This notebook contains an excerpt from the Python Data Science Handbook by Jake VanderPlas; the content is available on GitHub.

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High-Performance Pandas: eval() and query()¶

As we've already seen in previous sections, the power of the PyData stack is built upon the ability of NumPy and Pandas to push basic operations into C via an intuitive syntax: examples are vectorized/broadcasted operations in NumPy, and grouping-type operations in Pandas. While these abstractions are efficient and effective for many common use cases, they often rely on the creation of temporary intermediate objects, which can cause undue overhead in computational time and memory use.

As of version 0.13 (released January 2014), Pandas includes some experimental tools that allow you to directly access C-speed operations without costly allocation of intermediate arrays. These are the eval() and query() functions, which rely on the Numexpr package. In this notebook we will walk through their use and give some rules-of-thumb about when you might think about using them.

Motivating query() and eval(): Compound Expressions¶

We've seen previously that NumPy and Pandas support fast vectorized operations; for example, when adding the elements of two arrays:

In [1]:

import numpy as np
rng = np.random.RandomState(42)
x = rng.rand(1000000)
y = rng.rand(1000000)
%timeit x + y

import numpy as np rng = np.random.RandomState(42) x = rng.rand(1000000) y = rng.rand(1000000) %timeit x + y

100 loops, best of 3: 3.39 ms per loop

As discussed in Computation on NumPy Arrays: Universal Functions, this is much faster than doing the addition via a Python loop or comprehension:

In [2]:

%timeit np.fromiter((xi + yi for xi, yi in zip(x, y)), dtype=x.dtype, count=len(x))

%timeit np.fromiter((xi + yi for xi, yi in zip(x, y)), dtype=x.dtype, count=len(x))

1 loop, best of 3: 266 ms per loop

But this abstraction can become less efficient when computing compound expressions. For example, consider the following expression:

In [3]:

mask = (x > 0.5) & (y < 0.5)

mask = (x > 0.5) & (y < 0.5)

Because NumPy evaluates each subexpression, this is roughly equivalent to the following:

In [4]:

tmp1 = (x > 0.5)
tmp2 = (y < 0.5)
mask = tmp1 & tmp2

tmp1 = (x > 0.5) tmp2 = (y < 0.5) mask = tmp1 & tmp2

In other words, every intermediate step is explicitly allocated in memory. If the x and y arrays are very large, this can lead to significant memory and computational overhead. The Numexpr library gives you the ability to compute this type of compound expression element by element, without the need to allocate full intermediate arrays. The Numexpr documentation has more details, but for the time being it is sufficient to say that the library accepts a string giving the NumPy-style expression you'd like to compute:

In [5]:

import numexpr
mask_numexpr = numexpr.evaluate('(x > 0.5) & (y < 0.5)')
np.allclose(mask, mask_numexpr)

import numexpr mask_numexpr = numexpr.evaluate('(x > 0.5) & (y < 0.5)') np.allclose(mask, mask_numexpr)

Out[5]:

True

The benefit here is that Numexpr evaluates the expression in a way that does not use full-sized temporary arrays, and thus can be much more efficient than NumPy, especially for large arrays. The Pandas eval() and query() tools that we will discuss here are conceptually similar, and depend on the Numexpr package.

pandas.eval() for Efficient Operations¶

The eval() function in Pandas uses string expressions to efficiently compute operations using DataFrames. For example, consider the following DataFrames:

In [6]:

import pandas as pd
nrows, ncols = 100000, 100
rng = np.random.RandomState(42)
df1, df2, df3, df4 = (pd.DataFrame(rng.rand(nrows, ncols))
                      for i in range(4))

import pandas as pd nrows, ncols = 100000, 100 rng = np.random.RandomState(42) df1, df2, df3, df4 = (pd.DataFrame(rng.rand(nrows, ncols)) for i in range(4))

To compute the sum of all four DataFrames using the typical Pandas approach, we can just write the sum:

In [7]:

%timeit df1 + df2 + df3 + df4

%timeit df1 + df2 + df3 + df4

10 loops, best of 3: 87.1 ms per loop

The same result can be computed via pd.eval by constructing the expression as a string:

In [8]:

%timeit pd.eval('df1 + df2 + df3 + df4')

%timeit pd.eval('df1 + df2 + df3 + df4')

10 loops, best of 3: 42.2 ms per loop

The eval() version of this expression is about 50% faster (and uses much less memory), while giving the same result:

In [9]:

np.allclose(df1 + df2 + df3 + df4,
            pd.eval('df1 + df2 + df3 + df4'))

np.allclose(df1 + df2 + df3 + df4, pd.eval('df1 + df2 + df3 + df4'))

Out[9]:

True

Operations supported by pd.eval()¶

As of Pandas v0.16, pd.eval() supports a wide range of operations. To demonstrate these, we'll use the following integer DataFrames:

