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Pandas column of lists, create a row for each list element

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How do I turn a column into a row in pandas?

Method #2: Using pivot() method. In order to convert a column to row name/index in dataframe, Pandas has a built-in function Pivot. Now, let's say we want Result to be the rows/index, and columns be name in our dataframe, to achieve this pandas has provided a method called Pivot.

How do you convert a list into a row in python?

Method 1: Using T function This is known as the Transpose function, this will convert the list into a row. Here each value is stored in one column. Example: Python3.

How do I create a row list in pandas?

By using df. loc[index]=list you can append a list as a row to the DataFrame at a specified Index, In order to add at the end get the index of the last record using len(df) function. The below example adds the list ["Hyperion",27000,"60days",2000] to the end of the pandas DataFrame. Yields below output.


A bit longer than I expected:

>>> df
                samples  subject  trial_num
0  [-0.07, -2.9, -2.44]        1          1
1   [-1.52, -0.35, 0.1]        1          2
2  [-0.17, 0.57, -0.65]        1          3
3  [-0.82, -1.06, 0.47]        2          1
4   [0.79, 1.35, -0.09]        2          2
5   [1.17, 1.14, -1.79]        2          3
>>>
>>> s = df.apply(lambda x: pd.Series(x['samples']),axis=1).stack().reset_index(level=1, drop=True)
>>> s.name = 'sample'
>>>
>>> df.drop('samples', axis=1).join(s)
   subject  trial_num  sample
0        1          1   -0.07
0        1          1   -2.90
0        1          1   -2.44
1        1          2   -1.52
1        1          2   -0.35
1        1          2    0.10
2        1          3   -0.17
2        1          3    0.57
2        1          3   -0.65
3        2          1   -0.82
3        2          1   -1.06
3        2          1    0.47
4        2          2    0.79
4        2          2    1.35
4        2          2   -0.09
5        2          3    1.17
5        2          3    1.14
5        2          3   -1.79

If you want sequential index, you can apply reset_index(drop=True) to the result.

update:

>>> res = df.set_index(['subject', 'trial_num'])['samples'].apply(pd.Series).stack()
>>> res = res.reset_index()
>>> res.columns = ['subject','trial_num','sample_num','sample']
>>> res
    subject  trial_num  sample_num  sample
0         1          1           0    1.89
1         1          1           1   -2.92
2         1          1           2    0.34
3         1          2           0    0.85
4         1          2           1    0.24
5         1          2           2    0.72
6         1          3           0   -0.96
7         1          3           1   -2.72
8         1          3           2   -0.11
9         2          1           0   -1.33
10        2          1           1    3.13
11        2          1           2   -0.65
12        2          2           0    0.10
13        2          2           1    0.65
14        2          2           2    0.15
15        2          3           0    0.64
16        2          3           1   -0.10
17        2          3           2   -0.76

Pandas >= 0.25

Series and DataFrame methods define a .explode() method that explodes lists into separate rows. See the docs section on Exploding a list-like column.

df = pd.DataFrame({
    'var1': [['a', 'b', 'c'], ['d', 'e',], [], np.nan], 
    'var2': [1, 2, 3, 4]
})
df
        var1  var2
0  [a, b, c]     1
1     [d, e]     2
2         []     3
3        NaN     4

df.explode('var1')

  var1  var2
0    a     1
0    b     1
0    c     1
1    d     2
1    e     2
2  NaN     3  # empty list converted to NaN
3  NaN     4  # NaN entry preserved as-is

# to reset the index to be monotonically increasing...
df.explode('var1').reset_index(drop=True)

  var1  var2
0    a     1
1    b     1
2    c     1
3    d     2
4    e     2
5  NaN     3
6  NaN     4

Note that this also handles mixed columns of lists and scalars, as well as empty lists and NaNs appropriately (this is a drawback of repeat-based solutions).

However, you should note that explode only works on a single column (for now).

P.S.: if you are looking to explode a column of strings, you need to split on a separator first, then use explode. See this (very much) related answer by me.


