In the process of answering this question for myself, I learned many things, and I wanted to put together a catalog of examples and some explanation.
The specific answer to the point of the levels argument will come towards the end.
pandas.concat: The Missing Manual
Link To Current Documentation
Imports and defining objects
import pandas as pd
d1 = pd.DataFrame(dict(A=.1, B=.2, C=.3), index=[2, 3])
d2 = pd.DataFrame(dict(B=.4, C=.5, D=.6), index=[1, 2])
d3 = pd.DataFrame(dict(A=.7, B=.8, D=.9), index=[1, 3])
s1 = pd.Series([1, 2], index=[2, 3])
s2 = pd.Series([3, 4], index=[1, 2])
s3 = pd.Series([5, 6], index=[1, 3])
Arguments
objs
The first argument we come across is objs:
objs: a sequence or mapping of Series, DataFrame, or Panel objects
If a dict is passed, the sorted keys will be used as the keys argument, unless it is passed, in which case the values will be selected (see below). Any None objects will be dropped silently unless they are all None in which case a ValueError will be raised
- We typically see this used with a list of
Series or DataFrame objects.
- I'll show that
dict can be very useful as well.
- Generators may also be used and can be useful when using
map as in map(f, list_of_df)
For now, we'll stick with a list of some of the DataFrame and Series objects defined above.
I'll show how dictionaries can be leveraged to give very useful MultiIndex results later.
pd.concat([d1, d2])
A B C D
2 0.1 0.2 0.3 NaN
3 0.1 0.2 0.3 NaN
1 NaN 0.4 0.5 0.6
2 NaN 0.4 0.5 0.6
axis
The second argument we encounter is axis whose default value is 0:
axis: {0/’index’, 1/’columns’}, default 0
The axis to concatenate along.
Two DataFrames with axis=0 (stacked)
For values of 0 or index we mean to say: "Align along the columns and add to the index".
As shown above where we used axis=0, because 0 is the default value, and we see that the index of d2 extends the index of d1 despite there being overlap of the value 2:
pd.concat([d1, d2], axis=0)
A B C D
2 0.1 0.2 0.3 NaN
3 0.1 0.2 0.3 NaN
1 NaN 0.4 0.5 0.6
2 NaN 0.4 0.5 0.6
Two DataFrames with axis=1 (side by side)
For values 1 or columns we mean to say: "Align along the index and add to the columns",
pd.concat([d1, d2], axis=1)
A B C B C D
1 NaN NaN NaN 0.4 0.5 0.6
2 0.1 0.2 0.3 0.4 0.5 0.6
3 0.1 0.2 0.3 NaN NaN NaN
We can see that the resulting index is the union of indices and the resulting columns are the extension of columns from d1 by the columns of d2.
Two (or Three) Series with axis=0 (stacked)
When combining pandas.Series along axis=0, we get back a pandas.Series. The name of the resulting Series will be None unless all Series being combined have the same name. Pay attention to the 'Name: A' when we print out the resulting Series. When it isn't present, we can assume the Series name is None.
| | | pd.concat(
| pd.concat( | pd.concat( | [s1.rename('A'),
pd.concat( | [s1.rename('A'), | [s1.rename('A'), | s2.rename('B'),
[s1, s2]) | s2]) | s2.rename('A')]) | s3.rename('A')])
-------------- | --------------------- | ---------------------- | ----------------------
2 1 | 2 1 | 2 1 | 2 1
3 2 | 3 2 | 3 2 | 3 2
1 3 | 1 3 | 1 3 | 1 3
2 4 | 2 4 | 2 4 | 2 4
dtype: int64 | dtype: int64 | Name: A, dtype: int64 | 1 5
| | | 3 6
| | | dtype: int64
Two (or Three) Series with axis=1 (side by side)
When combining pandas.Series along axis=1, it is the name attribute that we refer to in order to infer a column name in the resulting pandas.DataFrame.
| | pd.concat(
| pd.concat( | [s1.rename('X'),
pd.concat( | [s1.rename('X'), | s2.rename('Y'),
[s1, s2], axis=1) | s2], axis=1) | s3.rename('Z')], axis=1)
---------------------- | --------------------- | ------------------------------
0 1 | X 0 | X Y Z
1 NaN 3.0 | 1 NaN 3.0 | 1 NaN 3.0 5.0
2 1.0 4.0 | 2 1.0 4.0 | 2 1.0 4.0 NaN
3 2.0 NaN | 3 2.0 NaN | 3 2.0 NaN 6.0
Mixed Series and DataFrame with axis=0 (stacked)
When performing a concatenation of a Series and DataFrame along axis=0, we convert all Series to single column DataFrames.