In [10]:

df1, df2, df3, df4, df5 = (pd.DataFrame(rng.randint(0, 1000, (100, 3)))
                           for i in range(5))

df1, df2, df3, df4, df5 = (pd.DataFrame(rng.randint(0, 1000, (100, 3))) for i in range(5))

Arithmetic operators¶

pd.eval() supports all arithmetic operators. For example:

In [11]:

result1 = -df1 * df2 / (df3 + df4) - df5
result2 = pd.eval('-df1 * df2 / (df3 + df4) - df5')
np.allclose(result1, result2)

result1 = -df1 * df2 / (df3 + df4) - df5 result2 = pd.eval('-df1 * df2 / (df3 + df4) - df5') np.allclose(result1, result2)

Out[11]:

True

Comparison operators¶

pd.eval() supports all comparison operators, including chained expressions:

In [12]:

result1 = (df1 < df2) & (df2 <= df3) & (df3 != df4)
result2 = pd.eval('df1 < df2 <= df3 != df4')
np.allclose(result1, result2)

result1 = (df1 < df2) & (df2 <= df3) & (df3 != df4) result2 = pd.eval('df1 < df2 <= df3 != df4') np.allclose(result1, result2)

Out[12]:

True

Bitwise operators¶

pd.eval() supports the & and | bitwise operators:

In [13]:

result1 = (df1 < 0.5) & (df2 < 0.5) | (df3 < df4)
result2 = pd.eval('(df1 < 0.5) & (df2 < 0.5) | (df3 < df4)')
np.allclose(result1, result2)

result1 = (df1 < 0.5) & (df2 < 0.5) | (df3 < df4) result2 = pd.eval('(df1 < 0.5) & (df2 < 0.5) | (df3 < df4)') np.allclose(result1, result2)

Out[13]:

True

In addition, it supports the use of the literal and and or in Boolean expressions:

In [14]:

result3 = pd.eval('(df1 < 0.5) and (df2 < 0.5) or (df3 < df4)')
np.allclose(result1, result3)

result3 = pd.eval('(df1 < 0.5) and (df2 < 0.5) or (df3 < df4)') np.allclose(result1, result3)

Out[14]:

True

Object attributes and indices¶

pd.eval() supports access to object attributes via the obj.attr syntax, and indexes via the obj[index] syntax:

In [15]:

result1 = df2.T[0] + df3.iloc[1]
result2 = pd.eval('df2.T[0] + df3.iloc[1]')
np.allclose(result1, result2)

result1 = df2.T[0] + df3.iloc[1] result2 = pd.eval('df2.T[0] + df3.iloc[1]') np.allclose(result1, result2)

Out[15]:

True

Other operations¶

Other operations such as function calls, conditional statements, loops, and other more involved constructs are currently not implemented in pd.eval(). If you'd like to execute these more complicated types of expressions, you can use the Numexpr library itself.

DataFrame.eval() for Column-Wise Operations¶

Just as Pandas has a top-level pd.eval() function, DataFrames have an eval() method that works in similar ways. The benefit of the eval() method is that columns can be referred to by name. We'll use this labeled array as an example:

In [16]:

df = pd.DataFrame(rng.rand(1000, 3), columns=['A', 'B', 'C'])
df.head()

df = pd.DataFrame(rng.rand(1000, 3), columns=['A', 'B', 'C']) df.head()

Out[16]:

	A	B	C
0	0.375506	0.406939	0.069938
1	0.069087	0.235615	0.154374
2	0.677945	0.433839	0.652324
3	0.264038	0.808055	0.347197
4	0.589161	0.252418	0.557789

Using pd.eval() as above, we can compute expressions with the three columns like this:

In [17]:

result1 = (df['A'] + df['B']) / (df['C'] - 1)
result2 = pd.eval("(df.A + df.B) / (df.C - 1)")
np.allclose(result1, result2)

result1 = (df['A'] + df['B']) / (df['C'] - 1) result2 = pd.eval("(df.A + df.B) / (df.C - 1)") np.allclose(result1, result2)

Out[17]:

True

The DataFrame.eval() method allows much more succinct evaluation of expressions with the columns:

In [18]:

result3 = df.eval('(A + B) / (C - 1)')
np.allclose(result1, result3)

result3 = df.eval('(A + B) / (C - 1)') np.allclose(result1, result3)

Out[18]:

True

Notice here that we treat column names as variables within the evaluated expression, and the result is what we would wish.

