如何解决是否有可能的替代方案代码来重塑当前使用熔化功能进行重塑的给定熊猫数据框?
Category Sector 1 Blanks Sector 2 Sector 3 Sector 4 Sector 5 Sector 6 Other Sector 7
Blank 0 1 0 0 0 0 0 0 0
3D 1 0 0 0 0 1 1 0 0
3DPrinting 0 0 1 0 0 1 1 0 0
3DTechnology 0 0 0 1 0 1 1 0 0
B2B 0 0 0 0 1 0 0 0 1
B2B Express Delivery 0 0 0 0 0 0 0 0 1
K-12 Education 0 0 0 0 0 0 0 1 0
M2M 0 0 0 0 0 0 0 1 0
P2P Money Transfer 0 0 0 0 1 0 0 0 1
cols = list(df.columns.difference(['Category']))
cols
df = pd.melt(df,id_vars=['Category'],value_vars=cols)
df = df[~(df.value == 0)]
df = df.drop('value',axis = 1)
df = df.rename(columns = {"variable" : "Sectors"})
df.head()
哪个可以完成这项工作,我可以根据需要重塑数据框,如下所示:
Category Sectors
3D Sector 1
3DPrinting Sector 2
3DTechnology Sector 3
B2B Sector 4
B3B Express Delivery Sector 7
....
....
不知何故,我不满意,正在寻找上述多行代码的替代方案。
有没有可能?
解决方法
尝试使用 melt
:
(df.melt('Category',var_name='Sectors')
.query('value>0 & Sectors != "Blanks"')
)
输出:
Category Sectors value
1 3D Sector 1 1
20 3DPrinting Sector 2 1
30 3DTechnology Sector 3 1
40 B2B Sector 4 1
44 P2P Money Transfer Sector 4 1
46 3D Sector 5 1
47 3DPrinting Sector 5 1
48 3DTechnology Sector 5 1
55 3D Sector 6 1
56 3DPrinting Sector 6 1
57 3DTechnology Sector 6 1
69 K-12 Education Other 1
70 M2M Other 1
76 B2B Sector 7 1
77 B2B Express Delivery Sector 7 1
80 P2P Money Transfer Sector 7 1
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