how to calculate if statement for relative value rebalancing / Error: “The truth value of a Series is ambiguous”

Question

Below you find the code I wrote to calculate a relative change in value of df.a and df.b while df is a dataframe. What has to be calculated is basically df[\"c\"] = df

Answer 1

Ok, I get the result you want, but this is still way too complicated and unefficient. I would be interested to see a superior solution:

import pandas as pd
import numpy as np
import datetime

randn = np.random.randn


rng = pd.date_range('1/1/2011', periods=10, freq='D')

df = pd.DataFrame({'a': [1.1, 1.2, 2.3, 1.4, 1.5, 1.8, 0.7, 1.8, 1.9, 2.0], 'b': [1.1, 1.5, 1.3, 1.6, 1.5, 1.1, 1.5, 1.7, 2.1, 2.1],'c':[None] * 10},index=rng)



df["d"]= [0,0,0,0,0,0,0,0,0,0]



df["t"]= np.arange(len(df))
tolerance = 0.3

df['d1'] = df.a/df.a.iloc[df.d].values > df.b/df.b.iloc[df.d].values * (1+tolerance)

df['d2'] = df.a/df.a.iloc[df.d].values * (1+tolerance) < df.b/df.b.iloc[df.d].values



df['e'] = df.d1*df.t
df['f'] = df.d2*df.t
df['g'] = df.e +df.f
df.ix[df.g > df.g.shift(1),"h"] = df.g * 1; df
df.h = df.h + 1
df.h = df.h.shift(1)
df['h'][0] = 0

df.h.fillna(method='ffill',inplace=True)
df["d"] = df.h
df["c"] = df.a/df.a.iloc[df.d].values

and that's the result:

              a    b         c  d  t     d1     d2  e  f  g  h
2011-01-01  1.1  1.1  1.000000  0  0  False  False  0  0  0  0
2011-01-02  1.2  1.5  1.090909  0  1  False  False  0  0  0  0
2011-01-03  2.3  1.3  2.090909  0  2   True  False  2  0  2  0
2011-01-04  1.4  1.6  1.000000  3  3  False  False  0  0  0  3
2011-01-05  1.5  1.5  1.071429  3  4  False  False  0  0  0  3
2011-01-06  1.8  1.1  1.285714  3  5   True  False  5  0  5  3
2011-01-07  0.7  1.5  1.000000  6  6  False   True  0  6  6  6
2011-01-08  1.8  1.7  1.000000  7  7  False  False  0  0  0  7
2011-01-09  1.9  2.1  1.055556  7  8  False  False  0  0  0  7
2011-01-10  2.0  2.1  1.111111  7  9  False  False  0  0  0  7

from here you can easily delete rows with e.g. del df['g']

Answer 2

This is not a 100% solution, but should at least get you down a better path and fix a primary problem. The core problem I'm seeing here from a syntax side is that you are trying to mix vectorized and non-vectorized code. You could instead do something more like this:

>>> df['d1'] = df.a/df.a.iloc[df.d].values > df.b/df.b.iloc[df.d].values * (1+tolerance)

>>> df['d2'] = df.a/df.a.iloc[df.d].values * (1+tolerance) < df.b/df.b.iloc[df.d].values

>>> df['d'] = df['d1'] | df['d2']

>>> df

              a    b     c      d  t     d1     d2
2011-01-01  1.1  1.1  None  False  0  False  False
2011-01-02  1.2  1.5  None  False  1  False  False
2011-01-03  2.3  1.3  None   True  2   True  False
2011-01-04  1.4  1.6  None  False  3  False  False
2011-01-05  1.5  1.5  None  False  4  False  False
2011-01-06  1.8  1.1  None   True  5   True  False
2011-01-07  0.7  1.5  None   True  6  False   True
2011-01-08  1.8  1.7  None  False  7  False  False
2011-01-09  1.9  2.1  None  False  8  False  False
2011-01-10  2.0  2.1  None  False  9  False  False

That's not quite the answer you want, but hopefully shows you what is going on with the code and how you can fix it to get what you want (i.e. you don't need or want to be using a function and applying it here, just use standard pandas vectorized code).

If you can get that to work, the cleaner way to do that would be with np.where (either two of them sequentially, or nested).