Select rows in pandas MultiIndex DataFrame
Recently I came across a use case where I had a 3+ level multi-index dataframe in which I couldn't make any of the solutions above produce the results I was looking for. It's quite possible that the above solutions do of course work for my use case, and I tried several, however I was unable to get them to work with the time I had available.
I am far from expert, but I stumbled across a solution that was not listed in the comprehensive answers above. I offer no guarantee that the solutions are in any way optimal.
This is a different way to get a slightly different result to Question #6 above. (and likely other questions as well)
Specifically I was looking for:
- A way to choose two+ values from one level of the index and a single value from another level of the index, and
- A way to leave the index values from the previous operation in the dataframe output.
As a monkey wrench in the gears (however totally fixable):
- The indexes were unnamed.
On the toy dataframe below:
index = pd.MultiIndex.from_product([['a','b'],
['stock1','stock2','stock3'],
['price','volume','velocity']])
df = pd.DataFrame([1,2,3,4,5,6,7,8,9,
10,11,12,13,14,15,16,17,18],
index)
0
a stock1 price 1
volume 2
velocity 3
stock2 price 4
volume 5
velocity 6
stock3 price 7
volume 8
velocity 9
b stock1 price 10
volume 11
velocity 12
stock2 price 13
volume 14
velocity 15
stock3 price 16
volume 17
velocity 18
Using the below works, of course:
df.xs(('stock1', 'velocity'), level=(1,2))
0
a 3
b 12
But I wanted a different result, so my method to get that result was:
df.iloc[df.index.isin(['stock1'], level=1) &
df.index.isin(['velocity'], level=2)]
0
a stock1 velocity 3
b stock1 velocity 12
And if I wanted two+ values from one level and a single (or 2+) value from another level:
df.iloc[df.index.isin(['stock1','stock3'], level=1) &
df.index.isin(['velocity'], level=2)]
0
a stock1 velocity 3
stock3 velocity 9
b stock1 velocity 12
stock3 velocity 18
The above method is probably a bit clunky, however I found it filled my needs and as a bonus was easier for me to understand and read.
This looks like a great case for dfsql
df.sql(<SQL select statement>)
https://github.com/mindsdb/dfsql
A complete article about it here:
https://medium.com/riselab/why-every-data-scientist-using-pandas-needs-modin-bringing-sql-to-dataframes-3b216b29a7c0
MultiIndex / Advanced Indexing
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.IndexSlicefor 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 withIndex.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,loccan be used to select and slice along both axes (axis=0oraxis=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,
locaccepts anaxisparameter:df.loc(axis=0)[pd.IndexSlice[:, 't']]Without the
axisparameter (i.e., just by doingdf.loc[pd.IndexSlice[:, 't']]), slicing is assumed to be on the columns, and aKeyErrorwill 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'])]
Question 3
How do I retrieve a cross section, i.e., a single row having a specific values for the index from
df? Specifically, how do I retrieve the cross section of('c', 'u'), given bycol one two c u 9
Use loc by specifying a tuple of keys:
df.loc[('c', 'u'), :]
Or,
df.loc[pd.IndexSlice[('c', 'u')]]
Note
At this point, you may run into aPerformanceWarningthat looks like this:PerformanceWarning: indexing past lexsort depth may impact performance.This just means that your index is not sorted. pandas depends on the index being sorted (in this case, lexicographically, since we are dealing with string values) for optimal search and retrieval. A quick fix would be to sort your DataFrame in advance using
DataFrame.sort_index. This is especially desirable from a performance standpoint if you plan on doing multiple such queries in tandem:df_sort = df.sort_index() df_sort.loc[('c', 'u')]You can also use
MultiIndex.is_lexsorted()to check whether the index is sorted or not. This function returnsTrueorFalseaccordingly. You can call this function to determine whether an additional sorting step is required or not.
With xs, this is again simply passing a single tuple as the first argument, with all other arguments set to their appropriate defaults:
df.xs(('c', 'u'))
With query, things become a bit clunky:
df.query("one == 'c' and two == 'u'")
You can see now that this is going to be relatively difficult to generalize. But is still OK for this particular problem.
With accesses spanning multiple levels, get_level_values can still be used, but is not recommended:
m1 = (df.index.get_level_values('one') == 'c')
m2 = (df.index.get_level_values('two') == 'u')
df[m1 & m2]
Question 4
How do I select the two rows corresponding to
('c', 'u'), and('a', 'w')?col one two c u 9 a w 3
With loc, this is still as simple as:
df.loc[[('c', 'u'), ('a', 'w')]]
# df.loc[pd.IndexSlice[[('c', 'u'), ('a', 'w')]]]
With query, you will need to dynamically generate a query string by iterating over your cross sections and levels:
cses = [('c', 'u'), ('a', 'w')]
levels = ['one', 'two']
# This is a useful check to make in advance.
assert all(len(levels) == len(cs) for cs in cses)
query = '(' + ') or ('.join([
' and '.join([f"({l} == {repr(c)})" for l, c in zip(levels, cs)])
for cs in cses
]) + ')'
print(query)
# ((one == 'c') and (two == 'u')) or ((one == 'a') and (two == 'w'))
df.query(query)
100% DO NOT RECOMMEND! But it is possible.
