def getFilename(aggregationInfo, inputFile):
<EOL>inputFile = resource_filename("<STR_LIT>", inputFile)<EOL>a = defaultdict(lambda: <NUM_LIT:0>, aggregationInfo)<EOL>outputDir = os.path.dirname(inputFile)<EOL>outputFile = '<STR_LIT>' % os.path.splitext(os.path.basename(inputFile))[<NUM_LIT:0>]<EOL>noAggregation = True<EOL>timePeriods = '<STR_LIT>''<STR_LIT>'<EOL>for k in timePeriods.split():<EOL><INDENT>if a[k] > <NUM_LIT:0>:<EOL><INDENT>noAggregation = False<EOL>outputFile += '<STR_LIT>' % (k, a[k])<EOL><DEDENT><DEDENT>if noAggregation:<EOL><INDENT>return inputFile<EOL><DEDENT>outputFile += '<STR_LIT>'<EOL>outputFile = os.path.join(outputDir, outputFile)<EOL>return outputFile<EOL>
Generate the filename for aggregated dataset

    The filename is based on the input filename and the
    aggregation period.

    Returns the inputFile if no aggregation required (aggregation
    info has all 0's)