從列表中獲取唯一值作為Pandas python中的值

Question

我有一個包含更多行和列的數據框，但是這里有一個示例：

id    values 
1   [v1, v2, v1]

如何從熊貓列中的列表中獲取唯一值？ 我已經嘗試使用df ['values']。unique（）在第二列v1，v2中提供所需的輸出，但顯然它無法正常工作。

Answer 1

一個簡單的解決方案是agg pd.unique即

df = pd.DataFrame({'x' : [['v','w','x','v','x']]})

df['x'].agg(pd.unique) # Also np.unique

0    [v, w, x]
Name: x, dtype: object

要么

df['x'].agg(set).agg(list)

0    [v, w, x]
Name: x, dtype: object

Answer 2

再次

df['new']=list(map(set,df['values'].values))

定時

%timeit df['values'].agg(np.unique)
The slowest run took 6.78 times longer than the fastest. This could mean that an intermediate result is being cached.
100 loops, best of 3: 6.99 ms per loop
%timeit list(map(set,df['values'].values))
The slowest run took 55.36 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 228 µs per loop
%timeit df['values'].apply(lambda x: list(set(x)))
1000 loops, best of 3: 743 µs per loop

Answer 3

嘗試

df['values'] = df['values'].apply(lambda x: list(set(x)))


    id  values
0   1   [v2, v1]

注意：values是pandas屬性，因此最好避免將其用作列名。

時間比較：

df= pd.DataFrame({'id':[1]*1000,    'values' :[['v1', 'v2', 'v1']]*1000})
%timeit df['values'].agg(np.unique)

34.7 ms ± 2.01 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)


%timeit df['values'].apply(lambda x: list(set(x)))

1.98 ms ± 259 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)