How can I pivot a dataframe?
- What is pivot?
- How do I pivot?
- Is this a pivot?
- Long format to wide format?
I've seen a lot of questions that ask about pivot tables. Even if they don't know that they are asking about pivot tables, they usually are. It is virtually impossible to write a canonical question and answer that encompasses all aspects of pivoting...
... But I'm going to give it a go.
The problem with existing questions and answers is that often the question is focused on a nuance that the OP has trouble generalizing in order to use a number of the existing good answers. However, none of the answers attempt to give a comprehensive explanation (because it's a daunting task)
Look a few examples from my Google Search
- How to pivot a dataframe in Pandas?
- Good question and answer. But the answer only answers the specific question with little explanation.
- pandas pivot table to data frame
- In this question, the OP is concerned with the output of the pivot. Namely how the columns look. OP wanted it to look like R. This isn't very helpful for pandas users.
- pandas pivoting a dataframe, duplicate rows
- Another decent question but the answer focuses on one method, namely
pd.DataFrame.pivot
So whenever someone searches for pivot they get sporadic results that are likely not going to answer their specific question.
Setup
You may notice that I conspicuously named my columns and relevant column values to correspond with how I'm going to pivot in the answers below.
import numpy as np
import pandas as pd
from numpy.core.defchararray import add
np.random.seed([3,1415])
n = 20
cols = np.array(['key', 'row', 'item', 'col'])
arr1 = (np.random.randint(5, size=(n, 4)) // [2, 1, 2, 1]).astype(str)
df = pd.DataFrame(
add(cols, arr1), columns=cols
).join(
pd.DataFrame(np.random.rand(n, 2).round(2)).add_prefix('val')
)
print(df)
key row item col val0 val1
0 key0 row3 item1 col3 0.81 0.04
1 key1 row2 item1 col2 0.44 0.07
2 key1 row0 item1 col0 0.77 0.01
3 key0 row4 item0 col2 0.15 0.59
4 key1 row0 item2 col1 0.81 0.64
5 key1 row2 item2 col4 0.13 0.88
6 key2 row4 item1 col3 0.88 0.39
7 key1 row4 item1 col1 0.10 0.07
8 key1 row0 item2 col4 0.65 0.02
9 key1 row2 item0 col2 0.35 0.61
10 key2 row0 item2 col1 0.40 0.85
11 key2 row4 item1 col2 0.64 0.25
12 key0 row2 item2 col3 0.50 0.44
13 key0 row4 item1 col4 0.24 0.46
14 key1 row3 item2 col3 0.28 0.11
15 key0 row3 item1 col1 0.31 0.23
16 key0 row0 item2 col3 0.86 0.01
17 key0 row4 item0 col3 0.64 0.21
18 key2 row2 item2 col0 0.13 0.45
19 key0 row2 item0 col4 0.37 0.70
Question(s)
-
Why do I get
ValueError: Index contains duplicate entries, cannot reshape -
How do I pivot
dfsuch that thecolvalues are columns,rowvalues are the index, and mean ofval0are the values?col col0 col1 col2 col3 col4 row row0 0.77 0.605 NaN 0.860 0.65 row2 0.13 NaN 0.395 0.500 0.25 row3 NaN 0.310 NaN 0.545 NaN row4 NaN 0.100 0.395 0.760 0.24 -
How do I pivot
dfsuch that thecolvalues are columns,rowvalues are the index, mean ofval0are the values, and missing values are0?col col0 col1 col2 col3 col4 row row0 0.77 0.605 0.000 0.860 0.65 row2 0.13 0.000 0.395 0.500 0.25 row3 0.00 0.310 0.000 0.545 0.00 row4 0.00 0.100 0.395 0.760 0.24 -
Can I get something other than
mean, like maybesum?col col0 col1 col2 col3 col4 row row0 0.77 1.21 0.00 0.86 0.65 row2 0.13 0.00 0.79 0.50 0.50 row3 0.00 0.31 0.00 1.09 0.00 row4 0.00 0.10 0.79 1.52 0.24 -
Can I do more that one aggregation at a time?
sum mean col col0 col1 col2 col3 col4 col0 col1 col2 col3 col4 row row0 0.77 1.21 0.00 0.86 0.65 0.77 0.605 0.000 0.860 0.65 row2 0.13 0.00 0.79 0.50 0.50 0.13 0.000 0.395 0.500 0.25 row3 0.00 0.31 0.00 1.09 0.00 0.00 0.310 0.000 0.545 0.00 row4 0.00 0.10 0.79 1.52 0.24 0.00 0.100 0.395 0.760 0.24 -
Can I aggregate over multiple value columns?
