如何在多列上将DataFrame在PySpark中枢轴？

Question

我有以下格式的数据集：

import numpy as np
import pandas as pd

# Create the data set

np.random.seed(42)

records = list()
for i in range(2):
    for j in range(2):        
        for k in range(500):
            t = np.random.randint(pd.Timestamp('2000-01-01').value, pd.Timestamp('2018-01-01').value)
            if np.random.rand() > .95: continue                
            ts = pd.Timestamp(t).strftime('%Y-%m-%d %H:%M:%S.%f')
            records.append( (i, j, np.random.rand(), ts) )

df = pd.DataFrame.from_records(records)
df.columns =['a_id', 'b_id', 'value', 'time']

看起来是这样的：

      a_id  b_id     value                        time
0        0     0  0.156019  2007-09-28 15:12:24.260596
1        0     0  0.601115  2015-09-08 01:59:18.043399
2        0     0  0.969910  2012-01-10 07:51:29.662492
3        0     0  0.181825  2011-08-28 19:58:33.281289
4        0     0  0.524756  2015-11-15 14:18:17.398715
5        0     0  0.611853  2015-01-07 23:44:37.034322
6        0     0  0.366362  2008-06-21 11:56:10.529679
7        0     0  0.199674  2010-11-08 18:24:18.794838
8        0     0  0.046450  2008-04-27 02:36:46.026876

在这里，a_id和b_id是传感器的关键。这意味着必须将数据框架转换为：

df_ = pd.pivot_table(df, index='time', columns=['a_id', 'b_id'], values='value')
df_.index = [pd.to_datetime(v) for v in df_.index]
df_ = df_.resample('1W').mean().ffill().bfill()

在重新采样和填补空白之后，数据采用了所需的格式：

a_id               0                   1          
b_id               0         1         0         1
2000-01-09  0.565028  0.560434  0.920740  0.458825
2000-01-16  0.565028  0.146963  0.920740  0.217588
2000-01-23  0.565028  0.840872  0.920740  0.209690
2000-01-30  0.565028  0.046852  0.920740  0.209690
2000-02-06  0.565028  0.046852  0.704871  0.209690

每一列现在都包含传感器的数据。

问题是，我不知道如何在PySpark中做到这一点。

df_test = spark.createDataFrame(df) \
    .withColumn('time', F.to_utc_timestamp('time', '%Y-%m-%d %H:%M:%S.%f'))
df_test.printSchema()

拥有

root
 |-- a_id: long (nullable = true)
 |-- b_id: long (nullable = true)
 |-- value: double (nullable = true)
 |-- time: timestamp (nullable = true)

如何转换df_test，使其具有与df_相同的形式

Answer 1

正如在评论中提到的那样，这里有一种解决方案可以将您的数据枢轴化：

您应该将您的列a_id和b_id连接到一个新的列c_id下，然后按date分组，然后在c_id上支点，并使用如何查看fit的值。

至于重采样，我要指出@zero323 这里提供的解决方案。

Answer 2

您可以使用pyspark.ml.feature.Bucketizer“重采样”数据。

# Truncate timestamp to day precision, and convert to unixtime
df = df.withColumn("tt",
                   F.unix_timestamp(F.date_trunc("day", "time")))
df.show(5)

# +----+----+--------------------+--------------------+---+----------+
# |a_id|b_id|               value|                time| id|        tt|
# +----+----+--------------------+--------------------+---+----------+
# |   0|   0| 0.15601864044243652|2007-09-28 15:12:...| 00|1190962800|
# |   0|   0|  0.6011150117432088|2015-09-08 01:59:...| 00|1441695600|
# |   0|   0|  0.9699098521619943|2012-01-10 07:51:...| 00|1326182400|
# |   0|   0| 0.18182496720710062|2011-08-28 19:58:...| 00|1314514800|
# |   0|   0|  0.5247564316322378|2015-11-15 14:18:...| 00|1447574400|
# +----+----+--------------------+--------------------+---+----------+

