我的培训数据train
SFrame
看起来像4列(&#34; Store&#34;列在此SFrame中非唯一):< / p>
+-------+------------+---------+-----------+
| Store | Date | Sales | Customers |
+-------+------------+---------+-----------+
| 1 | 2015-07-31 | 5263.0 | 555.0 |
| 2 | 2015-07-31 | 6064.0 | 625.0 |
| 3 | 2015-07-31 | 8314.0 | 821.0 |
| 4 | 2015-07-31 | 13995.0 | 1498.0 |
| 3 | 2015-07-20 | 4822.0 | 559.0 |
| 2 | 2015-07-10 | 5651.0 | 589.0 |
| 4 | 2015-07-11 | 15344.0 | 1414.0 |
| 5 | 2015-07-23 | 8492.0 | 833.0 |
| 2 | 2015-07-19 | 8565.0 | 687.0 |
| 10 | 2015-07-09 | 7185.0 | 681.0 |
+-------+------------+---------+-----------+
[986159 rows x 4 columns]
给定第二个store
SFrame
(&#34; Store&#34;列在此SFrame中是唯一的):
+-------+-----------+
| Store | StoreType |
+-------+-----------+
| 1 | c |
| 2 | a |
| 3 | a |
| 4 | c |
| 5 | a |
| 6 | a |
| 7 | a |
| 8 | a |
| 9 | a |
| 10 | a |
+-------+-----------+
我可以通过浏览StoreType
中的每一行,将相应的train
附加到我的SFrame
train
,然后从{{1}找到相应的StoreType
然后保留列和ise store
:
SFrame.add_column()
获得:
store_type_col = []
for row in train:
row_store = row['Store']
row_storetype = next(i for i in store if i['Store'] == row_store)['StoreType']
store_type_col.append(row_storetype)
train.add_column(graphlab.SArray(store_type_col, dtype=str), name='StoreType')
但我确信使用+-------+------------+---------+-----------+-----------+
| Store | Date | Sales | Customers | StoreType |
+-------+------------+---------+-----------+-----------+
| 1 | 2015-07-31 | 5263.0 | 555.0 | c
| 2 | 2015-07-31 | 6064.0 | 625.0 | a
| 3 | 2015-07-31 | 8314.0 | 821.0 | a
| 4 | 2015-07-31 | 13995.0 | 1498.0 | c
| 3 | 2015-07-20 | 4822.0 | 559.0 | a
| 2 | 2015-07-10 | 5651.0 | 589.0 | a
| 4 | 2015-07-11 | 15344.0 | 1414.0 | c
| 5 | 2015-07-23 | 8492.0 | 833.0 | a
| 2 | 2015-07-19 | 8565.0 | 687.0 | a
| 10 | 2015-07-09 | 7185.0 | 681.0 | a
+-------+------------+---------+-----------+-----------+
[986159 rows x 5 columns]
可以更简单快捷地完成此操作。当前方法的最差情况是Graphlab
,其中n = no。 O(n*m)
中的行数和m =否。 train
中的行数。
想象一下,我的m
store
有8列,我想将其附加到SFrame
。上面的代码效率极低。
我还可以将数据列从一个SFrame附加到另一个SFrame吗?(也欢迎Pandas解决方案)
答案 0 :(得分:1)
您可以使用join
操作执行此操作。
out = train.join(store, on = 'Store')