val0 val1 col col0 col1 col2 col3 col4 col0 col1 col2 col3 col4 row row0 0.77 0.605 0.000 0.860 0.65 0.01 0.745 0.00 0.010 0.02 row2 0.13 0.000 0.395 0.500 0.25 0.45 0.000 0.34 0.440 0.79 row3 0.00 0.310 0.000 0.545 0.00 0.00 0.230 0.00 0.075 0.00 row4 0.00 0.100 0.395 0.760 0.24 0.00 0.070 0.42 0.300 0.46 -
Can Subdivide by multiple columns?
item item0 item1 item2 col col2 col3 col4 col0 col1 col2 col3 col4 col0 col1 col3 col4 row row0 0.00 0.00 0.00 0.77 0.00 0.00 0.00 0.00 0.00 0.605 0.86 0.65 row2 0.35 0.00 0.37 0.00 0.00 0.44 0.00 0.00 0.13 0.000 0.50 0.13 row3 0.00 0.00 0.00 0.00 0.31 0.00 0.81 0.00 0.00 0.000 0.28 0.00 row4 0.15 0.64 0.00 0.00 0.10 0.64 0.88 0.24 0.00 0.000 0.00 0.00 -
Or
item item0 item1 item2 col col2 col3 col4 col0 col1 col2 col3 col4 col0 col1 col3 col4 key row key0 row0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.86 0.00 row2 0.00 0.00 0.37 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.50 0.00 row3 0.00 0.00 0.00 0.00 0.31 0.00 0.81 0.00 0.00 0.00 0.00 0.00 row4 0.15 0.64 0.00 0.00 0.00 0.00 0.00 0.24 0.00 0.00 0.00 0.00 key1 row0 0.00 0.00 0.00 0.77 0.00 0.00 0.00 0.00 0.00 0.81 0.00 0.65 row2 0.35 0.00 0.00 0.00 0.00 0.44 0.00 0.00 0.00 0.00 0.00 0.13 row3 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.28 0.00 row4 0.00 0.00 0.00 0.00 0.10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 key2 row0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.40 0.00 0.00 row2 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.13 0.00 0.00 0.00 row4 0.00 0.00 0.00 0.00 0.00 0.64 0.88 0.00 0.00 0.00 0.00 0.00 -
Can I aggregate the frequency in which the column and rows occur together, aka "cross tabulation"?
col col0 col1 col2 col3 col4 row row0 1 2 0 1 1 row2 1 0 2 1 2 row3 0 1 0 2 0 row4 0 1 2 2 1 -
How do I convert a DataFrame from long to wide by pivoting on ONLY two columns? Given,
np.random.seed([3, 1415]) df2 = pd.DataFrame({'A': list('aaaabbbc'), 'B': np.random.choice(15, 8)}) df2 A B 0 a 0 1 a 11 2 a 2 3 a 11 4 b 10 5 b 10 6 b 14 7 c 7The expected should look something like
a b c 0 0.0 10.0 7.0 1 11.0 10.0 NaN 2 2.0 14.0 NaN 3 11.0 NaN NaN -
How do I flatten the multiple index to single index after
pivot?From
1 2 1 1 2 a 2 1 1 b 2 1 0 c 1 0 0To
1|1 2|1 2|2 a 2 1 1 b 2 1 0 c 1 0 0
Solution 1:
We start by answering the first question:
Question 1
Why do I get
ValueError: Index contains duplicate entries, cannot reshape
This occurs because pandas is attempting to reindex either a columns or index object with duplicate entries. There are varying methods to use that can perform a pivot. Some of them are not well suited to when there are duplicates of the keys in which it is being asked to pivot on. For example. Consider pd.DataFrame.pivot. I know there are duplicate entries that share the row and col values:
df.duplicated(['row', 'col']).any()
True
So when I pivot using
df.pivot(index='row', columns='col', values='val0')
I get the error mentioned above. In fact, I get the same error when I try to perform the same task with:
df.set_index(['row', 'col'])['val0'].unstack()
Here is a list of idioms we can use to pivot
-
pd.DataFrame.groupby+pd.DataFrame.unstack- Good general approach for doing just about any type of pivot
- You specify all columns that will constitute the pivoted row levels and column levels in one group by. You follow that by selecting the remaining columns you want to aggregate and the function(s) you want to perform the aggregation. Finally, you
unstackthe levels that you want to be in the column index.