What if I have multiple levels?
One option in this scenario would be to use droplevel to drop the levels you're not checking, then use isin to test membership, and then boolean index on the final result.
df[df.index.droplevel(unused_level).isin([('c', 'u'), ('a', 'w')])]
Question 5
How can I retrieve all rows corresponding to "a" in level "one" or "t" in level "two"?
col one two a t 0 u 1 v 2 w 3 b t 4 t 8 d t 12
This is actually very difficult to do with loc while ensuring correctness and still maintaining code clarity. df.loc[pd.IndexSlice['a', 't']] is incorrect, it is interpreted as df.loc[pd.IndexSlice[('a', 't')]] (i.e., selecting a cross section). You may think of a solution with pd.concat to handle each label separately:
pd.concat([
df.loc[['a'],:], df.loc[pd.IndexSlice[:, 't'],:]
])
col
one two
a t 0
u 1
v 2
w 3
t 0 # Does this look right to you? No, it isn't!
b t 4
t 8
d t 12
But you'll notice one of the rows is duplicated. This is because that row satisfied both slicing conditions, and so appeared twice. You will instead need to do
v = pd.concat([
df.loc[['a'],:], df.loc[pd.IndexSlice[:, 't'],:]
])
v[~v.index.duplicated()]
But if your DataFrame inherently contains duplicate indices (that you want), then this will not retain them. Use with extreme caution.
With query, this is stupidly simple:
df.query("one == 'a' or two == 't'")
With get_level_values, this is still simple, but not as elegant:
m1 = (df.index.get_level_values('one') == 'a')
m2 = (df.index.get_level_values('two') == 't')
df[m1 | m2]
Question 6
How can I slice specific cross sections? For "a" and "b", I would like to select all rows with sub-levels "u" and "v", and for "d", I would like to select rows with sub-level "w".
col one two a u 1 v 2 b u 5 v 6 d w 11 w 15
This is a special case that I've added to help understand the applicability of the Four Idioms—this is one case where none of them will work effectively, since the slicing is very specific, and does not follow any real pattern.
Usually, slicing problems like this will require explicitly passing a list of keys to loc. One way of doing this is with:
keys = [('a', 'u'), ('a', 'v'), ('b', 'u'), ('b', 'v'), ('d', 'w')]
df.loc[keys, :]
If you want to save some typing, you will recognise that there is a pattern to slicing "a", "b" and its sublevels, so we can separate the slicing task into two portions and concat the result:
pd.concat([
df.loc[(('a', 'b'), ('u', 'v')), :],
df.loc[('d', 'w'), :]
], axis=0)
Slicing specification for "a" and "b" is slightly cleaner (('a', 'b'), ('u', 'v')) because the same sub-levels being indexed are the same for each level.
Question 7
How do I get all rows where values in level "two" are greater than 5?
col one two b 7 4 9 5 c 7 10 d 6 11 8 12 8 13 6 15
This can be done using query,
df2.query("two > 5")
And get_level_values.
df2[df2.index.get_level_values('two') > 5]
Note
Similar to this example, we can filter based on any arbitrary condition using these constructs. In general, it is useful to remember thatlocandxsare specifically for label-based indexing, whilequeryandget_level_valuesare helpful for building general conditional masks for filtering.
Bonus Question
What if I need to slice a
MultiIndexcolumn?
Actually, most solutions here are applicable to columns as well, with minor changes. Consider:
np.random.seed(0)
mux3 = pd.MultiIndex.from_product([
list('ABCD'), list('efgh')
], names=['one','two'])
df3 = pd.DataFrame(np.random.choice(10, (3, len(mux))), columns=mux3)
print(df3)
one A B C D
two e f g h e f g h e f g h e f g h
0 5 0 3 3 7 9 3 5 2 4 7 6 8 8 1 6
1 7 7 8 1 5 9 8 9 4 3 0 3 5 0 2 3
2 8 1 3 3 3 7 0 1 9 9 0 4 7 3 2 7
These are the following changes you will need to make to the Four Idioms to have them working with columns.
To slice with
loc, usedf3.loc[:, ....] # Notice how we slice across the index with `:`.or,
df3.loc[:, pd.IndexSlice[...]]To use
xsas appropriate, just pass an argumentaxis=1.You can access the column level values directly using
df.columns.get_level_values. You will then need to do something likedf.loc[:, {condition}]Where
{condition}represents some condition built usingcolumns.get_level_values.To use
query, your only option is to transpose, query on the index, and transpose again:df3.T.query(...).TNot recommended, use one of the other 3 options.