How to compile all lists in a column into one unique list

Question

I have a pandas dataframe as below:

How can I combine all the lists (in the \'val\' column) into a unique list (set), e.g. [val1, val2, val33, val9, v

Answer 1

Another solution with exporting Series to nested lists and then apply set to flatten list:

df = pd.DataFrame({'id':['a','b', 'c'], 'val':[['val1','val2'],
                                               ['val33','val9','val6'],
                                               ['val2','val6','val7']]})

print (df)
  id                  val
0  a         [val1, val2]
1  b  [val33, val9, val6]
2  c   [val2, val6, val7]

print (type(df.val.ix[0]))
<class 'list'>

print (df.val.tolist())
[['val1', 'val2'], ['val33', 'val9', 'val6'], ['val2', 'val6', 'val7']]

print (list(set([a for b in df.val.tolist() for a in b])))
['val7', 'val1', 'val6', 'val33', 'val2', 'val9']

Timings:

df = pd.concat([df]*1000).reset_index(drop=True)

In [307]: %timeit (df['val'].apply(pd.Series).stack().unique()).tolist()
1 loop, best of 3: 410 ms per loop

In [355]: %timeit (pd.Series(sum(df.val.tolist(),[])).unique().tolist())
10 loops, best of 3: 31.9 ms per loop

In [308]: %timeit np.unique(np.hstack(df.val)).tolist()
100 loops, best of 3: 10.7 ms per loop

In [309]: %timeit (list(set([a for b in df.val.tolist() for a in b])))
1000 loops, best of 3: 558 µs per loop

---

If types is not list but string use str.strip and str.split:

df = pd.DataFrame({'id':['a','b', 'c'], 'val':["[val1,val2]",
                                               "[val33,val9,val6]",
                                               "[val2,val6,val7]"]})

print (df)
  id                val
0  a        [val1,val2]
1  b  [val33,val9,val6]
2  c   [val2,val6,val7]

print (type(df.val.ix[0]))
<class 'str'>

print (df.val.str.strip('[]').str.split(','))
0           [val1, val2]
1    [val33, val9, val6]
2     [val2, val6, val7]
Name: val, dtype: object

print (list(set([a for b in df.val.str.strip('[]').str.split(',') for a in b])))
['val7', 'val1', 'val6', 'val33', 'val2', 'val9']

Answer 2

One way would be to extract those elements into an array using np.hstack and then using np.unique to give us an array of such unique elements, like so -

np.unique(np.hstack(df.val))

If you want a list as output, append with .tolist() -

np.unique(np.hstack(df.val)).tolist()

Answer 3

Convert that column into a DataFrame with .apply(pd.Series). If you stack the columns, you can call the unique method on the returned Series.

df
Out[123]: 
            val
0      [v1, v2]
1      [v3, v2]
2  [v4, v3, v2]

---

df['val'].apply(pd.Series).stack().unique()
Out[124]: array(['v1', 'v2', 'v3', 'v4'], dtype=object)

Answer 4

You can use str.concat followed by some string manipulations to obtain the desired list.

In [60]: import re
    ...: from collections import OrderedDict

In [62]: s = df['val'].str.cat()

In [63]: L = re.sub('[[]|[]]',' ', s).strip().replace("  ",',').split(',')

In [64]: list(OrderedDict.fromkeys(L))
Out[64]: ['val1', 'val2', 'val33', 'val9', 'val6', 'val7']