-
pd.DataFrame.pivot_table- A glorified version of
groupbywith more intuitive API. For many people, this is the preferred approach. And is the intended approach by the developers. - Specify row level, column levels, values to be aggregated, and function(s) to perform aggregations.
- A glorified version of
-
pd.DataFrame.set_index+pd.DataFrame.unstack- Convenient and intuitive for some (myself included). Cannot handle duplicate grouped keys.
- Similar to the
groupbyparadigm, we specify all columns that will eventually be either row or column levels and set those to be the index. We thenunstackthe levels we want in the columns. If either the remaining index levels or column levels are not unique, this method will fail.
-
pd.DataFrame.pivot- Very similar to
set_indexin that it shares the duplicate key limitation. The API is very limited as well. It only takes scalar values forindex,columns,values. - Similar to the
pivot_tablemethod in that we select rows, columns, and values on which to pivot. However, we cannot aggregate and if either rows or columns are not unique, this method will fail.
- Very similar to
-
pd.crosstab- This a specialized version of
pivot_tableand in its purest form is the most intuitive way to perform several tasks.
- This a specialized version of
-
pd.factorize+np.bincount- This is a highly advanced technique that is very obscure but is very fast. It cannot be used in all circumstances, but when it can be used and you are comfortable using it, you will reap the performance rewards.
-
pd.get_dummies+pd.DataFrame.dot- I use this for cleverly performing cross tabulation.
Examples
What I'm going to do for each subsequent answer and question is to answer it using pd.DataFrame.pivot_table. Then I'll provide alternatives to perform the same task.
Question 3
How do I pivot
dfsuch that thecolvalues are columns,rowvalues are the index, mean ofval0are the values, and missing values are0?
-
pd.DataFrame.pivot_table-
fill_valueis not set by default. I tend to set it appropriately. In this case I set it to0. Notice I skipped question 2 as it's the same as this answer without thefill_value -
aggfunc='mean'is the default and I didn't have to set it. I included it to be explicit.df.pivot_table( values='val0', index='row', columns='col', fill_value=0, aggfunc='mean') col col0 col1 col2 col3 col4 row row0 0.77 0.605 0.000 0.860 0.65 row2 0.13 0.000 0.395 0.500 0.25 row3 0.00 0.310 0.000 0.545 0.00 row4 0.00 0.100 0.395 0.760 0.24
-
-
pd.DataFrame.groupbydf.groupby(['row', 'col'])['val0'].mean().unstack(fill_value=0) -
pd.crosstabpd.crosstab( index=df['row'], columns=df['col'], values=df['val0'], aggfunc='mean').fillna(0)
Question 4
Can I get something other than
mean, like maybesum?
-
pd.DataFrame.pivot_tabledf.pivot_table( values='val0', index='row', columns='col', fill_value=0, aggfunc='sum') col col0 col1 col2 col3 col4 row row0 0.77 1.21 0.00 0.86 0.65 row2 0.13 0.00 0.79 0.50 0.50 row3 0.00 0.31 0.00 1.09 0.00 row4 0.00 0.10 0.79 1.52 0.24 -
pd.DataFrame.groupbydf.groupby(['row', 'col'])['val0'].sum().unstack(fill_value=0) -
pd.crosstabpd.crosstab( index=df['row'], columns=df['col'], values=df['val0'], aggfunc='sum').fillna(0)
Question 5
Can I do more that one aggregation at a time?
Notice that for pivot_table and crosstab I needed to pass list of callables. On the other hand, groupby.agg is able to take strings for a limited number of special functions. groupby.agg would also have taken the same callables we passed to the others, but it is often more efficient to leverage the string function names as there are efficiencies to be gained.
-
pd.DataFrame.pivot_tabledf.pivot_table( values='val0', index='row', columns='col', fill_value=0, aggfunc=[np.size, np.mean]) size mean col col0 col1 col2 col3 col4 col0 col1 col2 col3 col4 row row0 1 2 0 1 1 0.77 0.605 0.000 0.860 0.65 row2 1 0 2 1 2 0.13 0.000 0.395 0.500 0.25 row3 0 1 0 2 0 0.00 0.310 0.000 0.545 0.00 row4 0 1 2 2 1 0.00 0.100 0.395 0.760 0.24 -
pd.DataFrame.groupbydf.groupby(['row', 'col'])['val0'].agg(['size', 'mean']).unstack(fill_value=0) -
pd.crosstabpd.crosstab( index=df['row'], columns=df['col'], values=df['val0'], aggfunc=[np.size, np.mean]).fillna(0, downcast='infer')
Question 6
Can I aggregate over multiple value columns?
