Logical operators for boolean indexing in Pandas
When you say
(a['x']==1) and (a['y']==10)
You are implicitly asking Python to convert (a['x']==1) and (a['y']==10) to Boolean values.
NumPy arrays (of length greater than 1) and Pandas objects such as Series do not have a Boolean value -- in other words, they raise
ValueError: The truth value of an array is ambiguous. Use a.empty, a.any() or a.all().
when used as a Boolean value. That's because it's unclear when it should be True or False. Some users might assume they are True if they have non-zero length, like a Python list. Others might desire for it to be True only if all its elements are True. Others might want it to be True if any of its elements are True.
Because there are so many conflicting expectations, the designers of NumPy and Pandas refuse to guess, and instead raise a ValueError.
Instead, you must be explicit, by calling the empty(), all() or any() method to indicate which behavior you desire.
In this case, however, it looks like you do not want Boolean evaluation, you want element-wise logical-and. That is what the & binary operator performs:
(a['x']==1) & (a['y']==10)
returns a boolean array.
By the way, as alexpmil notes,
the parentheses are mandatory since & has a higher operator precedence than ==.
Without the parentheses, a['x']==1 & a['y']==10 would be evaluated as a['x'] == (1 & a['y']) == 10 which would in turn be equivalent to the chained comparison (a['x'] == (1 & a['y'])) and ((1 & a['y']) == 10). That is an expression of the form Series and Series.
The use of and with two Series would again trigger the same ValueError as above. That's why the parentheses are mandatory.
TLDR; Logical Operators in Pandas are &, | and ~, and parentheses (...) is important!
Python's and, or and not logical operators are designed to work with scalars. So Pandas had to do one better and override the bitwise operators to achieve vectorized (element-wise) version of this functionality.
So the following in python (exp1 and exp2 are expressions which evaluate to a boolean result)...
exp1 and exp2 # Logical AND
exp1 or exp2 # Logical OR
not exp1 # Logical NOT
...will translate to...
exp1 & exp2 # Element-wise logical AND
exp1 | exp2 # Element-wise logical OR
~exp1 # Element-wise logical NOT
for pandas.
If in the process of performing logical operation you get a ValueError, then you need to use parentheses for grouping:
(exp1) op (exp2)
For example,
(df['col1'] == x) & (df['col2'] == y)
And so on.
Boolean Indexing: A common operation is to compute boolean masks through logical conditions to filter the data. Pandas provides three operators: & for logical AND, | for logical OR, and ~ for logical NOT.
Consider the following setup:
np.random.seed(0)
df = pd.DataFrame(np.random.choice(10, (5, 3)), columns=list('ABC'))
df
A B C
0 5 0 3
1 3 7 9
2 3 5 2
3 4 7 6
4 8 8 1
Logical AND
For df above, say you'd like to return all rows where A < 5 and B > 5. This is done by computing masks for each condition separately, and ANDing them.
Overloaded Bitwise & Operator
Before continuing, please take note of this particular excerpt of the docs, which state
Another common operation is the use of boolean vectors to filter the data. The operators are:
|foror,&forand, and~fornot. These must be grouped by using parentheses, since by default Python will evaluate an expression such asdf.A > 2 & df.B < 3asdf.A > (2 & df.B) < 3, while the desired evaluation order is(df.A > 2) & (df.B < 3).
So, with this in mind, element wise logical AND can be implemented with the bitwise operator &:
df['A'] < 5
0 False
1 True
2 True
3 True
4 False
Name: A, dtype: bool
df['B'] > 5
0 False
1 True
2 False
3 True
4 True
Name: B, dtype: bool
(df['A'] < 5) & (df['B'] > 5)
0 False
1 True
2 False
3 True
4 False
dtype: bool
And the subsequent filtering step is simply,
df[(df['A'] < 5) & (df['B'] > 5)]
A B C
1 3 7 9
3 4 7 6
The parentheses are used to override the default precedence order of bitwise operators, which have higher precedence over the conditional operators < and >. See the section of Operator Precedence in the python docs.
If you do not use parentheses, the expression is evaluated incorrectly. For example, if you accidentally attempt something such as
df['A'] < 5 & df['B'] > 5
It is parsed as
df['A'] < (5 & df['B']) > 5
Which becomes,
df['A'] < something_you_dont_want > 5
Which becomes (see the python docs on chained operator comparison),
(df['A'] < something_you_dont_want) and (something_you_dont_want > 5)
Which becomes,
# Both operands are Series...
something_else_you_dont_want1 and something_else_you_dont_want2Which throws
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
So, don't make that mistake!1
Avoiding Parentheses Grouping
The fix is actually quite simple. Most operators have a corresponding bound method for DataFrames. If the individual masks are built up using functions instead of conditional operators, you will no longer need to group by parens to specify evaluation order:
df['A'].lt(5)
0 True
1 True
2 True
3 True
4 False
Name: A, dtype: bool
df['B'].gt(5)
0 False
1 True
2 False
3 True
4 True
Name: B, dtype: bool
df['A'].lt(5) & df['B'].gt(5)
0 False
1 True
2 False
3 True
4 False
dtype: bool
See the section on Flexible Comparisons.. To summarise, we have
╒════╤════════════╤════════════╕
│ │ Operator │ Function │
╞════╪════════════╪════════════╡
│ 0 │ > │ gt │
├────┼────────────┼────────────┤
│ 1 │ >= │ ge │
├────┼────────────┼────────────┤
│ 2 │ < │ lt │
├────┼────────────┼────────────┤
│ 3 │ <= │ le │
├────┼────────────┼────────────┤
│ 4 │ == │ eq │
├────┼────────────┼────────────┤
│ 5 │ != │ ne │
╘════╧════════════╧════════════╛
Another option for avoiding parentheses is to use DataFrame.query (or eval):
df.query('A < 5 and B > 5')
A B C
1 3 7 9
3 4 7 6
I have extensively documented query and eval in Dynamic Expression Evaluation in pandas using pd.eval().