Take special note that this is a concatenation along axis=0; that means extending the index (rows) while aligning the columns. In the examples below, we see the index becomes [2, 3, 2, 3] which is an indiscriminate appending of indices. The columns do not overlap unless I force the naming of the Series column with the argument to to_frame:
pd.concat( |
[s1.to_frame(), d1]) | pd.concat([s1, d1])
------------------------- | ---------------------
0 A B C | 0 A B C
2 1.0 NaN NaN NaN | 2 1.0 NaN NaN NaN
3 2.0 NaN NaN NaN | 3 2.0 NaN NaN NaN
2 NaN 0.1 0.2 0.3 | 2 NaN 0.1 0.2 0.3
3 NaN 0.1 0.2 0.3 | 3 NaN 0.1 0.2 0.3
You can see the results of pd.concat([s1, d1]) are the same as if I had perfromed the to_frame myself.
However, I can control the name of the resulting column with a parameter to to_frame. Renaming the Series with the rename method does not control the column name in the resulting DataFrame.
# Effectively renames | |
# `s1` but does not align | # Does not rename. So | # Renames to something
# with columns in `d1` | # Pandas defaults to `0` | # that does align with `d1`
pd.concat( | pd.concat( | pd.concat(
[s1.to_frame('X'), d1]) | [s1.rename('X'), d1]) | [s1.to_frame('B'), d1])
---------------------------- | -------------------------- | ----------------------------
A B C X | 0 A B C | A B C
2 NaN NaN NaN 1.0 | 2 1.0 NaN NaN NaN | 2 NaN 1.0 NaN
3 NaN NaN NaN 2.0 | 3 2.0 NaN NaN NaN | 3 NaN 2.0 NaN
2 0.1 0.2 0.3 NaN | 2 NaN 0.1 0.2 0.3 | 2 0.1 0.2 0.3
3 0.1 0.2 0.3 NaN | 3 NaN 0.1 0.2 0.3 | 3 0.1 0.2 0.3
Mixed Series and DataFrame with axis=1 (side by side)
This is fairly intuitive. Series column name defaults to an enumeration of such Series objects when a name attribute is not available.
| pd.concat(
pd.concat( | [s1.rename('X'),
[s1, d1], | s2, s3, d1],
axis=1) | axis=1)
------------------- | -------------------------------
0 A B C | X 0 1 A B C
2 1 0.1 0.2 0.3 | 1 NaN 3.0 5.0 NaN NaN NaN
3 2 0.1 0.2 0.3 | 2 1.0 4.0 NaN 0.1 0.2 0.3
| 3 2.0 NaN 6.0 0.1 0.2 0.3
join
The third argument is join that describes whether the resulting merge should be an outer merge (default) or an inner merge.
join: {‘inner’, ‘outer’}, default ‘outer’
How to handle indexes on other axis(es).
It turns out, there is no left or right option as pd.concat can handle more than just two objects to merge.
In the case of d1 and d2, the options look like:
outer
pd.concat([d1, d2], axis=1, join='outer')
A B C B C D
1 NaN NaN NaN 0.4 0.5 0.6
2 0.1 0.2 0.3 0.4 0.5 0.6
3 0.1 0.2 0.3 NaN NaN NaN
inner
pd.concat([d1, d2], axis=1, join='inner')
A B C B C D
2 0.1 0.2 0.3 0.4 0.5 0.6
join_axes
Fourth argument is the thing that allows us to do our left merge and more.
join_axes: list of Index objects
Specific indexes to use for the other n - 1 axes instead of performing inner/outer set logic.
