给定这样的sframe：
+------+-----------+-----------+-----------+-----------+-----------+-----------+
| X1 | X2 | X3 | X4 | X5 | X6 | X7 |
+------+-----------+-----------+-----------+-----------+-----------+-----------+
| the | -0.060292 | 0.06763 | -0.036891 | 0.066684 | 0.024045 | 0.099091 |
| , | 0.026625 | 0.073101 | -0.027073 | -0.019504 | 0.04173 | 0.038811 |
| . | -0.005893 | 0.093791 | 0.015333 | 0.046226 | 0.032791 | 0.110069 |
| of | -0.050371 | 0.031452 | 0.04091 | 0.033255 | -0.009195 | 0.061086 |
| and | 0.005456 | 0.063237 | -0.075793 | -0.000819 | 0.003407 | 0.053554 |
| to | 0.01347 | 0.043712 | -0.087122 | 0.015258 | 0.08834 | 0.139644 |
| in | -0.019466 | 0.077509 | -0.102543 | 0.034337 | 0.130886 | 0.032195 |
| a | -0.072288 | -0.017494 | -0.018383 | 0.001857 | -0.04645 | 0.133424 |
| is | 0.052726 | 0.041903 | 0.163781 | 0.006887 | -0.07533 | 0.108394 |
| for | -0.004082 | -0.024244 | 0.042166 | 0.007032 | -0.081243 | 0.026162 |
| on | -0.023709 | -0.038306 | -0.16072 | -0.171599 | 0.150983 | 0.042044 |
| that | 0.062037 | 0.100348 | -0.059753 | -0.041444 | 0.041156 | 0.166704 |
| ) | 0.052312 | 0.072473 | -0.02067 | -0.015581 | 0.063368 | -0.017216 |
| ( | 0.051408 | 0.186162 | 0.03028 | -0.048425 | 0.051376 | 0.004989 |
| with | 0.091825 | -0.081649 | -0.087926 | -0.061273 | 0.043528 | 0.107864 |
| was | 0.046042 | -0.058529 | 0.040581 | 0.067748 | 0.053724 | 0.041067 |
| as | 0.025248 | -0.012519 | -0.054685 | -0.040581 | 0.051061 | 0.114956 |
| it | 0.028606 | 0.106391 | 0.025065 | 0.023486 | 0.011184 | 0.016715 |
| by | -0.096704 | 0.150165 | -0.01775 | -0.07178 | 0.004458 | 0.098807 |
| be | -0.109489 | -0.025908 | 0.025608 | 0.076263 | -0.047246 | 0.100489 |
+------+-----------+-----------+-----------+-----------+-----------+-----------+
如何将sframe转换为字典，使
X1列成为键，并将X2转换为X7作为np.array()？我尝试逐行遍历原始的sframe并执行如下操作：
>>> import graphlab as gl
>>> import numpy as np
>>> x = gl.SFrame()
>>> a = np.array([1,2,3])
>>> w = 'foo'
>>> x.append(gl.SFrame({'word':[w], 'vector':[a]}))
Columns:
vector array
word str
Rows: 1
Data:
+-----------------+------+
| vector | word |
+-----------------+------+
| [1.0, 2.0, 3.0] | foo |
+-----------------+------+
[1 rows x 2 columns]
有没有其他方法可以做到这一点？
编辑
在尝试了@papayawarrior解决方案之后，如果我可以将整个数据帧加载到内存中，它就会工作，但是有一些qurik会让它变得奇怪。
假设我对sframe的原始输入如上所述（有501列），但在
.csv文件中，我有代码将它们读入所需的字典：def get_embeddings(embedding_gzip, size):
coltypes = [str] + [float] * size
sf = gl.SFrame.read_csv('compose-vectors/' + embedding_gzip, delimiter='\t', column_type_hints=coltypes, header=False, quote_char='\0')
sf = sf.pack_columns(['X'+str(i) for i in range(2, size+1)])
df = sf.to_dataframe().set_index('X1')
print list(df)
return df.to_dict(orient='dict')['X2']
但奇怪的是，它给出了这个错误：
File "sts_compose.py", line 28, in get_embeddings
return df.to_dict(orient='dict')['X2']
KeyError: 'X2'
所以当我在转换到字典之前检查列名时，我发现我的列名不是'x1'和'x2'，而是
list(df)打印['X501', 'X3']。我转换
graphlab.SFrame -> pandas.DataFrame -> dict的方式有问题吗？我知道我可以通过这样做来解决问题，但问题仍然是，“列名是怎么变得如此奇怪的？”以下内容：
def get_embeddings(embedding_gzip, size):
coltypes = [str] + [float] * size
sf = gl.SFrame.read_csv('compose-vectors/' + embedding_gzip, delimiter='\t', column_type_hints=coltypes, header=False, quote_char='\0')
sf = sf.pack_columns(['X'+str(i) for i in range(2, size+1)])
df = sf.to_dataframe().set_index('X1')
col_names = list(df)
return df.to_dict(orient='dict')[col_names[1]]
最佳答案
编辑以匹配文章中的新问题。
@Adrien Renaud非常熟悉SFrame.pack_columns方法，但如果数据集适合内存，我建议在最后一个问题中使用Pandas数据框to_dict。
>>> import graphlab as gl
>>> sf = gl.SFrame({'X1': ['cat', 'dog'], 'X2': [1, 2], 'X3': [3, 4]})
>>> sf
+-----+----+----+
| X1 | X2 | X3 |
+-----+----+----+
| cat | 1 | 3 |
| dog | 2 | 4 |
+-----+----+----+
>>> sf2 = sf.rename({'X1': 'word'})
>>> sf2 = sf.pack_columns(column_prefix='X', new_column_name='vector')
>>> sf2
+------+--------+
| word | vector |
+------+--------+
| cat | [1, 3] |
| dog | [2, 4] |
+------+--------+
>>> df = sf2.to_dataframe().set_index('word')
>>> result = df.to_dict(orient='dict')['vector']
>>> result
{'cat': [1, 3], 'dog': [2, 4]}