operator.and_
Allows you to perform this operation in a functional manner. Internally calls Series.__and__ which corresponds to the bitwise operator.
import operator
operator.and_(df['A'] < 5, df['B'] > 5)
# Same as,
# (df['A'] < 5).__and__(df['B'] > 5)
0 False
1 True
2 False
3 True
4 False
dtype: bool
df[operator.and_(df['A'] < 5, df['B'] > 5)]
A B C
1 3 7 9
3 4 7 6
You won't usually need this, but it is useful to know.
Generalizing: np.logical_and (and logical_and.reduce)
Another alternative is using np.logical_and, which also does not need parentheses grouping:
np.logical_and(df['A'] < 5, df['B'] > 5)
0 False
1 True
2 False
3 True
4 False
Name: A, dtype: bool
df[np.logical_and(df['A'] < 5, df['B'] > 5)]
A B C
1 3 7 9
3 4 7 6
np.logical_and is a ufunc (Universal Functions), and most ufuncs have a reduce method. This means it is easier to generalise with logical_and if you have multiple masks to AND. For example, to AND masks m1 and m2 and m3 with &, you would have to do
m1 & m2 & m3
However, an easier option is
np.logical_and.reduce([m1, m2, m3])
This is powerful, because it lets you build on top of this with more complex logic (for example, dynamically generating masks in a list comprehension and adding all of them):
import operator
cols = ['A', 'B']
ops = [np.less, np.greater]
values = [5, 5]
m = np.logical_and.reduce([op(df[c], v) for op, c, v in zip(ops, cols, values)])
m
# array([False, True, False, True, False])
df[m]
A B C
1 3 7 9
3 4 7 6
1 - I know I'm harping on this point, but please bear with me. This is a very, very common beginner's mistake, and must be explained very thoroughly.
Logical OR
For the df above, say you'd like to return all rows where A == 3 or B == 7.
Overloaded Bitwise |
df['A'] == 3
0 False
1 True
2 True
3 False
4 False
Name: A, dtype: bool
df['B'] == 7
0 False
1 True
2 False
3 True
4 False
Name: B, dtype: bool
(df['A'] == 3) | (df['B'] == 7)
0 False
1 True
2 True
3 True
4 False
dtype: bool
df[(df['A'] == 3) | (df['B'] == 7)]
A B C
1 3 7 9
2 3 5 2
3 4 7 6
If you haven't yet, please also read the section on Logical AND above, all caveats apply here.
Alternatively, this operation can be specified with
df[df['A'].eq(3) | df['B'].eq(7)]
A B C
1 3 7 9
2 3 5 2
3 4 7 6
operator.or_
Calls Series.__or__ under the hood.
operator.or_(df['A'] == 3, df['B'] == 7)
# Same as,
# (df['A'] == 3).__or__(df['B'] == 7)
0 False
1 True
2 True
3 True
4 False
dtype: bool
df[operator.or_(df['A'] == 3, df['B'] == 7)]
A B C
1 3 7 9
2 3 5 2
3 4 7 6
np.logical_or
For two conditions, use logical_or:
np.logical_or(df['A'] == 3, df['B'] == 7)
0 False
1 True
2 True
3 True
4 False
Name: A, dtype: bool
df[np.logical_or(df['A'] == 3, df['B'] == 7)]
A B C
1 3 7 9
2 3 5 2
3 4 7 6
For multiple masks, use logical_or.reduce:
np.logical_or.reduce([df['A'] == 3, df['B'] == 7])
# array([False, True, True, True, False])
df[np.logical_or.reduce([df['A'] == 3, df['B'] == 7])]
A B C
1 3 7 9
2 3 5 2
3 4 7 6
Logical NOT
Given a mask, such as
mask = pd.Series([True, True, False])
If you need to invert every boolean value (so that the end result is [False, False, True]), then you can use any of the methods below.
Bitwise ~
~mask
0 False
1 False
2 True
dtype: bool
Again, expressions need to be parenthesised.
~(df['A'] == 3)
0 True
1 False
2 False
3 True
4 True
Name: A, dtype: bool
This internally calls
mask.__invert__()
0 False
1 False
2 True
dtype: bool
But don't use it directly.
operator.inv
Internally calls __invert__ on the Series.
operator.inv(mask)
0 False
1 False
2 True
dtype: bool
np.logical_not
This is the numpy variant.
np.logical_not(mask)
0 False
1 False
2 True
dtype: bool
Note, np.logical_and can be substituted for np.bitwise_and, logical_or with bitwise_or, and logical_not with invert.