Related
How do I add a color column to the following dataframe so that color='green' if Set == 'Z', and color='red' otherwise?
Type Set
1 A Z
2 B Z
3 B X
4 C Y
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
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
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
Another way in which this could be achieved is
df['color'] = df.Set.map( lambda x: 'red' if x == 'Z' else 'green')
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']]
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:
Logical operators for boolean indexing in Pandas
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
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
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
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
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.
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)
Here is an easy one-liner you can use when you have one or several conditions:
df['color'] = np.select(condlist=[df['Set']=="Z", df['Set']=="Y"], choicelist=["green", "yellow"], default="red")
Easy and good to go!
See more here: https://numpy.org/doc/stable/reference/generated/numpy.select.html
How do I add a color column to the following dataframe so that color='green' if Set == 'Z', and color='red' otherwise?
Type Set
1 A Z
2 B Z
3 B X
4 C Y
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
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
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
Another way in which this could be achieved is
df['color'] = df.Set.map( lambda x: 'red' if x == 'Z' else 'green')
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']]
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:
Logical operators for boolean indexing in Pandas
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
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
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
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
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.
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)
Here is an easy one-liner you can use when you have one or several conditions:
df['color'] = np.select(condlist=[df['Set']=="Z", df['Set']=="Y"], choicelist=["green", "yellow"], default="red")
Easy and good to go!
See more here: https://numpy.org/doc/stable/reference/generated/numpy.select.html
How do I add a color column to the following dataframe so that color='green' if Set == 'Z', and color='red' otherwise?
Type Set
1 A Z
2 B Z
3 B X
4 C Y
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
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
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
Another way in which this could be achieved is
df['color'] = df.Set.map( lambda x: 'red' if x == 'Z' else 'green')
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']]
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:
Logical operators for boolean indexing in Pandas
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
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
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
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
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.
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)
Here is an easy one-liner you can use when you have one or several conditions:
df['color'] = np.select(condlist=[df['Set']=="Z", df['Set']=="Y"], choicelist=["green", "yellow"], default="red")
Easy and good to go!
See more here: https://numpy.org/doc/stable/reference/generated/numpy.select.html
I created three figures using seaborn relplot. Each one is a single row with a different y variable, but the same x, column, and hue variables. I want to visualize these three rows all on the same figure. I could make a new figure and copy the contents over, but that would lose the formatting and I am hoping seaborn offers this functionality, but I cannot find it.
My data is in a pandas dataframe named sample:
In: sample
Out:
X Val Col Val Hue Val Y Val Row 1 Y Val Row 2 Y Val Row 3
0 0.95 0.4 0.4 0.638958 0.635887 1061
1 0.96 0.4 0.4 0.639547 0.654841 976
2 0.97 0.4 0.4 0.638076 0.675838 916
3 0.98 0.4 0.4 0.633368 0.689067 831
4 0.99 0.4 0.4 0.629690 0.711538 719
.. ... ... ... ... ... ...
120 0.95 0.8 0.8 0.634979 0.624095 1066
121 0.96 0.8 0.8 0.635565 0.642270 986
122 0.97 0.8 0.8 0.635272 0.665792 916
123 0.98 0.8 0.8 0.633221 0.687564 826
124 0.99 0.8 0.8 0.629852 0.712159 706
[125 rows x 6 columns]
I have created three relplot figures:
import seaborn as sns
g1 = sns.relplot(data=sample, x="X Val", y = "Y Val Row 1", col="Col Val", hue="Hue Val", kind="line")
g3 = sns.relplot(data=sample, x="X Val", y = "Y Val Row 3", col="Col Val", hue="Hue Val", kind="line")
g2 = sns.relplot(data=sample, x="X Val", y = "Y Val Row 2", col="Col Val", hue="Hue Val", kind="line")
Each figure (g1, g2, g3) is a single row of five axes objects, each with the same column values, hue values, and x values. I want to combine these into one figure of these three rows stacked. I combined the figures in power point to show what I want for the final result:
Complete figure, with g1 as row 1, g2 as row 2, and g3 as row 3:
Thanks to #JohanC's comments, I now have the solution.
# melt the dataframe so all the "Y Val Row *" values are in the same column
samplemelt = pd.melt(sample, id_vars=['X Val', 'Col Val', 'Hue Val'],
value_vars=['Y Val Row 1', 'Y Val Row 2', 'Y Val Row 3'],
var_name="variable", value_name="value")
# make the relplot, limiting the sharey flag to each row
gm = sns.relplot(data=samplemelt, row="variable", y="value", x="X Val",
col="Col Val", hue="Hue Val", facet_kws={"sharey":"row"})
# pull the "Y Val Row *" label from each graph title onto the row y label
# this assumes the current default behavior seaborn.relplot for graph titling
for row in gm.axes:
for ax in row:
title = ax.title.get_text()
ax.set_title(title.split(" | ")[1])
row[0].set_ylabel(title.split(" | ")[0].replace("variable = ", ""))
I want to divide a dataframe by a number:
df = df/10
Is there a way to do this in a method chain?
# idea:
df = df.filter(['a','b']).query("a>100").assign(**divide by 10)
We can use DataFrame.div here:
df = df[['a','b']].query("a>100").div(10)
a b
0 40.0 0.7
1 50.0 0.8
5 70.0 0.3
Use DataFrame.pipe with lambda function for use some function for all data of DataFrame:
df = pd.DataFrame({
'a':[400,500,40,50,5,700],
'b':[7,8,9,4,2,3],
'c':[1,3,5,7,1,0],
'd':[5,3,6,9,2,4]
})
df = df.filter(['a','b']).query("a>100").pipe(lambda x: x / 10)
print (df)
a b
0 40.0 0.7
1 50.0 0.8
5 70.0 0.3
Here if use apply all columns are divided separately:
df = df.filter(['a','b']).query("a>100").apply(lambda x: x / 10)
You can see difference with print:
df1 = df.filter(['a','b']).query("a>100").pipe(lambda x: print (x))
a b
0 400 7
1 500 8
5 700 3
df2 = df.filter(['a','b']).query("a>100").apply(lambda x: print (x))
0 400
1 500
5 700
Name: a, dtype: int64
0 7
1 8
5 3
Name: b, dtype: int64