另一个解决方案，将 Series 导出到嵌套的 lists ，然后应用 set 扁平化列表:

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']

时间:

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

如果类型不是 list 而是 string ，请使用 str.split :

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']