Convert graphlab sframe into a dictionary of {key: values}

Question

Given an SFrame as such:

+------+-----------+-----------+-----------+-----------+-----------+-----------+
|  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 |
+------+-----------+-----------+-----------+-----------+-----------+-----------+

How can I convert the SFrame into a dictionary such that X1 column is the key and X2 to X7 as the np.array()?

I have tried iterating through the original SFrame row-by-row and do something like this:

>>> 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]

Is there another way to do the same?

EDITED

After trying @papayawarrior solution, it works if I can load the whole dataframe into memory but there's a few quriks that makes it odd.

Assuming that my original input to the SFrame is as presented above (with 501 columns) but in .csv file, I have the code to read them into the desired dictionary:

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']

But oddly it gives this error:

  File "sts_compose.py", line 28, in get_embeddings
    return df.to_dict(orient='dict')['X2']
KeyError: 'X2'

So when I check for the column names before the conversion to dictionary, I found that my column names are not 'X1' and 'X2' but list(df) prints ['X501', 'X3'].

Is there something wrong with how I have converting the graphlab.SFrame -> pandas.DataFrame -> dict?

I know I can resolve the problem by doing this instead, but the question remains, "How did the column names become so strange?":

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]]

Regarding the question in the edit, Adrien Renaud's answer makes use of the `new_column_name` parameter in the `SFrame.pack_columns` function, which I think is a good idea in this case to make sure you know your data column. — papayawarrior, Jan 14 '16 at 00:27
Another little tip is to rename your column of words then use the `column_prefix` parameter in `pack_columns` to avoid the list comprehension. I just updated my answer to show this. — papayawarrior, Jan 14 '16 at 00:27

Adrien Renaud · Answer 1 · 2016-01-11T21:56:29.997

Is there another way to do the same? Yes, you can use the pack_columns method from the SFrame class.

import graphlab as gl
data = gl.SFrame()
data.add_column(gl.SArray(['foo', 'bar']), 'X1')
data.add_column(gl.SArray([1., 3.]), 'X2')
data.add_column(gl.SArray([2., 4.]), 'X3')
print data
+-----+-----+-----+
|  X1 |  X2 |  X3 |
+-----+-----+-----+
| foo | 1.0 | 2.0 |
| bar | 3.0 | 4.0 |
+-----+-----+-----+
[2 rows x 3 columns]

import array
data = data.pack_columns(['X2', 'X3'], dtype=array.array, new_column_name='vector')
data = data.rename({'X1':'word'})
print data
+------+------------+
| word |   vector   |
+------+------------+
| foo  | [1.0, 2.0] |
| bar  | [3.0, 4.0] |
+------+------------+
[2 rows x 2 columns]

b=data['vector'][0]
print type(b)
<type 'array.array'>

How can I convert the SFrame into a dictionary such that X1 column is the key and X2 to X7 as the np.array()?

I didn't find any built-in method to convert an SFrame to a dict. You could try the following (it might be very slow):

a={}
def dump_sframe_to_dict(row, a):
    a[row['word']]=row['vector']
data.apply(lambda x: dump_sframe_to_dict(x, a))
print a
{'foo': array('d', [1.0, 2.0]), 'bar': array('d', [3.0, 4.0])}

Short question, when you read a row, let's say, `data[0]`, then take the vector column, `data[0][vector]` and then do a `type(data[0][vector])` did you get a list or a numpy array? — alvas, Jan 11 '16 at 20:54
By doing `data['vector'][0]`, I get an instance of `array.array`. — Adrien Renaud, Jan 11 '16 at 21:51

papayawarrior · Accepted Answer · 2016-01-14T00:31:17.050

2

Edited to match new questions in the post.

@Adrien Renaud is spot on with the SFrame.pack_columns method, but I would suggest using the Pandas dataframe to_dict for the last question if your dataset fits in memory.

>>> 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]}

edited Jan 14 '16 at 00:31

answered Jan 13 '16 at 00:24

papayawarrior

1,027
7
10

Thanks! This helps a lot. Quick question, `type(array[0][results])` returns a list, I have explicitly cast them into a `np.array()` for vector manipulation, right? – alvas Jan 13 '16 at 06:35
I'm not totally sure what you mean by `array[0]`, but assuming you mean `result['cat']`, then yes, it will be a list and you'd have to explicitly cast it to a numpy array. – papayawarrior Jan 14 '16 at 00:18
Yep! Thanks again! Casting the them into numpy array achieves the desired dictionary I'll need for my purposes. Sadly, having results in a python dictionary means I have to have enough RAM to store the whole dictionary. Is it possible to cast it back into a SFrame or DataFrame such that graphlab or pandas uses tcmalloc to handle the large dictionary? – alvas Jan 14 '16 at 00:28
1

You can definitely re-make the SFrame, but then you're moving away from your original goal of having the data in a dictionary. One suggestion would be to do feature engineering in the original SFrame, or even after the columns are packed, then convert only a small subset to the dictionary. – papayawarrior Jan 14 '16 at 00:37
Is `dict[key]` faster than `SFrame.filter_by(values=[key], column_name='word')`? The python dict is `O(1)` since it's a hash search and the other is `O(n)` right? Or is graphlab smart enough to shard the SFrame? – alvas Jan 14 '16 at 01:02

Convert graphlab sframe into a dictionary of {key: values}

EDITED

2 Answers2