Develop Reference

Why is .loc slicing in pandas inclusive of stop, contrary to typical python slicing?

I am slicing a pandas dataframe and I seem to be getting unexpected slices using .loc, at least as compared to numpy and ordinary python slicing. See the example below.
>>> import pandas as pd
>>> a = pd.DataFrame([[0,1,2],[3,4,5],[4,5,6],[9,10,11],[34,2,1]])
>>> a
0 1 2
0 0 1 2
1 3 4 5
2 4 5 6
3 9 10 11
4 34 2 1
>>> a.loc[1:3, :]
0 1 2
1 3 4 5
2 4 5 6
3 9 10 11
>>> a.values[1:3, :]
array([[3, 4, 5],
[4, 5, 6]])
Interestingly, this only happens with .loc, not .iloc.
>>> a.iloc[1:3, :]
0 1 2
1 3 4 5
2 4 5 6
Thus, .loc appears to be inclusive of the terminating index, but numpy and .iloc are not.
By the comments, it seems this is not a bug and we are well warned. But why is it the case?

Remember .loc is primarily label based indexing. The decision to include the stop endpoint becomes far more obvious when working with a non-RangeIndex:
df = pd.DataFrame([1,2,3,4], index=list('achz'))
# 0
#a 1
#c 2
#h 3
#z 4
If I want to select all rows between 'a' and 'h' (inclusive) I only know about 'a' and 'h'. In order to be consistent with other python slicing, you'd need to also know what index follows 'h', which in this case is 'z' but could have been anything.
There's also a section of the documents hidden away that explains this design choice Endpoints are Inclusive

Additionally to the point in the docs, pandas slice indexing using .loc is not cell index based. It is in fact value based indexing (in the pandas docs it is called "label based", but for numerical data I prefer the term "value based"), whereas with .iloc it is traditional numpy-style cell indexing.
Furthermore, value based indexing is right-inclusive, whereas cell indexing is not. Just try the following:
a = pd.DataFrame([[0,1,2],[3,4,5],[4,5,6],[9,10,11],[34,2,1]])
a.index = [0, 1, 2, 3.1, 4] # add a float index
# value based slicing: the following will output all value up to the slice value
a.loc[1:3.1]
# Out:
# 0 1 2
# 1.0 3 4 5
# 2.0 4 5 6
# 3.1 9 10 11
# index based slicing: will raise an error, since only integers are allowed
a.iloc[1:3.1]
# Out: TypeError: cannot do slice indexing on <class 'pandas.core.indexes.numeric.Float64Index'> with these indexers [3.2] of <class 'float'>
To give an explicit answer to your question why it is right-inclusive:
When using values/labels as indices, it is, at least in my opinion, intuitive, that the last index is included. This is as far as I know a design decision of how the implemented function is meant to work.

Dynamically evaluate an expression from a formula in Pandas

I would like to perform arithmetic on one or more dataframes columns using pd.eval. Specifically, I would like to port the following code that evaluates a formula:
x = 5
df2['D'] = df1['A'] + (df1['B'] * x)
...to code using pd.eval. The reason for using pd.eval is that I would like to automate many workflows, so creating them dynamically will be useful to me.
My two input DataFrames are:
import pandas as pd
import numpy as np
np.random.seed(0)
df1 = pd.DataFrame(np.random.choice(10, (5, 4)), columns=list('ABCD'))
df2 = pd.DataFrame(np.random.choice(10, (5, 4)), columns=list('ABCD'))
df1
A B C D
0 5 0 3 3
1 7 9 3 5
2 2 4 7 6
3 8 8 1 6
4 7 7 8 1
df2
A B C D
0 5 9 8 9
1 4 3 0 3
2 5 0 2 3
3 8 1 3 3
4 3 7 0 1
I am trying to better understand pd.eval's engine and parser arguments to determine how best to solve my problem. I have gone through the documentation, but the difference was not made clear to me.
What arguments should be used to ensure my code is working at the maximum performance?
Is there a way to assign the result of the expression back to df2?
Also, to make things more complicated, how do I pass x as an argument inside the string expression?

