如何通过一个数组列展平pySpark数据帧?

时间:2016-04-21 03:04:26

标签: python apache-spark pyspark

我有一个像这样的火花数据框:

+------+--------+--------------+--------------------+
|   dbn|    boro|total_students|                sBus|
+------+--------+--------------+--------------------+
|17K548|Brooklyn|           399|[B41, B43, B44-SB...|
|09X543|   Bronx|           378|[Bx13, Bx15, Bx17...|
|09X327|   Bronx|           543|[Bx1, Bx11, Bx13,...|
+------+--------+--------------+--------------------+

如何对它进行展平,以便为sBus中的每个元素复制每一行,并且sBus将是一个普通的字符串列?

所以结果会是这样的:

+------+--------+--------------+--------------------+
|   dbn|    boro|total_students|                sBus|
+------+--------+--------------+--------------------+
|17K548|Brooklyn|           399| B41                |
|17K548|Brooklyn|           399| B43                |
|17K548|Brooklyn|           399| B44-SB             |
+------+--------+--------------+--------------------+

依旧......

1 个答案:

答案 0 :(得分:2)

我无法想到这样做的方法而不将其变成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()

棘手的部分是将其他列与新展平的值保持在一起。我这样做是将每一行映射到(dict of other columns, list to flatten)的元组,然后调用flatMapValues。这会将值列表的每个元素拆分成一个单独的行,但保持附加的键,即

(key, ['A', 'B', 'C'])

变为

(key, 'A')
(key, 'B')
(key, 'C')

然后,我将展平的值移回其他列的字典中,然后将其重新转换回DataFrame。