如何将一列中的所有列表编译为一个唯一列表

问题描述

我有一个熊猫数据框，如下所示:

I have a pandas dataframe as below:

如何将所有列表(在"val"列中)组合成唯一列表(集合)，例如 [val1，val2，val33，val9，val6，val7] ?

How can I combine all the lists (in the 'val' column) into a unique list (set), e.g. [val1, val2, val33, val9, val6, val7]?

我可以使用以下代码解决此问题.我想知道是否有一种更简单的方法来从列中获取所有唯一值而又不迭代数据框行?

I can solve this with the following code. I wonder if there is an easier way to get all unique values from a column without iterating the dataframe rows?

def_contributors=[]
for index, row in df.iterrows():
    contri = ast.literal_eval(row['val'])
    def_contributors.extend(contri)
def_contributors = list(set(def_contributors))

推荐答案

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.strip 和 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']

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