The or and and python statements require truth-values. For pandas these are considered ambiguous so you should use “bitwise” | (or) or & (and) operations:

result = result[(result[‘var’]>0.25) | (result[‘var’]<-0.25)]
These are overloaded for these kind of datastructures to yield the element-wise or (or and).
Just to add some more explanation to this statement:
The exception is thrown when you want to get the bool of a pandas.Series:
>>> import pandas as pd

>>> x = pd.Series([1])

>>> bool(x)

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

What you hit was a place where the operator implicitly converted the operands to bool (you used or but it also happens for and, if and while):

>>> x or x

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

>>> x and x

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

>>> if x:

… print(‘fun’)

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

>>> while x:

… print(‘fun’)

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Besides these 4 statements there are several python functions that hide some bool calls (like any, all, filter, …) these are normally not problematic with pandas.Series but for completeness I wanted to mention these.

In your case the exception isn’t really helpful, because it doesn’t mention the right alternatives. For and and or you can use (if you want element-wise comparisons):

numpy.logical_or:

>>> import numpy as np

>>> np.logical_or(x, y)

or simply the | operator:

>>> x | y

numpy.logical_and:

>>> np.logical_and(x, y)

or simply the & operator:

>>> x & y

If you’re using the operators then make sure you set your parenthesis correctly because of the operator precedence.

There are several logical numpy functions which should work on pandas.Series.

The alternatives mentioned in the Exception are more suited if you encountered it when doing if or while. I’ll shortly explain each of these:

If you want to check if your Series is empty:

>>> x = pd.Series([])

>>> x.empty

True

>>> x = pd.Series([1])

>>> x.empty

False

Python normally interprets the length of containers (like list, tuple, …) as truth-value if it has no explicit boolean interpretation. So if you want the python-like check, you could do: if x.size or if not x.empty instead of if x.

If your Series contains one and only one boolean value:

>>> x = pd.Series([100])

>>> (x > 50).bool()

True

>>> (x < 50).bool()
False
If you want to check the first and only item of your Series (like .bool() but works even for not boolean contents):
>>> x = pd.Series([100])

>>> x.item()

100

If you want to check if all or any item is not-zero, not-empty or not-False:

>>> x = pd.Series([0, 1, 2])

>>> x.all() # because one element is zero

False

>>> x.any() # because one (or more) elements are non-zero

True

For boolean logic, use & and |.

np.random.seed(0)

df = pd.DataFrame(np.random.randn(5,3), columns=list(‘ABC’))

>>> df

A B C

0 1.764052 0.400157 0.978738

1 2.240893 1.867558 -0.977278

2 0.950088 -0.151357 -0.103219

3 0.410599 0.144044 1.454274

4 0.761038 0.121675 0.443863

>>> df.loc[(df.C > 0.25) | (df.C < -0.25)]
A B C
0 1.764052 0.400157 0.978738
1 2.240893 1.867558 -0.977278
3 0.410599 0.144044 1.454274
4 0.761038 0.121675 0.443863
To see what is happening, you get a column of booleans for each comparison, e.g.
df.C > 0.25

0 True

1 False

2 False

3 True

4 True

Name: C, dtype: bool

When you have multiple criteria, you will get multiple columns returned. This is why the join logic is ambiguous. Using and or or treats each column separately, so you first need to reduce that column to a single boolean value. For example, to see if any value or all values in each of the columns is True.

# Any value in either column is True?

(df.C > 0.25).any() or (df.C < -0.25).any()
True
# All values in either column is True?
(df.C > 0.25).all() or (df.C < -0.25).all()
False
One convoluted way to achieve the same thing is to zip all of these columns together, and perform the appropriate logic.
>>> df[[any([a, b]) for a, b in zip(df.C > 0.25, df.C < -0.25)]]
A B C
0 1.764052 0.400157 0.978738
1 2.240893 1.867558 -0.977278
3 0.410599 0.144044 1.454274
4 0.761038 0.121675 0.443863
For more details, refer to Boolean Indexing in the docs.