熊貓寬到長，附加字典

Question

我的數據框看起來像這樣

>df
ds           A  B  C
01/01/2010   4  2  1
02/01/2010   2  9  3
03/01/2010   1  3  0

A和B屬於第1類，C是第2類。

我想將其轉換為：

ds           Category  Company  Value
01/01/2010      1         A      4
01/01/2010      1         B      2
01/01/2010      2         C      1

等等，以后的情節。

Answer 1

使用DataFrame.melt ：

df['ds'] = pd.to_datetime(df['ds'], format='%d/%m/%Y')

df = df.melt('ds', var_name='Company')

如果可以使用多個類別，請創建字典並通過Series.map創建新列：

d = {1:['A','B'], 2:['C']}
#swap key values in dict
#http://stackoverflow.com/a/31674731/2901002
d1 = {k: oldk for oldk, oldv in d.items() for k in oldv}

df['Category'] = df['Company'].map(d1)

#alternative1
#df['Category'] = np.where(df['Company'] == 'C', 2, 1)
#alternative2
#df['Category'] = np.where(df['Company'].isin(['A','B']), 1, 2)

---

df = df.sort_values(['ds','Company']).reset_index(drop=True)

或DataFrame.set_index與DataFrame.stack ：

df['ds'] = pd.to_datetime(df['ds'], format='%d/%m/%Y')

df = df.set_index('ds').stack().rename_axis(('ds','Company')).reset_index(name='value')
df['Category'] = np.where(df['Company'] == 'C', 2, 1)

---

print (df)
          ds Company  value  Category
0 2010-01-01       A      4         1
1 2010-01-01       B      2         1
2 2010-01-01       C      1         2
3 2010-01-02       A      2         1
4 2010-01-02       B      9         1
5 2010-01-02       C      3         2
6 2010-01-03       A      1         1
7 2010-01-03       B      3         1
8 2010-01-03       C      0         2

Answer 2

我們可以使用pd.melt后跟np.where ：

df2 = pd.melt(df, id_vars=['ds'], value_vars=['A', 'B', 'C'])

df2['Category'] = np.where((df2['variable'] == 'A') | (df2['variable'] == 'B'), 1, 2)