`numpy`

Question

我有一个这样的数据框：

In [1]: df = pd.DataFrame([['jon snow', 'jon-snow'], ['jon snow', 'jon+snow'], [jon snow, 'jonsnow']], columns=['name', 'name_variation'])

我想要的是：

df_want = pd.DataFrame([['jon snow', 'jon-snow', 'jon-snow'], 
              ['jon snow', 'jon-snow', 'jon+snow'], 
              ['jon snow', 'jon-snow', 'jonsnow'], 
              ['jon snow', 'jon-snow', np.nan], 
              ['jon snow', 'jon+snow', 'jon-snow'], 
              ['jon snow', 'jon+snow', 'jon+snow'], 
              ['jon snow', 'jon+snow', 'jonsnow'], 
              ['jon snow', 'jon+snow', np.nan],  
              ['jon snow', 'jonsnow', 'jon-snow'], 
              ['jon snow', 'jonsnow', 'jon+snow'], 
              ['jon snow', 'jonsnow', 'jon-snow'], 
              ['jon snow', 'jonsnow', np.nan], 
              ['jon snow', np.nan, 'jon-snow'], 
              ['jon snow', np.nan, 'jon+snow'], 
              ['jon snow', np.nan, 'jonsnow'], 
              ['jon snow', np.nan, np.nan]], columns=['name', 'name_variation', 'name_variation_2'])

我正在尝试这个有效，但感觉很长时间啰嗦：

def combinations(df):
    df = df.drop_duplicates()
    df = df.dropna()

    df['k'] = df['brand_variation']
    df['val'] = 1

    df_final = pd.DataFrame(columns=['brand', 'k', 'brand_variation',])
    for res in df['brand'].unique():
        #print(res, len(df[df['brand'] == res]))
        dfm = df[df['brand'] == res]
        dfk = pd.pivot_table(dfm, index=['brand', 'k'], columns=['brand_variation'], values=['val'], fill_value=0, aggfunc=[np.sum]).stack().reset_index()
        dfk.columns = dfk.columns.get_level_values(level=0)
        dfk = dfk[['brand', 'k', 'brand_variation']]
        df_final = df_final.append(dfk)

    df_final = df_final.reset_index(drop=True)
    return df_final

更好的方法吗？

Answer 1

`numpy`

u = pd.unique(df.values.ravel())
r = np.arange(u.size)
i, j = r.repeat(u.size), np.tile(r, u.size)

pd.DataFrame(dict(
        name=['jon snow' for _ in range(i.size)],
        name_variation=u[i],
        name_variation2=u[j]
    ))

        name name_variation name_variation2
0   jon snow       jon snow        jon snow
1   jon snow       jon snow        jon-snow
2   jon snow       jon snow        jon+snow
3   jon snow       jon snow         jonsnow
4   jon snow       jon-snow        jon snow
5   jon snow       jon-snow        jon-snow
6   jon snow       jon-snow        jon+snow
7   jon snow       jon-snow         jonsnow
8   jon snow       jon+snow        jon snow
9   jon snow       jon+snow        jon-snow
10  jon snow       jon+snow        jon+snow
11  jon snow       jon+snow         jonsnow
12  jon snow        jonsnow        jon snow
13  jon snow        jonsnow        jon-snow
14  jon snow        jonsnow        jon+snow
15  jon snow        jonsnow         jonsnow

`pandas`

u = pd.unique(df.values.ravel())
pd.Series(
    'jon snow',
    pd.MultiIndex.from_product(
        [u, u], names=['name_variation', 'name_variation2']
    ),
    name='name'
).reset_index()

   name_variation name_variation2      name
0        jon snow        jon snow  jon snow
1        jon snow        jon-snow  jon snow
2        jon snow        jon+snow  jon snow
3        jon snow         jonsnow  jon snow
4        jon-snow        jon snow  jon snow
5        jon-snow        jon-snow  jon snow
6        jon-snow        jon+snow  jon snow
7        jon-snow         jonsnow  jon snow
8        jon+snow        jon snow  jon snow
9        jon+snow        jon-snow  jon snow
10       jon+snow        jon+snow  jon snow
11       jon+snow         jonsnow  jon snow
12        jonsnow        jon snow  jon snow
13        jonsnow        jon-snow  jon snow
14        jonsnow        jon+snow  jon snow
15        jonsnow         jonsnow  jon snow

Answer 2

实际上有一个非常简单的解决方案，我只是迟到了：

df = pd.merge(df, df, left_on='brand', right_on='brand', how='inner').drop_duplicates()

arrrrgh！

创建由另一列pandas分组的一列的排列

2 个答案:

`numpy`

`pandas`