Left Merge
pd.concat([d1, d2, d3], axis=1, join_axes=[d1.index])
A B C B C D A B D
2 0.1 0.2 0.3 0.4 0.5 0.6 NaN NaN NaN
3 0.1 0.2 0.3 NaN NaN NaN 0.7 0.8 0.9
Right Merge
pd.concat([d1, d2, d3], axis=1, join_axes=[d3.index])
A B C B C D A B D
1 NaN NaN NaN 0.4 0.5 0.6 0.7 0.8 0.9
3 0.1 0.2 0.3 NaN NaN NaN 0.7 0.8 0.9
ignore_index
ignore_index: boolean, default False
If True, do not use the index values along the concatenation axis. The resulting axis will be labeled 0, ..., n - 1. This is useful if you are concatenating objects where the concatenation axis does not have meaningful indexing information. Note the index values on the other axes are still respected in the join.
Like when I stack d1 on top of d2, if I don't care about the index values, I could reset them or ignore them.
| pd.concat( | pd.concat(
| [d1, d2], | [d1, d2]
pd.concat([d1, d2]) | ignore_index=True) | ).reset_index(drop=True)
--------------------- | ----------------------- | -------------------------
A B C D | A B C D | A B C D
2 0.1 0.2 0.3 NaN | 0 0.1 0.2 0.3 NaN | 0 0.1 0.2 0.3 NaN
3 0.1 0.2 0.3 NaN | 1 0.1 0.2 0.3 NaN | 1 0.1 0.2 0.3 NaN
1 NaN 0.4 0.5 0.6 | 2 NaN 0.4 0.5 0.6 | 2 NaN 0.4 0.5 0.6
2 NaN 0.4 0.5 0.6 | 3 NaN 0.4 0.5 0.6 | 3 NaN 0.4 0.5 0.6
And when using axis=1:
| pd.concat(
| [d1, d2], axis=1,
pd.concat([d1, d2], axis=1) | ignore_index=True)
------------------------------- | -------------------------------
A B C B C D | 0 1 2 3 4 5
1 NaN NaN NaN 0.4 0.5 0.6 | 1 NaN NaN NaN 0.4 0.5 0.6
2 0.1 0.2 0.3 0.4 0.5 0.6 | 2 0.1 0.2 0.3 0.4 0.5 0.6
3 0.1 0.2 0.3 NaN NaN NaN | 3 0.1 0.2 0.3 NaN NaN NaN
keys
We can pass a list of scalar values or tuples in order to assign tuple or scalar values to corresponding MultiIndex. The length of the passed list must be the same length as the number of items being concatenated.
keys: sequence, default None
If multiple levels passed, should contain tuples. Construct hierarchical index using the passed keys as the outermost level
axis=0
When concatenating Series objects along axis=0 (extending the index).
Those keys, become a new initial level of a MultiIndex object in the index attribute.
# length 3 length 3 # length 2 length 2
# /--------\ /-----------\ # /----\ /------\
pd.concat([s1, s2, s3], keys=['A', 'B', 'C']) pd.concat([s1, s2], keys=['A', 'B'])
---------------------------------------------- -------------------------------------
A 2 1 A 2 1
3 2 3 2
B 1 3 B 1 3
2 4 2 4
C 1 5 dtype: int64
3 6
dtype: int64
However, we can use more than scalar values in the keys argument to create an even deeper MultiIndex. Here we pass tuples of length 2 the prepend two new levels of a MultiIndex:
pd.concat(
[s1, s2, s3],
keys=[('A', 'X'), ('A', 'Y'), ('B', 'X')])
-----------------------------------------------
A X 2 1
3 2
Y 1 3
2 4
B X 1 5
3 6
dtype: int64
axis=1
It's a bit different when extending along columns. When we used axis=0 (see above) our keys acted as MultiIndex levels in addition to the existing index. For axis=1, we are referring to an axis that Series objects don't have, namely the columns attribute.
