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+--- Thread: Use pandas to obtain cartesian product between a dataframe of int and equations? (/thread-39134.html)
Use pandas to obtain cartesian product between a dataframe of int and equations? - haihal - Jan-06-2023
I've attached an image further detailing what I'm trying to achieve.
The following is what I came up with; but I'm confident it can be done much simpler with built pandas modules. I'm hoping I won't have to resort to nested 4 loops.
The following is the code I used in the python interpreter:
nestDict = { 'A': {'Entity 1': 2, 'Entity 2': 1, 'Entity 3': 2},
'B': {'Entity 1': 4, 'Entity 2': 6, 'Entity 3': 3},
'C': {'Entity 1': 5, 'Entity 2': 8, 'Entity 3': 4}
}
frame1 = pd.DataFrame(nestDict)
dict1 = { 'Squared' : 'x**2',
'Sqrt' : 'round(math.sqrt(x), 2)',
'isPrime' : 'is_Prime_Function(x)'
}
series1 = pd.Series(dict1)
def is_Prime_Function(n):
if n == 1:
return 0
for i in range(2,n):
if (n%i) == 0:
return 0
return 1
frame1_sqrd = frame1.applymap(lambda x:eval(series1['Squared']))
frame1_sqrt = frame1.applymap(lambda x:eval(series1['Sqrt']))
frame1_isPrime = frame1.applymap(lambda x:eval(series1['isPrime']))
# Use nested for loop to retrieve each row from the evaluated frames make results 'readable' from 1 single 'cartesian-like' dataframe.This is the best or closes I've gotten to getting what I'm looking for.def triple_Evaluation(x):
return pd.Series([x**2, round(math.sqrt(x), 2), is_Prime_Function(x)], index=['Squared','Sqrt','Is Prime'])
frame1.iloc[[0]].append(frame1.iloc[0].apply(triple_Evaluation).T)