Bollinger Bands Part 2

The other operator that one can use is &, which will be the same as and in this case. So if we wanted to find all points outside the bounds, we could check the two conditions below:

In [16]:

#Check points within the bounds
print(df[(df < upper_band) & (df > lower_band)])

21      97.376812
22      98.278403
23      98.780897
24      99.681753
25      98.998025
          ...
195    116.459768
196    117.562087
197    118.482369
198    119.793321
199    121.337763
Length: 155, dtype: float64

You will notice that the first 21 points are not in here? Why is that? If you look into the bands it will become obvious.

In [17]:

print(upper_band.head(21))

0    NaN
1    NaN
2    NaN
3    NaN
4    NaN
5    NaN
6    NaN
7    NaN
8    NaN
9    NaN
10   NaN
11   NaN
12   NaN
13   NaN
14   NaN
15   NaN
16   NaN
17   NaN
18   NaN
19   NaN
20   NaN
dtype: float64

Because the windows are rolling over 22 days, the first 21 points are null. If we are testing to find the truth in both of these cases the null values will always return false because they are not valid numbers. One work around to this is that we can take an index and reverse the values with ~. So for example, let's grab an index for outside the bands and then reverse it.

In [18]:

#Get an index and then the opposite of the index
i1 = (df > upper_band) | (df < lower_band)
print(i1[:5])
print()
i2 = ~i1
print(i2[:5])

0    False
1    False
2    False
3    False
4    False
dtype: bool

0    True
1    True
2    True
3    True
4    True
dtype: bool

A useful application of this is to find the percent of the time that the price is outside the bounds.

In [19]:

#What percent is outside of the bands?
print(len(df[i1]) / len(df))

0.12

Data Science

Data Science

Bollinger Bands Part 2

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