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this code gives the result i want:

  • it takes the value n-1 and calculates n from it

take the previous value in column 'y' lets call it y-1 and calculate a value which becomes the new y, than in the next row take that new y as y-1 and calculate another new y aso

size = 10
x= range(1,size+1)
df = pd.DataFrame(data={'x': x,'y': size })

for n in range(1,len(x)):
    df['y'].iloc[n] = df['y'].iloc[n-1]*2

out:

    x   y
0   1   10
1   2   20
2   3   40
... ... ...
9   10  5120

I want to put it into a lambda but somehow fail to get it right:

b=2
df['y'].loc[1::] = df['y'].shift(-1).apply(lambda x: x*b)

out:

    x   y
0   1   10.0
1   2   20.0
2   3   20.0
... ... ...

the lambda function takes the pre-populated value (10) in each row instead of shifting 1 step back and taking the previous value as base for the multiplication

i looked at some threads, but its above my comprahension, if i am dealing here with recursion and this is not possible with lambdas?

recursive lambda-expressions possible?

Python Recursion on Pandas df

Can a lambda function call itself recursively in Python?

Edit: I want that subsequent entries in 'y' are calculated with previous 'y' entries, starting from idx 1

DataFrame at start:

idx | y |
 0   10


DataFrame after 1st calc:
y1 = y0 *2

# *2 is a placeholder could be mx+b, or something else

idx | y |
 0   10
 1   20
chubaka
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1 Answers1

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I'm not sure you need to do any recursion, unless I'm misunderstanding this is mathematical exponents.

Note sure what you're actual use case but something like one of these should work.

[v*(2**i) for i,v in enumerate(df.y)]    

df.apply(lambda j: j.y*(2**(j.x-1)), axis=1)
Quixotic22
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