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

1. 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 unstack the levels that you want to be in the column index.

2. pd.DataFrame.pivot_table

   • A glorified version of groupby with 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.

3. pd.DataFrame.set_index + pd.DataFrame.unstack

   • Convenient and intuitive for some (myself included). Cannot handle duplicate grouped keys.
   • Similar to the groupby paradigm, we specify all columns that will eventually be either row or column levels and set those to be the index. We then unstack the levels we want in the columns. If either the remaining index levels or column levels are not unique, this method will fail.

4. pd.DataFrame.pivot

   • Very similar to set_index in that it shares the duplicate key limitation. The API is very limited as well. It only takes scalar values for index, columns, values.
   • Similar to the pivot_table method 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.

5. pd.crosstab

   • This a specialized version of pivot_table and in its purest form is the most intuitive way to perform several tasks.

6. 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.

7. 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 df such that the col values are columns, row values are the index, mean of val0 are the values, and missing values are 0?

• pd.DataFrame.pivot_table

  • fill_value is not set by default. I tend to set it appropriately. In this case I set it to 0. Notice I skipped question 2 as it's the same as this answer without the fill_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.groupby

    df.groupby(['row', 'col'])['val0'].mean().unstack(fill_value=0)

• pd.crosstab

    pd.crosstab(      index=df['row'], columns=df['col'],      values=df['val0'], aggfunc='mean').fillna(0)

Question 4

Can I get something other than mean, like maybe sum?

• pd.DataFrame.pivot_table

    df.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.groupby

    df.groupby(['row', 'col'])['val0'].sum().unstack(fill_value=0)

• pd.crosstab

    pd.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_table

    df.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.groupby

    df.groupby(['row', 'col'])['val0'].agg(['size', 'mean']).unstack(fill_value=0)

• pd.crosstab

    pd.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_table we pass values=['val0', 'val1'] but we could've left that off completely

    df.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.groupby

    df.groupby(['row', 'col'])['val0', 'val1'].mean().unstack(fill_value=0)

Question 7

Can Subdivide by multiple columns?

• pd.DataFrame.pivot_table

    df.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.groupby

    df.groupby(      ['row', 'item', 'col']  )['val0'].mean().unstack(['item', 'col']).fillna(0).sort_index(1)

Question 8

Can Subdivide by multiple columns?

• pd.DataFrame.pivot_table

    df.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.groupby

    df.groupby(      ['key', 'row', 'item', 'col']  )['val0'].mean().unstack(['item', 'col']).fillna(0).sort_index(1)

• pd.DataFrame.set_index because the set of keys are unique for both rows and columns

    df.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_table

    df.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.groupby

    df.groupby(['row', 'col'])['val0'].size().unstack(fill_value=0)

• pd.crosstab

    pd.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_dummies

    pd.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 twocolumns?

• DataFrame.pivot

  The 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   7

  The 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_table

  Whereas DataFrame.pivot only accepts columns, DataFrame.pivot_table also accepts arrays, so the GroupBy.cumcount can be passed directly as the index without 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)

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  B0  1  a1  1  b2  1  c3  2  a4  2  b5  3  a6  5  c

Output:

   0     1     2A1  a     b     c2  a     b  None3  a  None  None5  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     2A1  a     b     c2  a     b  None3  a  None  None5  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)

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 zoo0  one   A    1   x1  one   B    2   y2  one   C    3   z3  two   A    4   q4  two   B    5   w5  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  Cfoo         one  1  2  3two  4  5  6