-
pd.DataFrame.pivot_tablewe passvalues=['val0', 'val1']but we could've left that off completelydf.pivot_table( values=['val0', 'val1'], index='row', columns='col', fill_value=0, aggfunc='mean') val0 val1 col col0 col1 col2 col3 col4 col0 col1 col2 col3 col4 row row0 0.77 0.605 0.000 0.860 0.65 0.01 0.745 0.00 0.010 0.02 row2 0.13 0.000 0.395 0.500 0.25 0.45 0.000 0.34 0.440 0.79 row3 0.00 0.310 0.000 0.545 0.00 0.00 0.230 0.00 0.075 0.00 row4 0.00 0.100 0.395 0.760 0.24 0.00 0.070 0.42 0.300 0.46 -
pd.DataFrame.groupbydf.groupby(['row', 'col'])['val0', 'val1'].mean().unstack(fill_value=0)
Question 7
Can Subdivide by multiple columns?
-
pd.DataFrame.pivot_tabledf.pivot_table( values='val0', index='row', columns=['item', 'col'], fill_value=0, aggfunc='mean') item item0 item1 item2 col col2 col3 col4 col0 col1 col2 col3 col4 col0 col1 col3 col4 row row0 0.00 0.00 0.00 0.77 0.00 0.00 0.00 0.00 0.00 0.605 0.86 0.65 row2 0.35 0.00 0.37 0.00 0.00 0.44 0.00 0.00 0.13 0.000 0.50 0.13 row3 0.00 0.00 0.00 0.00 0.31 0.00 0.81 0.00 0.00 0.000 0.28 0.00 row4 0.15 0.64 0.00 0.00 0.10 0.64 0.88 0.24 0.00 0.000 0.00 0.00 -
pd.DataFrame.groupbydf.groupby( ['row', 'item', 'col'] )['val0'].mean().unstack(['item', 'col']).fillna(0).sort_index(1)
Question 8
Can Subdivide by multiple columns?
-
pd.DataFrame.pivot_tabledf.pivot_table( values='val0', index=['key', 'row'], columns=['item', 'col'], fill_value=0, aggfunc='mean') item item0 item1 item2 col col2 col3 col4 col0 col1 col2 col3 col4 col0 col1 col3 col4 key row key0 row0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.86 0.00 row2 0.00 0.00 0.37 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.50 0.00 row3 0.00 0.00 0.00 0.00 0.31 0.00 0.81 0.00 0.00 0.00 0.00 0.00 row4 0.15 0.64 0.00 0.00 0.00 0.00 0.00 0.24 0.00 0.00 0.00 0.00 key1 row0 0.00 0.00 0.00 0.77 0.00 0.00 0.00 0.00 0.00 0.81 0.00 0.65 row2 0.35 0.00 0.00 0.00 0.00 0.44 0.00 0.00 0.00 0.00 0.00 0.13 row3 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.28 0.00 row4 0.00 0.00 0.00 0.00 0.10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 key2 row0 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.40 0.00 0.00 row2 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.13 0.00 0.00 0.00 row4 0.00 0.00 0.00 0.00 0.00 0.64 0.88 0.00 0.00 0.00 0.00 0.00 -
pd.DataFrame.groupbydf.groupby( ['key', 'row', 'item', 'col'] )['val0'].mean().unstack(['item', 'col']).fillna(0).sort_index(1) -
pd.DataFrame.set_indexbecause the set of keys are unique for both rows and columnsdf.set_index( ['key', 'row', 'item', 'col'] )['val0'].unstack(['item', 'col']).fillna(0).sort_index(1)
Question 9
Can I aggregate the frequency in which the column and rows occur together, aka "cross tabulation"?