Assignment in DataFrame.eval()¶

In addition to the options just discussed, DataFrame.eval() also allows assignment to any column. Let's use the DataFrame from before, which has columns 'A', 'B', and 'C':

In [19]:

df.head()

df.head()

Out[19]:

	A	B	C
0	0.375506	0.406939	0.069938
1	0.069087	0.235615	0.154374
2	0.677945	0.433839	0.652324
3	0.264038	0.808055	0.347197
4	0.589161	0.252418	0.557789

We can use df.eval() to create a new column 'D' and assign to it a value computed from the other columns:

In [20]:

df.eval('D = (A + B) / C', inplace=True)
df.head()

df.eval('D = (A + B) / C', inplace=True) df.head()

Out[20]:

	A	B	C	D
0	0.375506	0.406939	0.069938	11.187620
1	0.069087	0.235615	0.154374	1.973796
2	0.677945	0.433839	0.652324	1.704344
3	0.264038	0.808055	0.347197	3.087857
4	0.589161	0.252418	0.557789	1.508776

In the same way, any existing column can be modified:

In [21]:

df.eval('D = (A - B) / C', inplace=True)
df.head()

df.eval('D = (A - B) / C', inplace=True) df.head()

Out[21]:

	A	B	C	D
0	0.375506	0.406939	0.069938	-0.449425
1	0.069087	0.235615	0.154374	-1.078728
2	0.677945	0.433839	0.652324	0.374209
3	0.264038	0.808055	0.347197	-1.566886
4	0.589161	0.252418	0.557789	0.603708

Local variables in DataFrame.eval()¶

The DataFrame.eval() method supports an additional syntax that lets it work with local Python variables. Consider the following:

In [22]:

column_mean = df.mean(1)
result1 = df['A'] + column_mean
result2 = df.eval('A + @column_mean')
np.allclose(result1, result2)

column_mean = df.mean(1) result1 = df['A'] + column_mean result2 = df.eval('A + @column_mean') np.allclose(result1, result2)

Out[22]:

True

The @ character here marks a variable name rather than a column name, and lets you efficiently evaluate expressions involving the two "namespaces": the namespace of columns, and the namespace of Python objects. Notice that this @ character is only supported by the DataFrame.eval() method, not by the pandas.eval() function, because the pandas.eval() function only has access to the one (Python) namespace.

DataFrame.query() Method¶

The DataFrame has another method based on evaluated strings, called the query() method. Consider the following:

In [23]:

result1 = df[(df.A < 0.5) & (df.B < 0.5)]
result2 = pd.eval('df[(df.A < 0.5) & (df.B < 0.5)]')
np.allclose(result1, result2)

result1 = df[(df.A < 0.5) & (df.B < 0.5)] result2 = pd.eval('df[(df.A < 0.5) & (df.B < 0.5)]') np.allclose(result1, result2)

Out[23]:

True

As with the example used in our discussion of DataFrame.eval(), this is an expression involving columns of the DataFrame. It cannot be expressed using the DataFrame.eval() syntax, however! Instead, for this type of filtering operation, you can use the query() method:

In [24]:

result2 = df.query('A < 0.5 and B < 0.5')
np.allclose(result1, result2)

result2 = df.query('A < 0.5 and B < 0.5') np.allclose(result1, result2)

Out[24]:

True

In addition to being a more efficient computation, compared to the masking expression this is much easier to read and understand. Note that the query() method also accepts the @ flag to mark local variables:

In [25]:

Cmean = df['C'].mean()
result1 = df[(df.A < Cmean) & (df.B < Cmean)]
result2 = df.query('A < @Cmean and B < @Cmean')
np.allclose(result1, result2)

Cmean = df['C'].mean() result1 = df[(df.A < Cmean) & (df.B < Cmean)] result2 = df.query('A < @Cmean and B < @Cmean') np.allclose(result1, result2)

Out[25]:

True

Performance: When to Use These Functions¶

When considering whether to use these functions, there are two considerations: computation time and memory use. Memory use is the most predictable aspect. As already mentioned, every compound expression involving NumPy arrays or Pandas DataFrames will result in implicit creation of temporary arrays: For example, this:

In [26]:

x = df[(df.A < 0.5) & (df.B < 0.5)]

x = df[(df.A < 0.5) & (df.B < 0.5)]

Is roughly equivalent to this:

In [27]:

tmp1 = df.A < 0.5
tmp2 = df.B < 0.5
tmp3 = tmp1 & tmp2
x = df[tmp3]

tmp1 = df.A < 0.5 tmp2 = df.B < 0.5 tmp3 = tmp1 & tmp2 x = df[tmp3]

If the size of the temporary DataFrames is significant compared to your available system memory (typically several gigabytes) then it's a good idea to use an eval() or query() expression. You can check the approximate size of your array in bytes using this:

In [28]:

df.values.nbytes

df.values.nbytes

Out[28]:

32000

On the performance side, eval() can be faster even when you are not maxing-out your system memory. The issue is how your temporary DataFrames compare to the size of the L1 or L2 CPU cache on your system (typically a few megabytes in 2016); if they are much bigger, then eval() can avoid some potentially slow movement of values between the different memory caches. In practice, I find that the difference in computation time between the traditional methods and the eval/query method is usually not significant–if anything, the traditional method is faster for smaller arrays! The benefit of eval/query is mainly in the saved memory, and the sometimes cleaner syntax they offer.

We've covered most of the details of eval() and query() here; for more information on these, you can refer to the Pandas documentation. In particular, different parsers and engines can be specified for running these queries; for details on this, see the discussion within the "Enhancing Performance" section.