python Pandas : adding list of elements in a specific column to find all_elements

Question

假设我有一个像这样使用 pandas.dataframe 的列:

id  available_fruits  
1   ['apple', 'banana']   
1   []
2   ['apple', 'tomato']
1   ['banana']
2   ['kiwi']

我想创建没有重复的 all_available_fruits 列表，它应该是 ['apple', 'banana', 'kiwi', 'tomato']。

换句话说，我想在 pandas.dataframe 列中添加列表中的所有元素。我怎样才能做到这一点？

Answer 1

使用numpy.concatenate对于 flatennig，然后是 numpy.unique :

a = np.unique(np.concatenate(df['available_fruits'].values.tolist())).tolist()
print(a)

['apple', 'banana', 'kiwi', 'tomato']

另一种解决方案是通过chain.from_iterable扁平化，通过set获取唯一性，最后转换为list:

from  itertools import chain
a = list(set(chain.from_iterable(df.available_fruits.values.tolist())))
print(a)
['tomato', 'kiwi', 'apple', 'banana']

时间:

df = pd.concat([df]*10000).reset_index(drop=True)
#print (df)

In [62]: %timeit list(set(concat(df.available_fruits.values.tolist())))
100 loops, best of 3: 3.16 ms per loop

In [63]: %timeit np.unique(np.concatenate(df['available_fruits'].values.tolist())).tolist()
10 loops, best of 3: 99.2 ms per loop

#John Galt's solution
In [64]: %timeit list(set(df.available_fruits.sum()))
1 loop, best of 3: 4.12 s per loop

#pir's solution 0
In [65]: %timeit list(set(concat(df.available_fruits.values.tolist())))
100 loops, best of 3: 3.16 ms per loop

#pir's solution 1
In [66]: %timeit list({k: 1 for x in df.available_fruits.values.tolist() for k in x})
100 loops, best of 3: 4.59 ms per loop

#pir's solution 2
In [67]: %%timeit
    ...: from sklearn.preprocessing import MultiLabelBinarizer
    ...: 
    ...: mlb = MultiLabelBinarizer()
    ...: mlb.fit(df.available_fruits)
    ...: list(mlb.classes_)
    ...: 
100 loops, best of 3: 4.07 ms per loop

#perigon's solution
In [68]: %timeit list(set([val for lst in df.available_fruits for val in lst]))
100 loops, best of 3: 5.1 ms per loop