Performing calculation based off multiple rows in Pandas dataframe - pandas

Question

Set up an example dataframe:
import pandas as pd
import numpy as np
df = pd.DataFrame([[10, False, np.nan, np.nan], [5, False, np.nan, np.nan], [np.nan, True, 'a', 'b'],[np.nan,True,'b','a']],
columns=['value', 'IsRatio', 'numerator','denominator'],index=['a','b','c','d'])
index
value
IsRatio
numerator
denominator
a
10
False
nan
nan
b
5
False
nan
nan
c
nan
True
a
b
d
nan
True
b
a
For rows where IsRatio is True, I would like to lookup the values for the numerator and denominator, and calculate a ratio.
For a single row I can use .loc
numerator_name = df.loc['c','numerator']
denominator_name = df.loc['c','denominator']
df.loc['c','value'] = int(df.loc[numerator_name]['value'])/int(df.loc[denominator_name]['value'])
This will calculate the ratio for a single row
index
value
IsRatio
numerator
denominator
a
10
False
nan
nan
b
5
False
nan
nan
c
2
True
a
b
d
nan
True
b
a
How can I generalise this to all rows? I think I might need an apply function but I can't figure it out.

Answer 1

You can use apply to apply your computation to each row (mind the axis=1 input argument):
df['value'] = df.apply(
lambda x: int(df.loc[x.numerator]['value']) / int(df.loc[x.denominator]['value'])
if x.IsRatio else x.value,
axis=1
)
The result is the following:
value IsRatio numerator denominator
a 10 False nan nan
b 5 False nan nan
c 2 True a b
d 0.5 True b a
Note: you should remove np.array from the creation of the example DataFrame, otherwise the IsRatio column has type str. So df should be defined as follow:
import pandas as pd
import numpy as np
df = pd.DataFrame([[10, False, np.nan, np.nan], [5, False, np.nan, np.nan], [np.nan, True, 'a', 'b'],[np.nan,True,'b','a']],
columns=['value', 'IsRatio', 'numerator','denominator'],index=['a','b','c','d'])
Otherwise, if IsRatio column is actually of type str, you should modify the previous code as following:
df['value'] = df.apply(
lambda x: int(df.loc[x.numerator]['value']) / int(df.loc[x.denominator]['value'])
if x.IsRatio == 'True' else x.value,
axis=1
)
value IsRatio numerator denominator
a 10 False nan nan
b 5 False nan nan
c 2 True a b
d 0.5 True b a

Answer 2

To do as a vectorised solution numpy where() is a good solution.
df = pd.DataFrame(np.array([[10, False, np.nan, np.nan], [5, False, np.nan, np.nan], [np.nan, True, 10, 5],[np.nan,True,30,10]]),
columns=['value', 'IsRatio', 'numerator','denominator'],index=['a','b','c','d'])
# df.assign(v=np.where)
df.IsRatio = df.IsRatio.astype(bool)
df.assign(v=np.where(df.IsRatio, df.numerator/df.denominator, df.value))
value
IsRatio
numerator
denominator
v
a
10
False
nan
nan
10
b
5
False
nan
nan
5
c
nan
True
10
5
2
d
nan
True
30
10
3