You can use 1) pd.eval(), 2) df.query(), or 3) df.eval(). Their various features and functionality are discussed below.
Examples will involve these dataframes (unless otherwise specified).
np.random.seed(0)
df1 = pd.DataFrame(np.random.choice(10, (5, 4)), columns=list('ABCD'))
df2 = pd.DataFrame(np.random.choice(10, (5, 4)), columns=list('ABCD'))
df3 = pd.DataFrame(np.random.choice(10, (5, 4)), columns=list('ABCD'))
df4 = pd.DataFrame(np.random.choice(10, (5, 4)), columns=list('ABCD'))
1) pandas.eval
This is the "Missing Manual" that pandas doc should contain.
Note: of the three functions being discussed, pd.eval is the most important. df.eval and df.query call
pd.eval under the hood. Behaviour and usage is more or less
consistent across the three functions, with some minor semantic
variations which will be highlighted later. This section will
introduce functionality that is common across all the three functions - this includes, (but not limited to) allowed syntax, precedence rules, and keyword arguments.
pd.eval can evaluate arithmetic expressions which can consist of variables and/or literals. These expressions must be passed as strings. So, to answer the question as stated, you can do
x = 5
pd.eval("df1.A + (df1.B * x)")
Some things to note here:
The entire expression is a string
df1, df2, and x refer to variables in the global namespace, these are picked up by eval when parsing the expression
Specific columns are accessed using the attribute accessor index. You can also use "df1['A'] + (df1['B'] * x)" to the same effect.
I will be addressing the specific issue of reassignment in the section explaining the target=... attribute below. But for now, here are more simple examples of valid operations with pd.eval:
pd.eval("df1.A + df2.A") # Valid, returns a pd.Series object
pd.eval("abs(df1) ** .5") # Valid, returns a pd.DataFrame object
...and so on. Conditional expressions are also supported in the same way. The statements below are all valid expressions and will be evaluated by the engine.
pd.eval("df1 > df2")
pd.eval("df1 > 5")
pd.eval("df1 < df2 and df3 < df4")
pd.eval("df1 in [1, 2, 3]")
pd.eval("1 < 2 < 3")
A list detailing all the supported features and syntax can be found in the documentation. In summary,
Arithmetic operations except for the left shift (<<) and right shift (>>) operators, e.g., df + 2 * pi / s ** 4 % 42 - the_golden_ratio
Comparison operations, including chained comparisons, e.g., 2 < df < df2
Boolean operations, e.g., df < df2 and df3 < df4 or not df_bool
list and tuple literals, e.g., [1, 2] or (1, 2)
Attribute access, e.g., df.a
Subscript expressions, e.g., df[0]
Simple variable evaluation, e.g., pd.eval('df') (this is not very useful)
Math functions: sin, cos, exp, log, expm1, log1p, sqrt, sinh, cosh, tanh, arcsin, arccos, arctan, arccosh, arcsinh, arctanh, abs and
arctan2.
This section of the documentation also specifies syntax rules that are not supported, including set/dict literals, if-else statements, loops, and comprehensions, and generator expressions.
From the list, it is obvious you can also pass expressions involving the index, such as
pd.eval('df1.A * (df1.index > 1)')
1a) Parser Selection: The parser=... argument
pd.eval supports two different parser options when parsing the expression string to generate the syntax tree: pandas and python. The main difference between the two is highlighted by slightly differing precedence rules.
Using the default parser pandas, the overloaded bitwise operators & and | which implement vectorized AND and OR operations with pandas objects will have the same operator precedence as and and or. So,
pd.eval("(df1 > df2) & (df3 < df4)")
Will be the same as
pd.eval("df1 > df2 & df3 < df4")
# pd.eval("df1 > df2 & df3 < df4", parser='pandas')
And also the same as
pd.eval("df1 > df2 and df3 < df4")
Here, the parentheses are necessary. To do this conventionally, the parentheses would be required to override the higher precedence of bitwise operators:
(df1 > df2) & (df3 < df4)
Without that, we end up with
df1 > df2 & df3 < df4
ValueError: The truth value of a DataFrame is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
Use parser='python' if you want to maintain consistency with python's actual operator precedence rules while evaluating the string.
pd.eval("(df1 > df2) & (df3 < df4)", parser='python')
The other difference between the two types of parsers are the semantics of the == and != operators with list and tuple nodes, which have the similar semantics as in and not in respectively, when using the 'pandas' parser. For example,
pd.eval("df1 == [1, 2, 3]")
Is valid, and will run with the same semantics as
pd.eval("df1 in [1, 2, 3]")
OTOH, pd.eval("df1 == [1, 2, 3]", parser='python') will throw a NotImplementedError error.