Seriesaxis=1Notice that naming the s1 and s2 matters so long as no keys are passed, but it gets overridden if keys are passed.
| | | pd.concat(
| pd.concat( | pd.concat( | [s1.rename('U'),
pd.concat( | [s1, s2], | [s1.rename('U'), | s2.rename('V')],
[s1, s2], | axis=1, | s2.rename('V')], | axis=1,
axis=1) | keys=['X', 'Y']) | axis=1) | keys=['X', 'Y'])
-------------- | --------------------- | ---------------------- | ----------------------
0 1 | X Y | U V | X Y
1 NaN 3.0 | 1 NaN 3.0 | 1 NaN 3.0 | 1 NaN 3.0
2 1.0 4.0 | 2 1.0 4.0 | 2 1.0 4.0 | 2 1.0 4.0
3 2.0 NaN | 3 2.0 NaN | 3 2.0 NaN | 3 2.0 NaN
MultiIndexSeriesaxis=1 pd.concat(
[s1, s2],
axis=1,
keys=[('W', 'X'), ('W', 'Y')])
-----------------------------------
W
X Y
1 NaN 3.0
2 1.0 4.0
3 2.0 NaN
DataFrameaxis=1As with the axis=0 examples, keys add levels to a MultiIndex, but this time to the object stored in the columns attribute.
pd.concat( | pd.concat(
[d1, d2], | [d1, d2],
axis=1, | axis=1,
keys=['X', 'Y']) | keys=[('First', 'X'), ('Second', 'X')])
------------------------------- | --------------------------------------------
X Y | First Second
A B C B C D | X X
1 NaN NaN NaN 0.4 0.5 0.6 | A B C B C D
2 0.1 0.2 0.3 0.4 0.5 0.6 | 1 NaN NaN NaN 0.4 0.5 0.6
3 0.1 0.2 0.3 NaN NaN NaN | 2 0.1 0.2 0.3 0.4 0.5 0.6
| 3 0.1 0.2 0.3 NaN NaN NaN
SeriesDataFrameaxis=1This is tricky. In this case, a scalar key value cannot act as the only level of index for the Series object when it becomes a column while also acting as the first level of a MultiIndex for the DataFrame. So Pandas will again use the name attribute of the Series object as the source of the column name.
pd.concat( | pd.concat(
[s1, d1], | [s1.rename('Z'), d1],
axis=1, | axis=1,
keys=['X', 'Y']) | keys=['X', 'Y'])
--------------------- | --------------------------
X Y | X Y
0 A B C | Z A B C
2 1 0.1 0.2 0.3 | 2 1 0.1 0.2 0.3
3 2 0.1 0.2 0.3 | 3 2 0.1 0.2 0.3
keysMultiIndexPandas only seems to infer column names from Series name, but it will not fill in the blanks when doing an analogous concatenation among data frames with a different number of column levels.
d1_ = pd.concat(
[d1], axis=1,
keys=['One'])
d1_
One
A B C
2 0.1 0.2 0.3
3 0.1 0.2 0.3
Then concatenate this with another data frame with only one level in the columns object and Pandas will refuse to try and make tuples of the MultiIndex object and combine all data frames as if a single level of objects, scalars and tuples.
pd.concat([d1_, d2], axis=1)
(One, A) (One, B) (One, C) B C D
1 NaN NaN NaN 0.4 0.5 0.6
2 0.1 0.2 0.3 0.4 0.5 0.6
3 0.1 0.2 0.3 NaN NaN NaN
Passing a dict instead of a list
When passing a dictionary, pandas.concat will use the keys from the dictionary as the keys parameter.
# axis=0 | # axis=1
pd.concat( | pd.concat(
{0: d1, 1: d2}) | {0: d1, 1: d2}, axis=1)
----------------------- | -------------------------------
A B C D | 0 1
0 2 0.1 0.2 0.3 NaN | A B C B C D
3 0.1 0.2 0.3 NaN | 1 NaN NaN NaN 0.4 0.5 0.6
1 1 NaN 0.4 0.5 0.6 | 2 0.1 0.2 0.3 0.4 0.5 0.6
2 NaN 0.4 0.5 0.6 | 3 0.1 0.2 0.3 NaN NaN NaN
levels
This is used in conjunction with the keys argument.When levels is left as its default value of None, Pandas will take the unique values of each level of the resulting MultiIndex and use that as the object used in the resulting index.levels attribute.
levels: list of sequences, default None
Specific levels (unique values) to use for constructing a MultiIndex. Otherwise they will be inferred from the keys.
