I have a spark dataframe like this:
+------+--------+--------------+--------------------+
| dbn| boro|total_students| sBus|
+------+--------+--------------+--------------------+
|17K548|Brooklyn| 399|[B41, B43, B44-SB...|
|09X543| Bronx| 378|[Bx13, Bx15, Bx17...|
|09X327| Bronx| 543|[Bx1, Bx11, Bx13,...|
+------+--------+--------------+--------------------+
How do I flattern it so that each row is copied for each for each element in sBus, and sBus will be a normal string column?
So that result would be like this:
+------+--------+--------------+--------------------+
| dbn| boro|total_students| sBus|
+------+--------+--------------+--------------------+
|17K548|Brooklyn| 399| B41 |
|17K548|Brooklyn| 399| B43 |
|17K548|Brooklyn| 399| B44-SB |
+------+--------+--------------+--------------------+
and so on...
I can't think of a way to do this without turning it into an RDD.
# convert df to rdd
rdd = df.rdd
def extract(row, key):
"""Takes dictionary and key, returns tuple of (dict w/o key, dict[key])."""
_dict = row.asDict()
_list = _dict[key]
del _dict[key]
return (_dict, _list)
def add_to_dict(_dict, key, value):
_dict[key] = value
return _dict
# preserve rest of values in key, put list to flatten in value
rdd = rdd.map(lambda x: extract(x, 'sBus'))
# make a row for each item in value
rdd = rdd.flatMapValues(lambda x: x)
# add flattened value back into dictionary
rdd = rdd.map(lambda x: add_to_dict(x[0], 'sBus', x[1]))
# convert back to dataframe
df = sqlContext.createDataFrame(rdd)
df.show()
The tricky part is keeping the other columns together with the newly flattened values. I do this by mapping each row to a tuple of (dict of other columns, list to flatten) and then calling flatMapValues. This will split each element of the value list into a separate row, but keep the keys attached, i.e.
(key, ['A', 'B', 'C'])
becomes
(key, 'A')
(key, 'B')
(key, 'C')
Then, I move the flattened value back into the dictionary of other columns, and reconvert it back to a DataFrame.
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