: 'green' if x == 'Z' else 'red')

Output:

  Type Set  color
1    A   Z  green
2    B   Z  green
3    B   X    red
4    C   Y    red

Performance comparison from @chai:

import pandas as pd
import numpy as np
df = pd.DataFrame({'Type':list('ABBC')*1000000, 'Set':list('ZZXY')*1000000})
 
%timeit df['color1'] = 'red'; df['color1'].where(df['Set']=='Z','green')
%timeit df['color2'] = ['red' if x == 'Z' else 'green' for x in df['Set']]
%timeit df['color3'] = np.where(df['Set']=='Z', 'red', 'green')
%timeit df['color4'] = df.Set.map(lambda x: 'red' if x == 'Z' else 'green')

397 ms ± 101 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
976 ms ± 241 ms per loop
673 ms ± 139 ms per loop
796 ms ± 182 ms per loop

Solution 8 - Python

One liner with .apply() method is following:

df['color'] = df['Set'].apply(lambda set_: 'green' if set_=='Z' else 'red')

After that, df data frame looks like this:

>>> print(df)
  Type Set  color
0    A   Z  green
1    B   Z  green
2    B   X    red
3    C   Y    red

Solution 9 - Python

if you have only 2 choices, use np.where()

df = pd.DataFrame({'A':range(3)})
df['B'] = np.where(df.A>2, 'yes', 'no')

if you have over 2 choices, maybe apply() could work input

arr = pd.DataFrame({'A':list('abc'), 'B':range(3), 'C':range(3,6), 'D':range(6, 9)})

and arr is

  	A	B	C	D
0	a	0	3	6
1	b	1	4	7
2	c	2	5	8

if you want the column E tobe if arr.A =='a' then arr.B elif arr.A=='b' then arr.C elif arr.A == 'c' then arr.D else something_else

arr['E'] = arr.apply(lambda x: x['B'] if x['A']=='a' else(x['C'] if x['A']=='b' else(x['D'] if x['A']=='c' else 1234)), axis=1)

and finally the arr is

    A	B	C	D	E
0	a	0	3	6	0
1	b	1	4	7	4
2	c	2	5	8	8

Solution 10 - Python

If you're working with massive data, a memoized approach would be best:

# First create a dictionary of manually stored values
color_dict = {'Z':'red'}

# Second, build a dictionary of "other" values
color_dict_other = {x:'green' for x in df['Set'].unique() if x not in color_dict.keys()}

# Next, merge the two
color_dict.update(color_dict_other)

# Finally, map it to your column
df['color'] = df['Set'].map(color_dict)

This approach will be fastest when you have many repeated values. My general rule of thumb is to memoize when: data_size > 10**4 & n_distinct < data_size/4

E.x. Memoize in a case 10,000 rows with 2,500 or fewer distinct values.

Solution 11 - Python

The case_when function from pyjanitor is a wrapper around pd.Series.mask and offers a chainable/convenient form for multiple conditions:

For a single condition:

df.case_when(
    df.col1 == "Z",  # condition
    "green",         # value if True
    "red",           # value if False
    column_name = "color"
    )

  Type Set  color
1    A   Z  green
2    B   Z  green
3    B   X    red
4    C   Y    red

For multiple conditions:

df.case_when(
    df.Set.eq('Z') & df.Type.eq('A'), 'yellow', # condition, result
    df.Set.eq('Z') & df.Type.eq('B'), 'blue',   # condition, result
    df.Type.eq('B'), 'purple',                  # condition, result
    'black',              # default if none of the conditions evaluate to True
    column_name = 'color'  
)
  Type  Set   color
1    A   Z  yellow
2    B   Z    blue
3    B   X  purple
4    C   Y   black

More examples can be found here

Solution 12 - Python

A Less verbose approach using np.select:

a = np.array([['A','Z'],['B','Z'],['B','X'],['C','Y']])
df = pd.DataFrame(a,columns=['Type','Set'])

conditions = [    df['Set'] == 'Z'
]

outputs = [    'Green'    ]
             # conditions Z is Green, Red Otherwise.
res = np.select(conditions, outputs, 'Red')
res 
array(['Green', 'Green', 'Red', 'Red'], dtype='<U5')
df.insert(2, 'new_column',res)    

df
	Type	Set	new_column
0	A	Z	Green
1	B	Z	Green
2	B	X	Red
3	C	Y	Red

df.to_numpy()    
    
array([['A', 'Z', 'Green'],
       ['B', 'Z', 'Green'],
       ['B', 'X', 'Red'],
       ['C', 'Y', 'Red']], dtype=object)

%%timeit conditions = [df['Set'] == 'Z'] 
outputs = ['Green'] 
np.select(conditions, outputs, 'Red')

134 µs ± 9.71 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

df2 = pd.DataFrame({'Type':list('ABBC')*1000000, 'Set':list('ZZXY')*1000000})
%%timeit conditions = [df2['Set'] == 'Z'] 
outputs = ['Green'] 
np.select(conditions, outputs, 'Red')

188 ms ± 26.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

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