Note
This post will be structured in the following manner:
- The questions put forth in the OP will be addressed, one by one
- For each question, one or more methods applicable to solving this problem and getting the expected result will be demonstrated.
Notes (much like this one) will be included for readers interested in learning about additional functionality, implementation details,
and other info cursory to the topic at hand. These notes have been
compiled through scouring the docs and uncovering various obscure
features, and from my own (admittedly limited) experience.
All code samples have created and tested on pandas v0.23.4, python3.7. If something is not clear, or factually incorrect, or if you did not
find a solution applicable to your use case, please feel free to
suggest an edit, request clarification in the comments, or open a new
question, ....as applicable.
Here is an introduction to some common idioms (henceforth referred to as the Four Idioms) we will be frequently re-visiting
DataFrame.loc
- A general solution for selection by label (+ pd.IndexSlice
for more complex applications involving slices)
DataFrame.xs
- Extract a particular cross section from a Series/DataFrame.
DataFrame.query
- Specify slicing and/or filtering operations dynamically (i.e., as an expression that is evaluated dynamically. Is more applicable to some scenarios than others. Also see this section of the docs for querying on MultiIndexes.
Boolean indexing with a mask generated using MultiIndex.get_level_values
(often in conjunction with Index.isin
, especially when filtering with multiple values). This is also quite useful in some circumstances.
It will be beneficial to look at the various slicing and filtering problems in terms of the Four Idioms to gain a better understanding what can be applied to a given situation. It is very important to understand that not all of the idioms will work equally well (if at all) in every circumstance. If an idiom has not been listed as a potential solution to a problem below, that means that idiom cannot be applied to that problem effectively.
Question 1
How do I select rows having "a" in level "one"?
col
one two
a t 0
u 1
v 2
w 3
You can use loc
, as a general purpose solution applicable to most situations:
df.loc[['a']]
At this point, if you get
TypeError: Expected tuple, got str
That means you're using an older version of pandas. Consider upgrading! Otherwise, use df.loc[('a', slice(None)), :]
.
Alternatively, you can use xs
here, since we are extracting a single cross section. Note the levels
and axis
arguments (reasonable defaults can be assumed here).
df.xs('a', level=0, axis=0, drop_level=False)
# df.xs('a', drop_level=False)
Here, the drop_level=False
argument is needed to prevent xs
from dropping level "one" in the result (the level we sliced on).
Yet another option here is using query
:
df.query("one == 'a'")
If the index did not have a name, you would need to change your query string to be "ilevel_0 == 'a'"
.
Finally, using get_level_values
:
df[df.index.get_level_values('one') == 'a']
# If your levels are unnamed, or if you need to select by position (not label),
# df[df.index.get_level_values(0) == 'a']
Additionally, how would I be able to drop level "one" in the output?
col
two
t 0
u 1
v 2
w 3
This can be easily done using either
df.loc['a'] # Notice the single string argument instead the list.
Or,
df.xs('a', level=0, axis=0, drop_level=True)
# df.xs('a')
Notice that we can omit the drop_level
argument (it is assumed to be True
by default).
Note
You may notice that a filtered DataFrame may still have all the levels, even if they do not show when printing the DataFrame out. For example,
v = df.loc[['a']]
print(v)
col
one two
a t 0
u 1
v 2
w 3
print(v.index)
MultiIndex(levels=[['a', 'b', 'c', 'd'], ['t', 'u', 'v', 'w']],
labels=[[0, 0, 0, 0], [0, 1, 2, 3]],
names=['one', 'two'])
You can get rid of these levels using MultiIndex.remove_unused_levels
:
v.index = v.index.remove_unused_levels()
print(v.index)
MultiIndex(levels=[['a'], ['t', 'u', 'v', 'w']],
labels=[[0, 0, 0, 0], [0, 1, 2, 3]],
names=['one', 'two'])
Question 1b
How do I slice all rows with value "t" on level "two"?
col
one two
a t 0
b t 4
t 8
d t 12
Intuitively, you would want something involving slice()
:
df.loc[(slice(None), 't'), :]
It Just Works!? But it is clunky. We can facilitate a more natural slicing syntax using the pd.IndexSlice
API here.
idx = pd.IndexSlice
df.loc[idx[:, 't'], :]
This is much, much cleaner.
Note
Why is the trailing slice :
across the columns required? This is because, loc
can be used to select and slice along both axes (axis=0
or
axis=1
). Without explicitly making it clear which axis the slicing
is to be done on, the operation becomes ambiguous. See the big red box in the documentation on slicing.
If you want to remove any shade of ambiguity, loc
accepts an axis
parameter:
df.loc(axis=0)[pd.IndexSlice[:, 't']]
Without the axis
parameter (i.e., just by doing df.loc[pd.IndexSlice[:, 't']]
), slicing is assumed to be on the columns,
and a KeyError
will be raised in this circumstance.
This is documented in slicers. For the purpose of this post, however, we will explicitly specify all axes.
With xs
, it is
df.xs('t', axis=0, level=1, drop_level=False)
With query
, it is
df.query("two == 't'")
# Or, if the first level has no name,
# df.query("ilevel_1 == 't'")
And finally, with get_level_values
, you may do
df[df.index.get_level_values('two') == 't']
# Or, to perform selection by position/integer,
# df[df.index.get_level_values(1) == 't']
All to the same effect.
Question 2
How can I select rows corresponding to items "b" and "d" in level "one"?
col
one two
b t 4
u 5
v 6
w 7
t 8
d w 11
t 12
u 13
v 14
w 15
Using loc, this is done in a similar fashion by specifying a list.
df.loc[['b', 'd']]
To solve the above problem of selecting "b" and "d", you can also use query
:
items = ['b', 'd']
df.query("one in @items")
# df.query("one == @items", parser='pandas')
# df.query("one in ['b', 'd']")
# df.query("one == ['b', 'd']", parser='pandas')
Note
Yes, the default parser is 'pandas'
, but it is important to highlight this syntax isn't conventionally python. The
Pandas parser generates a slightly different parse tree from the
expression. This is done to make some operations more intuitive to
specify. For more information, please read my post on
Dynamic Expression Evaluation in pandas using pd.eval().
And, with get_level_values
+ Index.isin
:
df[df.index.get_level_values("one").isin(['b', 'd'])]
Question 2b
How would I get all values corresponding to "t" and "w" in level "two"?
col
one two
a t 0
w 3
b t 4
w 7
t 8
d w 11
t 12
w 15
With loc
, this is possible only in conjuction with pd.IndexSlice
.
df.loc[pd.IndexSlice[:, ['t', 'w']], :]
The first colon :
in pd.IndexSlice[:, ['t', 'w']]
means to slice across the first level. As the depth of the level being queried increases, you will need to specify more slices, one per level being sliced across. You will not need to specify more levels beyond the one being sliced, however.
With query
, this is
items = ['t', 'w']
df.query("two in @items")
# df.query("two == @items", parser='pandas')
# df.query("two in ['t', 'w']")
# df.query("two == ['t', 'w']", parser='pandas')
With get_level_values
and Index.isin
(similar to above):
df[df.index.get_level_values('two').isin(['t', 'w'])]
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