1b) Backend Selection: The engine=... argument
There are two options - numexpr (the default) and python. The numexpr option uses the numexpr backend which is optimized for performance.
With Python backend, your expression is evaluated similar to just passing the expression to Python's eval function. You have the flexibility of doing more inside expressions, such as string operations, for instance.
df = pd.DataFrame({'A': ['abc', 'def', 'abacus']})
pd.eval('df.A.str.contains("ab")', engine='python')
0 True
1 False
2 True
Name: A, dtype: bool
Unfortunately, this method offers no performance benefits over the numexpr engine, and there are very few security measures to ensure that dangerous expressions are not evaluated, so use at your own risk! It is generally not recommended to change this option to 'python' unless you know what you're doing.
1c) local_dict and global_dict arguments
Sometimes, it is useful to supply values for variables used inside expressions, but not currently defined in your namespace. You can pass a dictionary to local_dict
For example:
pd.eval("df1 > thresh")
UndefinedVariableError: name 'thresh' is not defined
This fails because thresh is not defined. However, this works:
pd.eval("df1 > thresh", local_dict={'thresh': 10})
This is useful when you have variables to supply from a dictionary. Alternatively, with the Python engine, you could simply do this:
mydict = {'thresh': 5}
# Dictionary values with *string* keys cannot be accessed without
# using the 'python' engine.
pd.eval('df1 > mydict["thresh"]', engine='python')
But this is going to possibly be much slower than using the 'numexpr' engine and passing a dictionary to local_dict or global_dict. Hopefully, this should make a convincing argument for the use of these parameters.
1d) The target (+ inplace) argument, and Assignment Expressions
This is not often a requirement because there are usually simpler ways of doing this, but you can assign the result of pd.eval to an object that implements __getitem__ such as dicts, and (you guessed it) DataFrames.
Consider the example in the question
x = 5
df2['D'] = df1['A'] + (df1['B'] * x)
To assign a column "D" to df2, we do
pd.eval('D = df1.A + (df1.B * x)', target=df2)
A B C D
0 5 9 8 5
1 4 3 0 52
2 5 0 2 22
3 8 1 3 48
4 3 7 0 42
This is not an in-place modification of df2 (but it can be... read on). Consider another example:
pd.eval('df1.A + df2.A')
0 10
1 11
2 7
3 16
4 10
dtype: int32
If you wanted to (for example) assign this back to a DataFrame, you could use the target argument as follows:
df = pd.DataFrame(columns=list('FBGH'), index=df1.index)
df
F B G H
0 NaN NaN NaN NaN
1 NaN NaN NaN NaN
2 NaN NaN NaN NaN
3 NaN NaN NaN NaN
4 NaN NaN NaN NaN
df = pd.eval('B = df1.A + df2.A', target=df)
# Similar to
# df = df.assign(B=pd.eval('df1.A + df2.A'))
df
F B G H
0 NaN 10 NaN NaN
1 NaN 11 NaN NaN
2 NaN 7 NaN NaN
3 NaN 16 NaN NaN
4 NaN 10 NaN NaN
If you wanted to perform an in-place mutation on df, set inplace=True.
pd.eval('B = df1.A + df2.A', target=df, inplace=True)
# Similar to
# df['B'] = pd.eval('df1.A + df2.A')
df
F B G H
0 NaN 10 NaN NaN
1 NaN 11 NaN NaN
2 NaN 7 NaN NaN
3 NaN 16 NaN NaN
4 NaN 10 NaN NaN
If inplace is set without a target, a ValueError is raised.
While the target argument is fun to play around with, you will seldom need to use it.
If you wanted to do this with df.eval, you would use an expression involving an assignment:
df = df.eval("B = #df1.A + #df2.A")
# df.eval("B = #df1.A + #df2.A", inplace=True)
df
F B G H
0 NaN 10 NaN NaN
1 NaN 11 NaN NaN
2 NaN 7 NaN NaN
3 NaN 16 NaN NaN
4 NaN 10 NaN NaN
Note
One of pd.eval's unintended uses is parsing literal strings in a manner very similar to ast.literal_eval:
pd.eval("[1, 2, 3]")
array([1, 2, 3], dtype=object)
It can also parse nested lists with the 'python' engine:
pd.eval("[[1, 2, 3], [4, 5], [10]]", engine='python')
[[1, 2, 3], [4, 5], [10]]
And lists of strings:
pd.eval(["[1, 2, 3]", "[4, 5]", "[10]"], engine='python')
[[1, 2, 3], [4, 5], [10]]
The problem, however, is for lists with length larger than 100:
pd.eval(["[1]"] * 100, engine='python') # Works
pd.eval(["[1]"] * 101, engine='python')
AttributeError: 'PandasExprVisitor' object has no attribute 'visit_Ellipsis'
More information can this error, causes, fixes, and workarounds can be found here.
2) DataFrame.eval:
As mentioned above, df.eval calls pd.eval under the hood, with a bit of juxtaposition of arguments. The v0.23 source code shows this:
def eval(self, expr, inplace=False, **kwargs):
from pandas.core.computation.eval import eval as _eval
inplace = validate_bool_kwarg(inplace, 'inplace')
resolvers = kwargs.pop('resolvers', None)
kwargs['level'] = kwargs.pop('level', 0) + 1
if resolvers is None:
index_resolvers = self._get_index_resolvers()
