pyspark sql functions instead of rdd distinct

Question

I have been attempting to replace strings in a data set for specific columns. Either with 1 or 0, 'Y' if 1, otherwise 0.

I have managed to identify which columns to target, using a dataframe to rdd conversion with a lambda, but it is taking a while to process.

A switch to an rdd is done for each column and then a distinct is performed, this is taking a while!

If a 'Y' exists in the distinct result set then the column is identified as requiring a transformation.

I was wondering if anyone can suggest how I can use pyspark sql functions exclusively to obtain the same result instead of having to switch for each column?

The code, on sample data, is as follows:

    import pyspark.sql.types as typ
    import pyspark.sql.functions as func

    col_names = [
        ('ALIVE', typ.StringType()),
        ('AGE', typ.IntegerType()),
        ('CAGE', typ.IntegerType()),
        ('CNT1', typ.IntegerType()),
        ('CNT2', typ.IntegerType()),
        ('CNT3', typ.IntegerType()),
        ('HE', typ.IntegerType()),
        ('WE', typ.IntegerType()),
        ('WG', typ.IntegerType()),
        ('DBP', typ.StringType()),
        ('DBG', typ.StringType()),
        ('HT1', typ.StringType()),
        ('HT2', typ.StringType()),
        ('PREV', typ.StringType())
        ]

    schema = typ.StructType([typ.StructField(c[0], c[1], False) for c in col_names])
    df = spark.createDataFrame([('Y',22,56,4,3,65,180,198,18,'N','Y','N','N','N'),
                                ('N',38,79,3,4,63,155,167,12,'N','N','N','Y','N'),
                                ('Y',39,81,6,6,60,128,152,24,'N','N','N','N','Y')]
                               ,schema=schema)

    cols = [(col.name, col.dataType) for col in df.schema]

    transform_cols = []

    for s in cols:
      if s[1] == typ.StringType():
        distinct_result = df.select(s[0]).distinct().rdd.map(lambda row: row[0]).collect()
        if 'Y' in distinct_result:
          transform_cols.append(s[0])

    print(transform_cols)

The output is :

['ALIVE', 'DBG', 'HT2', 'PREV']

Answer 1

I managed to use udf in order to do the task. First, pick the column with Y or N (here I use func.first in order to skim through the first row):

cols_sel = df.select([func.first(col).alias(col) for col in df.columns]).collect()[0].asDict()
cols = [col_name for (col_name, v) in cols_sel.items() if v in ['Y', 'N']]
# return ['HT2', 'ALIVE', 'DBP', 'HT1', 'PREV', 'DBG']

Next, You can create udf function in order to map Y , N to 1 , 0 .

def map_input(val):
    map_dict = dict(zip(['Y', 'N'], [1, 0]))
    return map_dict.get(val)
udf_map_input = func.udf(map_input, returnType=typ.IntegerType())

for col in cols:
    df = df.withColumn(col, udf_map_input(col))
df.show()

Finally, you can sum the column. I then transform output into dictionary and check which columns has value greater than 0 (ie contains Y )

out = df.select([func.sum(col).alias(col) for col in cols]).collect()
out = out[0]
print([col_name for (col_name, val) in out.asDict().items() if val > 0])

Output

['DBG', 'HT2', 'ALIVE', 'PREV']