# Pivot some rows to new columns in DataFrame

I'm after a pythonic and pandemic (from pandas, pun not intended =) way to pivot some rows in a dataframe into new columns.

My data has this format:

           dof  foo  bar  qux
idxA idxB
100  101     1   10   30   50
101     2   11   31   51
101     3   12   32   52
102     1   13   33   53
102     2   14   34   54
102     3   15   35   55
200  101     1   16   36   56
101     2   17   37   57
101     3   18   38   58
102     1   19   39   59
102     2   20   40   60
102     3   21   41   61


The variables foo, bar and qux actually have 3 dimensional coordinates, which I would like to call foo1, foo2, foo3, bar1, ..., qux3. These are identified by the column dof. Each row represents one axis in 3D, dof == 1 is the x axis, dof == 2 the y axis and dof == 3 is the z axis.

So, here is the final dataframe I want:

           foo1  bar1  qux1  foo2  bar2  qux2  foo3  bar3  qux3
idxA idxB
100  101     10    30    50    11    31    51    12    32    52
102     13    33    53    14    34    54    15    35    55
200  101     16    36    56    17    37    57    18    38    58
102     19    39    59    20    40    60    21    41    61


Here is what I have done.

import pandas as pd

data = [[100, 101, 1, 10, 30, 50],
[100, 101, 2, 11, 31, 51],
[100, 101, 3, 12, 32, 52],
[100, 102, 1, 13, 33, 53],
[100, 102, 2, 14, 34, 54],
[100, 102, 3, 15, 35, 55],
[200, 101, 1, 16, 36, 56],
[200, 101, 2, 17, 37, 57],
[200, 101, 3, 18, 38, 58],
[200, 102, 1, 19, 39, 59],
[200, 102, 2, 20, 40, 60],
[200, 102, 3, 21, 41, 61],
]

df = pd.DataFrame(data=data, columns=['idxA', 'idxB', 'dof', 'foo', 'bar', 'qux'])
df.set_index(['idxA', 'idxB'], inplace=True)

#
#

# Create an ampty dataframe with the same indexes
df2 = df[df.dof == 1].reset_index()[['idxA', 'idxB']]
df2.set_index(['idxA', 'idxB'], inplace=True)

# Loop through each DOF and add columns for bar, foo and qux manually.
for pivot in [1, 2, 3]:
df2.loc[:, 'foo%d' % pivot] = df[df.dof == pivot]['foo']
df2.loc[:, 'bar%d' % pivot] = df[df.dof == pivot]['bar']
df2.loc[:, 'qux%d' % pivot] = df[df.dof == pivot]['qux']


However I'm not too happy with these .loc calls and incremental column additions inside a loop. I thought that pandas being awesome as it is would have a neater way of doing that.

• I'm too lazy to try an implementation, but perhaps you should look into a multi-index where the innermost index has size 3. – Reinderien Jun 3 '20 at 1:55

# groupby

When iterating over the values in a column, it is bad practice to hardcode the values (for pivot in [1, 2, 3]). A better way would have been for pivot in df["dof"].unique(), but the best way is with DataFrame.groupby

To see what happens in the groupby, I try it first with an iteration, and printing the groups

for pivot, data in df.groupby("dof"):
print(pivot)
print(data)


Then I get to work with one of the data to mold it the way I want. In this case, we don't need the column dof any more, since we have it in the pivot variable, and we rename the columns using rename

for pivot, data in df.groupby("dof"):
print(pivot)
print(
data.drop(columns="dof").rename(
mapper={
column_name: f"{column_name}{pivot}"
for column_name in data.columns
},
axis=1,
)
)


Then we can use pd.concat to stitch it together

pd.concat(
[
data.drop(columns="dof").rename(
mapper={
column_name: f"{column_name}{pivot}"
for column_name in data.columns
},
axis=1,
)
for pivot, data in df.groupby("dof")
],
axis=1,
)


# unstack

An alternative is with unstack:

From you description, dof is part of the index, so add it there. Then you can use DataFrame.unstack to bring it to the columns.

df2 = df.set_index("dof", append=True).unstack("dof")

        foo foo foo bar bar bar qux qux qux
dof     1   2   3   1   2   3   1   2   3
idxA idxB
100 101 10  11  12  30  31  32  50  51  52
100 102 13  14  15  33  34  35  53  54  55
200 101 16  17  18  36  37  38  56  57  58
200 102 19  20  21  39  40  41  59  60  61


If you are okay with a MultiIndex, which will be handier then the concatenated strings in most cases, you can leave it at that. If you want it in the form you have it, you can do df2.columns = df2.columns.map(lambda x: f"{x[0]}{x[1]}").