resolvers = dict(self.iteritems()), index_resolvers
if 'target' not in kwargs:
kwargs['target'] = self
kwargs['resolvers'] = kwargs.get('resolvers', ()) + tuple(resolvers)
return _eval(expr, inplace=inplace, **kwargs)
eval creates arguments, does a little validation, and passes the arguments on to pd.eval.
For more, you can read on: When to use DataFrame.eval() versus pandas.eval() or Python eval()
2a) Usage Differences
2a1) Expressions with DataFrames vs. Series Expressions
For dynamic queries associated with entire DataFrames, you should prefer pd.eval. For example, there is no simple way to specify the equivalent of pd.eval("df1 + df2") when you call df1.eval or df2.eval.
2a2) Specifying Column Names
Another other major difference is how columns are accessed. For example, to add two columns "A" and "B" in df1, you would call pd.eval with the following expression:
pd.eval("df1.A + df1.B")
With df.eval, you need only supply the column names:
df1.eval("A + B")
Since, within the context of df1, it is clear that "A" and "B" refer to column names.
You can also refer to the index and columns using index (unless the index is named, in which case you would use the name).
df1.eval("A + index")
Or, more generally, for any DataFrame with an index having 1 or more levels, you can refer to the kth level of the index in an expression using the variable "ilevel_k" which stands for "index at level k". IOW, the expression above can be written as df1.eval("A + ilevel_0").
These rules also apply to df.query.
2a3) Accessing Variables in Local/Global Namespace
Variables supplied inside expressions must be preceded by the "#" symbol, to avoid confusion with column names.
A = 5
df1.eval("A > #A")
The same goes for query.
It goes without saying that your column names must follow the rules for valid identifier naming in Python to be accessible inside eval. See here for a list of rules on naming identifiers.
2a4) Multiline Queries and Assignment
A little known fact is that eval supports multiline expressions that deal with assignment (whereas query doesn't). For example, to create two new columns "E" and "F" in df1 based on some arithmetic operations on some columns, and a third column "G" based on the previously created "E" and "F", we can do
df1.eval("""
E = A + B
F = #df2.A + #df2.B
G = E >= F
""")
A B C D E F G
0 5 0 3 3 5 14 False
1 7 9 3 5 16 7 True
2 2 4 7 6 6 5 True
3 8 8 1 6 16 9 True
4 7 7 8 1 14 10 True
3) eval vs query
It helps to think of df.query as a function that uses pd.eval as a subroutine.
Typically, query (as the name suggests) is used to evaluate conditional expressions (i.e., expressions that result in True/False values) and return the rows corresponding to the True result. The result of the expression is then passed to loc (in most cases) to return the rows that satisfy the expression. According to the documentation,
The result of the evaluation of this expression is first passed to
DataFrame.loc and if that fails because of a multidimensional key
(e.g., a DataFrame) then the result will be passed to
DataFrame.__getitem__().
This method uses the top-level pandas.eval() function to evaluate the
passed query.
In terms of similarity, query and df.eval are both alike in how they access column names and variables.
This key difference between the two, as mentioned above is how they handle the expression result. This becomes obvious when you actually run an expression through these two functions. For example, consider
df1.A
0 5
1 7
2 2
3 8
4 7
Name: A, dtype: int32
df1.B
0 9
1 3
2 0
3 1
4 7
Name: B, dtype: int32
To get all rows where "A" >= "B" in df1, we would use eval like this:
m = df1.eval("A >= B")
m
0 True
1 False
2 False
3 True
4 True
dtype: bool
m represents the intermediate result generated by evaluating the expression "A >= B". We then use the mask to filter df1:
df1[m]
# df1.loc[m]
A B C D
0 5 0 3 3
3 8 8 1 6
4 7 7 8 1
However, with query, the intermediate result "m" is directly passed to loc, so with query, you would simply need to do
df1.query("A >= B")
A B C D
0 5 0 3 3
3 8 8 1 6
4 7 7 8 1
Performance wise, it is exactly the same.
df1_big = pd.concat([df1] * 100000, ignore_index=True)
%timeit df1_big[df1_big.eval("A >= B")]
%timeit df1_big.query("A >= B")
14.7 ms ± 33.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
14.7 ms ± 24.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
But the latter is more concise, and expresses the same operation in a single step.
Note that you can also do weird stuff with query like this (to, say, return all rows indexed by df1.index)
df1.query("index")
# Same as df1.loc[df1.index] # Pointless,... I know
A B C D
0 5 0 3 3
1 7 9 3 5
2 2 4 7 6
3 8 8 1 6
4 7 7 8 1
But don't.
Bottom line: Please use query when querying or filtering rows based on a conditional expression.