# Get the minimum and maximum dates
tmin = df.select(F.min("tt")).collect()[0][0]
tmax = df.select(F.max("tt")).collect()[0][0]

# Get the number of seconds in a week
week = 60 * 60 * 24 * 7

# Get a list of bucket splits (add infinity for last split if weeks don't evenly divide)
splits = list(range(tmin, tmax, week)) + [float("inf")]

# Create bucket and bucket your data 
bucketizer = Bucketizer(inputCol="tt", outputCol="num_week", splits=splits)
bucketed_df = Bucketizer.transform(df)
bucketed_df.show(5)

# +----+----+-------------------+--------------------+---+----------+---------+
# |a_id|b_id|              value|                time| id|        tt|num_weeks|
# +----+----+-------------------+--------------------+---+----------+---------+
# |   0|   0|0.15601864044243652|2007-09-28 15:12:...| 00|1190962800|403.0    |
# |   0|   0| 0.6011150117432088|2015-09-08 01:59:...| 00|1441695600|818.0    |
# |   0|   0| 0.9699098521619943|2012-01-10 07:51:...| 00|1326182400|627.0    |
# |   0|   0|0.18182496720710062|2011-08-28 19:58:...| 00|1314514800|607.0    |
# |   0|   0| 0.5247564316322378|2015-11-15 14:18:...| 00|1447574400|827.0    |
# +----+----+-------------------+--------------------+---+----------+---------+

# Convert the buckets to a timestamp (seconds in week * bucket value + min_date)
bucketed_df = bucketed_df.withColumn(
    "time",
    F.from_unixtime(col("weeks") * week + tmin).cast("date")))
bucketed_df.show(5)

# +----+----+-------------------+----------+---+----------+-----+
# |a_id|b_id|              value|      time| id|        tt|weeks|
# +----+----+-------------------+----------+---+----------+-----+
# |   0|   0|0.15601864044243652|2007-09-24| 00|1190962800|403.0|
# |   0|   0| 0.6011150117432088|2015-09-07| 00|1441695600|818.0|
# |   0|   0| 0.9699098521619943|2012-01-09| 00|1326182400|627.0|
# |   0|   0|0.18182496720710062|2011-08-22| 00|1314514800|607.0|
# |   0|   0| 0.5247564316322378|2015-11-09| 00|1447574400|827.0|
# +----+----+-------------------+----------+---+----------+-----+

# Finally, do the groupBy and pivot as already explained
# (I already concatenated "a_id" and "b_id" into the column "id"
final_df = bucketed_df.groupBy("time").pivot("id").agg(F.avg("value"))
final_df.show(10)


#    +----------+--------------------+--------------------+-------------------+-------------------+
#    |      time|                  00|                  01|                 10|                 11|
#    +----------+--------------------+--------------------+-------------------+-------------------+
#    |2015-03-09|0.045227288910538066|  0.8633336495718252| 0.8229838050417675|               null|
#    |2000-07-03|                null|                null| 0.7855315583735368|               null|
#    |2013-09-09|  0.6334037565104235|                null|0.14284196481433187|               null|
#    |2005-06-06|                null|  0.9095933818175037|               null|               null|
#    |2017-09-11|                null|  0.9684887775943838|               null|               null|
#    |2004-02-23|                null|  0.3782888656818202|               null|0.26674411859262276|
#    |2004-07-12|                null|                null| 0.2528581182501112| 0.4189697737795244|
#    |2000-12-25|                null|                null| 0.5473347601436167|               null|
#    |2016-04-25|                null|  0.9918099513493635|               null|               null|
#    |2016-10-03|                null|0.057844449447160606| 0.2770125243259788|               null|
#    +----------+--------------------+--------------------+-------------------+-------------------+

这会让你得到你所需要的东西。不幸的是，实现pandas.DataFrame.ffill()和pandas.DataFrame.bfill()方法并不像在熊猫中那样容易，因为数据是分布式的。有关建议，请参见这里和这里。