UPDATE: the solution below was helpful for older Pandas versions, because the DataFrame.explode() wasn’t available. Starting from Pandas 0.25.0 you can simply use DataFrame.explode().


lst_col = 'samples'

r = pd.DataFrame({
      col:np.repeat(df[col].values, df[lst_col].str.len())
      for col in df.columns.drop(lst_col)}
    ).assign(**{lst_col:np.concatenate(df[lst_col].values)})[df.columns]

Result:

In [103]: r
Out[103]:
    samples  subject  trial_num
0      0.10        1          1
1     -0.20        1          1
2      0.05        1          1
3      0.25        1          2
4      1.32        1          2
5     -0.17        1          2
6      0.64        1          3
7     -0.22        1          3
8     -0.71        1          3
9     -0.03        2          1
10    -0.65        2          1
11     0.76        2          1
12     1.77        2          2
13     0.89        2          2
14     0.65        2          2
15    -0.98        2          3
16     0.65        2          3
17    -0.30        2          3

PS here you may find a bit more generic solution


UPDATE: some explanations: IMO the easiest way to understand this code is to try to execute it step-by-step:

in the following line we are repeating values in one column N times where N - is the length of the corresponding list:

In [10]: np.repeat(df['trial_num'].values, df[lst_col].str.len())
Out[10]: array([1, 1, 1, 2, 2, 2, 3, 3, 3, 1, 1, 1, 2, 2, 2, 3, 3, 3], dtype=int64)

this can be generalized for all columns, containing scalar values:

In [11]: pd.DataFrame({
    ...:           col:np.repeat(df[col].values, df[lst_col].str.len())
    ...:           for col in df.columns.drop(lst_col)}
    ...:         )
Out[11]:
    trial_num  subject
0           1        1
1           1        1
2           1        1
3           2        1
4           2        1
5           2        1
6           3        1
..        ...      ...
11          1        2
12          2        2
13          2        2
14          2        2
15          3        2
16          3        2
17          3        2

[18 rows x 2 columns]

using np.concatenate() we can flatten all values in the list column (samples) and get a 1D vector:

In [12]: np.concatenate(df[lst_col].values)
Out[12]: array([-1.04, -0.58, -1.32,  0.82, -0.59, -0.34,  0.25,  2.09,  0.12,  0.83, -0.88,  0.68,  0.55, -0.56,  0.65, -0.04,  0.36, -0.31])

putting all this together:

In [13]: pd.DataFrame({
    ...:           col:np.repeat(df[col].values, df[lst_col].str.len())
    ...:           for col in df.columns.drop(lst_col)}
    ...:         ).assign(**{lst_col:np.concatenate(df[lst_col].values)})
Out[13]:
    trial_num  subject  samples
0           1        1    -1.04
1           1        1    -0.58
2           1        1    -1.32
3           2        1     0.82
4           2        1    -0.59
5           2        1    -0.34
6           3        1     0.25
..        ...      ...      ...
11          1        2     0.68
12          2        2     0.55
13          2        2    -0.56
14          2        2     0.65
15          3        2    -0.04
16          3        2     0.36
17          3        2    -0.31

[18 rows x 3 columns]

using pd.DataFrame()[df.columns] will guarantee that we are selecting columns in the original order...


you can also use pd.concat and pd.melt for this:

>>> objs = [df, pd.DataFrame(df['samples'].tolist())]
>>> pd.concat(objs, axis=1).drop('samples', axis=1)
   subject  trial_num     0     1     2
0        1          1 -0.49 -1.00  0.44
1        1          2 -0.28  1.48  2.01
2        1          3 -0.52 -1.84  0.02
3        2          1  1.23 -1.36 -1.06
4        2          2  0.54  0.18  0.51
5        2          3 -2.18 -0.13 -1.35
>>> pd.melt(_, var_name='sample_num', value_name='sample', 
...         value_vars=[0, 1, 2], id_vars=['subject', 'trial_num'])
    subject  trial_num sample_num  sample
0         1          1          0   -0.49
1         1          2          0   -0.28
2         1          3          0   -0.52
3         2          1          0    1.23
4         2          2          0    0.54
5         2          3          0   -2.18
6         1          1          1   -1.00
7         1          2          1    1.48
8         1          3          1   -1.84
9         2          1          1   -1.36
10        2          2          1    0.18
11        2          3          1   -0.13
12        1          1          2    0.44
13        1          2          2    2.01
14        1          3          2    0.02
15        2          1          2   -1.06
16        2          2          2    0.51
17        2          3          2   -1.35

last, if you need you can sort base on the first the first three columns.


