CODEWITHSUNDEEP
pythonpandas
aur.serv
aur.serv

Reputation: 59

Add one new value to the equation till the condition will be met

I'm stuck with the moment of appending new values to equation, so any help is welcome.

Assuming I have dataframe like this: 

   Name| A | B |
     A | 30| 0 |  
     A | 15| 5 |  
     B | 10| 5 |  
     A | 94| 40|
     B | 30| 40|  
     A | 15|125|
     A | 3 |150|

and c=0.01

I want to reproduce the next rule: A1[A]-A2[B]*c+A2[A]+...+An WHILE > 0. So, since A1[A]-A2[B]*c+A2[A] > 0 at the next step the A3[A] must be added to the initial equation and so on till the condition will be met But I can't get how to append An to the end of the equation.

For now my code looks like this: 

for a in df['Name']:
c = 0.1
shift = 1
sum = df[df['Name']=='A']['A'] - df[df['Name']=='A']['B'].shift(-1)*c + df[df['Name']=='A']['A'].shift(-shift)
while sum >0:
    shift=+1
    sum = sum+df[df['Name']=='A']['A'].shift(-shift)

The problem in the code that I have just computations for each pair of A-s and the output is

0 44.5
1 105.0
3 96.5
5 4.0
6 NaN

While I need the program to continue addition of the n-th A value to initial equation while the sum is >0

Upvotes: 3

Views: 110

Answers (2)

ansev
ansev

Reputation: 30930

c = 0.1
A =df.loc[df['Name'].eq('A')].reset_index(drop=True)

Case 1: If the first value of A[B] is 0 as in the example.

s = ( A['A'].sub(A['B'].mul(c))
            .where(A.index <2,A['A'])
            .cumsum() )
print(s)
0     30.0
1     44.5
2    138.5
3    153.5
4    156.5
dtype: float64

Case 2: first value of A[B] isn't 0

s = ( A['A'].sub(A['B'].mul(c).mask(A.index==0,0))
            .where(A.index <2,A['A'])
            .cumsum() )

s[s.le(0).cumsum().eq(0)].iloc[-1]

Output

156.5

Upvotes: 2

Quang Hoang
Quang Hoang

Reputation: 150785

IIUC, you can do:

# extract the name A
dfA = df[df.Name=='A']

# compute cumulative sum of A1[A] - A2[B]
s = dfA['A'].sub(dfA['B'].shift(-1).mul(-c).fillna(0)).cumsum()

# find the last positive value
# might need to check if there is one
s[s.le(0).cumsum().eq(0)].iloc[-1]

Output:

160.2

Upvotes: 1

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