There are great tutorials already, but bear in mind that before jumping wildly into the usage of eval/query attracted by its simpler syntax, it has severe performance issues if your dataset has less than 15,000 rows.
In that case, simply use df.loc[mask1, mask2].
Refer to: Expression Evaluation via eval()

Pandas dataframe: creating a new column that is a custom function using 2 other columns

Consider the following data set stored in a pandas DataFrame dfX:
A B
1 2
4 6
7 9
I have a function that is:
def someThingSpecial(x,y)
# z = do something special with x,y
return z
I now want to create a new column in df that bears the computed z value
Looking at other SO examples, I've tried several variants including:
dfX['C'] = dfX.apply(lambda x: someThingSpecial(x=x['A'], y=x['B']), axis=1)
Which returns errors. What is the right way to do this?

This seems to work for me on v0.21. Take a look -
df
A B
0 1 2
1 4 6
2 7 9
def someThingSpecial(x,y):
return x + y
df.apply(lambda x: someThingSpecial(x.A, x.B), 1)
0 3
1 10
2 16
dtype: int64
You might want to try upgrading your pandas version to the latest stable release (0.21 as of now).
Here's another option. You can vectorise your function.
v = np.vectorize(someThingSpecial)
v now accepts arrays, but operates on each pair of elements individually. Note that this just hides the loop, as apply does, but is much cleaner. Now, you can compute C as so -
df['C'] = v(df.A, df.B)