If Pandas already infers what these levels should be, what advantage is there to specify it ourselves? I'll show one example and leave it up to you to think up other reasons why this might be useful.
Example
Per the documentation, the levels argument is a list of sequences. This means that we can use another pandas.Index as one of those sequences.
Consider the data frame df that is the concatenation of d1, d2 and d3:
df = pd.concat(
[d1, d2, d3], axis=1,
keys=['First', 'Second', 'Fourth'])
df
First Second Fourth
A B C B C D A B D
1 NaN NaN NaN 0.4 0.5 0.6 0.7 0.8 0.9
2 0.1 0.2 0.3 0.4 0.5 0.6 NaN NaN NaN
3 0.1 0.2 0.3 NaN NaN NaN 0.7 0.8 0.9
The levels of the columns object are:
print(df, *df.columns.levels, sep='\n')
Index(['First', 'Second', 'Fourth'], dtype='object')
Index(['A', 'B', 'C', 'D'], dtype='object')
If we use sum within a groupby we get:
df.groupby(axis=1, level=0).sum()
First Fourth Second
1 0.0 2.4 1.5
2 0.6 0.0 1.5
3 0.6 2.4 0.0
But what if instead of ['First', 'Second', 'Fourth'] there were another missing categories named Third and Fifth? And I wanted them included in the results of a groupby aggregation? We can do this if we had a pandas.CategoricalIndex. And we can specify that ahead of time with the levels argument.
So instead, let's define df as:
cats = ['First', 'Second', 'Third', 'Fourth', 'Fifth']
lvl = pd.CategoricalIndex(cats, categories=cats, ordered=True)
df = pd.concat(
[d1, d2, d3], axis=1,
keys=['First', 'Second', 'Fourth'],
levels=[lvl]
)
df
First Fourth Second
1 0.0 2.4 1.5
2 0.6 0.0 1.5
3 0.6 2.4 0.0
But the first level of the columns object is:
df.columns.levels[0]
CategoricalIndex(
['First', 'Second', 'Third', 'Fourth', 'Fifth'],
categories=['First', 'Second', 'Third', 'Fourth', 'Fifth'],
ordered=True, dtype='category')
And our groupby summation looks like:
df.groupby(axis=1, level=0).sum()
First Second Third Fourth Fifth
1 0.0 1.5 0.0 2.4 0.0
2 0.6 1.5 0.0 0.0 0.0
3 0.6 0.0 0.0 2.4 0.0
names
This is used to name the levels of a resulting MultiIndex. The length of the names list should match the number of levels in the resulting MultiIndex.
names: list, default None
Names for the levels in the resulting hierarchical index
# axis=0 | # axis=1
pd.concat( | pd.concat(
[d1, d2], | [d1, d2],
keys=[0, 1], | axis=1, keys=[0, 1],
names=['lvl0', 'lvl1']) | names=['lvl0', 'lvl1'])
----------------------------- | ----------------------------------
A B C D | lvl0 0 1
lvl0 lvl1 | lvl1 A B C B C D
0 2 0.1 0.2 0.3 NaN | 1 NaN NaN NaN 0.4 0.5 0.6
3 0.1 0.2 0.3 NaN | 2 0.1 0.2 0.3 0.4 0.5 0.6
1 1 NaN 0.4 0.5 0.6 | 3 0.1 0.2 0.3 NaN NaN NaN
2 NaN 0.4 0.5 0.6 |
verify_integrity
Self explanatory documentation
verify_integrity: boolean, default False
Check whether the new concatenated axis contains duplicates. This can be very expensive relative to the actual data concatenation.
Because the resulting index from concatenating d1 and d2 is not unique, it would fail the integrity check.
pd.concat([d1, d2])
A B C D
2 0.1 0.2 0.3 NaN
3 0.1 0.2 0.3 NaN
1 NaN 0.4 0.5 0.6
2 NaN 0.4 0.5 0.6
And
pd.concat([d1, d2], verify_integrity=True)
>
ValueError: Indexes have overlapping values: [2]