Trying to work through Roman Pekar's solution step-by-step to understand it better, I came up with my own solution, which uses melt to avoid some of the confusing stacking and index resetting. I can't say that it's obviously a clearer solution though:

items_as_cols = df.apply(lambda x: pd.Series(x['samples']), axis=1)
# Keep original df index as a column so it's retained after melt
items_as_cols['orig_index'] = items_as_cols.index

melted_items = pd.melt(items_as_cols, id_vars='orig_index', 
                       var_name='sample_num', value_name='sample')
melted_items.set_index('orig_index', inplace=True)

df.merge(melted_items, left_index=True, right_index=True)

Output (obviously we can drop the original samples column now):

                 samples  subject  trial_num sample_num  sample
0    [1.84, 1.05, -0.66]        1          1          0    1.84
0    [1.84, 1.05, -0.66]        1          1          1    1.05
0    [1.84, 1.05, -0.66]        1          1          2   -0.66
1    [-0.24, -0.9, 0.65]        1          2          0   -0.24
1    [-0.24, -0.9, 0.65]        1          2          1   -0.90
1    [-0.24, -0.9, 0.65]        1          2          2    0.65
2    [1.15, -0.87, -1.1]        1          3          0    1.15
2    [1.15, -0.87, -1.1]        1          3          1   -0.87
2    [1.15, -0.87, -1.1]        1          3          2   -1.10
3   [-0.8, -0.62, -0.68]        2          1          0   -0.80
3   [-0.8, -0.62, -0.68]        2          1          1   -0.62
3   [-0.8, -0.62, -0.68]        2          1          2   -0.68
4    [0.91, -0.47, 1.43]        2          2          0    0.91
4    [0.91, -0.47, 1.43]        2          2          1   -0.47
4    [0.91, -0.47, 1.43]        2          2          2    1.43
5  [-1.14, -0.24, -0.91]        2          3          0   -1.14
5  [-1.14, -0.24, -0.91]        2          3          1   -0.24
5  [-1.14, -0.24, -0.91]        2          3          2   -0.91

For those looking for a version of Roman Pekar's answer that avoids manual column naming:

column_to_explode = 'samples'
res = (df
       .set_index([x for x in df.columns if x != column_to_explode])[column_to_explode]
       .apply(pd.Series)
       .stack()
       .reset_index())
res = res.rename(columns={
          res.columns[-2]:'exploded_{}_index'.format(column_to_explode),
          res.columns[-1]: '{}_exploded'.format(column_to_explode)})

I found the easiest way was to:

  1. Convert the samples column into a DataFrame
  2. Joining with the original df
  3. Melting

Shown here:

    df.samples.apply(lambda x: pd.Series(x)).join(df).\
melt(['subject','trial_num'],[0,1,2],var_name='sample')

        subject  trial_num sample  value
    0         1          1      0  -0.24
    1         1          2      0   0.14
    2         1          3      0  -0.67
    3         2          1      0  -1.52
    4         2          2      0  -0.00
    5         2          3      0  -1.73
    6         1          1      1  -0.70
    7         1          2      1  -0.70
    8         1          3      1  -0.29
    9         2          1      1  -0.70
    10        2          2      1  -0.72
    11        2          3      1   1.30
    12        1          1      2  -0.55
    13        1          2      2   0.10
    14        1          3      2  -0.44
    15        2          1      2   0.13
    16        2          2      2  -1.44
    17        2          3      2   0.73

It's worth noting that this may have only worked because each trial has the same number of samples (3). Something more clever may be necessary for trials of different sample sizes.


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