if your function only needs one column's value, then do this instead of coldspeed's answer:
dfX['A'].apply(your_func)
to store it:
dfX['C'] = dfX['A'].apply(your_func)

simplifying routine in python with numpy array or pandas

The initial problem is the following: I have an initial matrix with let say 10 lines and 12 rows. For all lines, I want to sum two rows together. At the end I must have 10 lines but with only 6 rows. Currently, I am doing the following for loop in python (using initial which is a pandas DataFrame)
for i in range(0,12,2):
coarse[i]=initial.iloc[:,i:i+1].sum(axis=1)
In fact, I am quite sure that something more efficient is possible. I am thinking something like list comprehension but for a DataFrame or a numpy array. Does anybody have an idea ?
Moreover I would want to know if it is better to manipulate large numpy arrays or pandas DataFrame.

Let's create a small sample dataframe to illustrate the solution:
np.random.seed(0)
df = pd.DataFrame(np.random.rand(6, 3))
>>> df
0 1 2
0 0.548814 0.715189 0.602763
1 0.544883 0.423655 0.645894
2 0.437587 0.891773 0.963663
3 0.383442 0.791725 0.528895
4 0.568045 0.925597 0.071036
5 0.087129 0.020218 0.832620
You can use slice notation to select every other row starting from the first row (::2) and starting from the second row (1::2). iloc is for integer indexing. You need to select the values at these locations, and add them together. The result is a numpy array that you could then convert back into a DataFrame if required.
>>> df.iloc[::2].values + df.iloc[1::2].values
array([[ 1.09369669, 1.13884417, 1.24865749],
[ 0.82102873, 1.68349804, 1.49255768],
[ 0.65517386, 0.94581504, 0.9036559 ]])
You use values to remove the indexing. This is what happens otherwise:
>>> df.iloc[::2] + df.iloc[1::2].values
0 1 2
0 1.093697 1.138844 1.248657
2 0.821029 1.683498 1.492558
4 0.655174 0.945815 0.903656
>>> df.iloc[::2].values + df.iloc[1::2]
0 1 2
1 1.093697 1.138844 1.248657
3 0.821029 1.683498 1.492558
5 0.655174 0.945815 0.903656
For a more general solution:
df = pd.DataFrame(np.random.rand(9, 3))
n = 3 # Number of consecutive rows to group.
df['group'] = [idx // n for idx in range(len(df.index))]
df.groupby('group').sum()
0 1 2
group
0 1.531284 2.030617 2.212320
1 1.038615 1.737540 1.432551
2 1.695590 1.971413 1.902501

is there any quick function to do looking-back calculating in pandas dataframe?

I wanna implement a calculate method like a simple scenario:
value computed as the sum of daily data during the previous N days (set N = 3 in the following example)
Dataframe df: (df.index is 'date')
date value
20140718 1
20140721 2
20140722 3
20140723 4
20140724 5
20140725 6
20140728 7
......
to do calculating like:
date value new
20140718 1 0
20140721 2 0
20140722 3 0
20140723 4 6 (3+2+1)
20140724 5 9 (4+3+2)
20140725 6 12 (5+4+3)
20140728 7 15 (6+5+4)
......
Now I have done this using for cycle like:
df['value']=[0]*len(df)
for idx in df.index
loc=df.index.get_loc(idx)
if((loc-N)>=0):
tmp=df.ix[df.index[loc-3]:df.index[loc-1]]
sum=tmp['value'].sum()
else:
sum=0
df['new'].ix(idx)=sum
But, when the length of dataframe or the value of N is very long / big, these calculating will be very slow....How I can implement this faster using a function or by other ways?
Besides, if the scenario is more complex? how ? Thanks.

Since you want the sum of the previous three excluding the current one, you can use rolling_apply over the a window of four and sum up all but the last value.
new = rolling_apply(df, 4, lambda x:sum(x[:-1]), min_periods=4)
This is the same as shifting afterwards with a window of three:
new = rolling_apply(df, 3, sum, min_periods=3).shift()
Then
df["new"] = new["value"].fillna(0)