How to query an HDF store using Pandas/Python

hdfs, pandas, python

Solution

You are pretty close.

In [1]: df = DataFrame({'A' : ['foo','foo','bar','bar','baz'],
                        'B' : [1,2,1,2,1], 
                        'C' : np.random.randn(5) })

In [2]: df
Out[2]: 
     A  B         C
0  foo  1 -0.909708
1  foo  2  1.321838
2  bar  1  0.368994
3  bar  2 -0.058657
4  baz  1 -1.159151

[5 rows x 3 columns]

Write the store as a table (note that in 0.12 you will use `table=True`, rather than `format='table'`). Remember to specify the `data_columns` that you want to query when creating the table (or you can do `data_columns=True`)

In [3]: df.to_hdf('test.h5','df',mode='w',format='table',data_columns=['A','B'])

In [4]: pd.read_hdf('test.h5','df')
Out[4]: 
     A  B         C
0  foo  1 -0.909708
1  foo  2  1.321838
2  bar  1  0.368994
3  bar  2 -0.058657
4  baz  1 -1.159151

[5 rows x 3 columns]

Syntax in master/0.13, isin is accomplished via `query_column=list_of_values`. This is presented as a string to where.

In [8]: pd.read_hdf('test.h5','df',where='A=["foo","bar"] & B=1')
Out[8]: 
     A  B         C
0  foo  1 -0.909708
2  bar  1  0.368994

[2 rows x 3 columns]

Syntax in 0.12, this must be a list (which ands the conditions).

In [11]: pd.read_hdf('test.h5','df',where=[pd.Term('A','=',["foo","bar"]),'B=1'])
Out[11]: 
     A  B         C
0  foo  1 -0.909708
2  bar  1  0.368994

[2 rows x 3 columns]

Problem

To manage the amount of RAM I consume in doing an analysis, I have a large dataset stored in hdf5 (.h5) and I need to query this dataset efficiently using Pandas. The data set contains user performance data for a suite of apps. I only want to pull a few fields out of the 40 possible, and then filter the resulting dataframe to only those users who are using a one of a few apps that interest me. ``` # list of apps I want to analyze apps = ['a','d','f'] # Users.h5 contains only one field_table called 'df' store = pd.HDFStore('Users.h5') # the following query works fine df = store.select('df',columns=['account','metric1','metric2'],where=['Month==10','IsMessager==1']) # the following pseudo-query fails df = store.select('df',columns=['account','metric1','metric2'],where=['Month==10','IsMessager==1', 'app in apps']) ``` I realize that the string 'app in apps' is not what I want. This is simply a SQL-like representation of what I hope to achieve. I cant seem to pass a list of strings in any way that I try, but there must be a way. For now I am simply running the query without this parameter and then I filter out the apps I don't want in a subsequent step thusly ``` df = df[df['app'].isin(apps)] ``` But this is much less efficient since ALL of the apps need to first be loaded into memory before I can remove them. In some cases, this is big problem because I don't have enough memory to support the whole unfiltered df.

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