-
pd.DataFrame.pivot_tabledf.pivot_table(index='row', columns='col', fill_value=0, aggfunc='size') col col0 col1 col2 col3 col4 row row0 1 2 0 1 1 row2 1 0 2 1 2 row3 0 1 0 2 0 row4 0 1 2 2 1 -
pd.DataFrame.groupbydf.groupby(['row', 'col'])['val0'].size().unstack(fill_value=0) -
pd.crosstabpd.crosstab(df['row'], df['col']) -
pd.factorize+np.bincount# get integer factorization `i` and unique values `r` # for column `'row'` i, r = pd.factorize(df['row'].values) # get integer factorization `j` and unique values `c` # for column `'col'` j, c = pd.factorize(df['col'].values) # `n` will be the number of rows # `m` will be the number of columns n, m = r.size, c.size # `i * m + j` is a clever way of counting the # factorization bins assuming a flat array of length # `n * m`. Which is why we subsequently reshape as `(n, m)` b = np.bincount(i * m + j, minlength=n * m).reshape(n, m) # BTW, whenever I read this, I think 'Bean, Rice, and Cheese' pd.DataFrame(b, r, c) col3 col2 col0 col1 col4 row3 2 0 0 1 0 row2 1 2 1 0 2 row0 1 0 1 2 1 row4 2 2 0 1 1 -
pd.get_dummiespd.get_dummies(df['row']).T.dot(pd.get_dummies(df['col'])) col0 col1 col2 col3 col4 row0 1 2 0 1 1 row2 1 0 2 1 2 row3 0 1 0 2 0 row4 0 1 2 2 1
Question 10
How do I convert a DataFrame from long to wide by pivoting on ONLY two columns?
-
DataFrame.pivotThe first step is to assign a number to each row - this number will be the row index of that value in the pivoted result. This is done using
GroupBy.cumcount:df2.insert(0, 'count', df2.groupby('A').cumcount()) df2 count A B 0 0 a 0 1 1 a 11 2 2 a 2 3 3 a 11 4 0 b 10 5 1 b 10 6 2 b 14 7 0 c 7The second step is to use the newly created column as the index to call
DataFrame.pivot.df2.pivot(*df2) # df2.pivot(index='count', columns='A', values='B') A a b c count 0 0.0 10.0 7.0 1 11.0 10.0 NaN 2 2.0 14.0 NaN 3 11.0 NaN NaN -
DataFrame.pivot_tableWhereas
DataFrame.pivotonly accepts columns,DataFrame.pivot_tablealso accepts arrays, so theGroupBy.cumcountcan be passed directly as theindexwithout creating an explicit column.df2.pivot_table(index=df2.groupby('A').cumcount(), columns='A', values='B') A a b c 0 0.0 10.0 7.0 1 11.0 10.0 NaN 2 2.0 14.0 NaN 3 11.0 NaN NaN
Question 11
How do I flatten the multiple index to single index after
pivot
If columns type object with string join
df.columns = df.columns.map('|'.join)
else format
df.columns = df.columns.map('{0[0]}|{0[1]}'.format)
Solution 2:
To extend @piRSquared's answer another version of Question 10
Question 10.1
DataFrame:
d = data = {'A': {0: 1, 1: 1, 2: 1, 3: 2, 4: 2, 5: 3, 6: 5},
'B': {0: 'a', 1: 'b', 2: 'c', 3: 'a', 4: 'b', 5: 'a', 6: 'c'}}
df = pd.DataFrame(d)
A B
0 1 a
1 1 b
2 1 c
3 2 a
4 2 b
5 3 a
6 5 c
Output:
0 1 2
A
1 a b c
2 a b None
3 a None None
5 c None None
Using df.groupby and pd.Series.tolist
t = df.groupby('A')['B'].apply(list)
out = pd.DataFrame(t.tolist(),index=t.index)
out
0 1 2
A
1 a b c
2 a b None
3 a None None
5 c None None
Or
A much better alternative using pd.pivot_table with df.squeeze.
t = df.pivot_table(index='A',values='B',aggfunc=list).squeeze()
out = pd.DataFrame(t.tolist(),index=t.index)
Solution 3:
To better understand how pivot works you can look at the example from Pandas documentation:
df = pd.DataFrame({
'foo': ['one', 'one', 'one', 'two', 'two', 'two'],
'bar': ['A', 'B', 'C', 'A', 'B', 'C'],
'baz': [1, 2, 3, 4, 5, 6],
'zoo': ['x', 'y', 'z', 'q', 'w', 't']
})
Input Table:
foo bar baz zoo
0 one A 1 x
1 one B 2 y
2 one C 3 z
3 two A 4 q
4 two B 5 w
5 two C 6 t
Pivot:
pd.pivot(
data=df,
index='foo', # Column to use to make new frame’s index. If None, uses existing index.
columns='bar', # Column to use to make new frame’s columns.
values='baz' # Column(s) to use for populating new frame’s values.
)
Output table:
bar A B C
foo
one 1 2 3
two 4 5 6