Question

当我在pyspark中使用bucketizer时，我试图获取拆分值。当前结果包含存储桶的索引：

data = [(0, -1.0), (1, 0.0), (2, 0.5), (3, 1.0), (4, 10.0),(5, 25.0),(6, 100.0),(7, 300.0),(8,float("nan"))]
df = spark.createDataFrame(data, ["id", "value"])
splits = [-float("inf"),0,0.001, 1, 5,10, 20, 30, 40, 50, 60, 70, 80, 90, 100, float("inf")]
result_bucketizer = Bucketizer(splits=splits, inputCol="value",outputCol="result").setHandleInvalid("keep").transform(df)
result_bucketizer.show()

结果是：

+---+-----+------+
| id|value|result|
+---+-----+------+
|  0| -1.0|   0.0|
|  1|  0.0|   1.0|
|  2|  0.5|   2.0|
|  3|  1.0|   3.0|
|  4| 10.0|   5.0|
|  5| 25.0|   6.0|
|  6|100.0|  14.0|
|  7|300.0|  14.0|
|  8|  NaN|  15.0|
+---+-----+------+

我希望结果是：

+---+-----+------+
| id|value|result|
+---+-----+------+
|  0| -1.0|  -inf|
|  1|  0.0|   0.0|
|  2|  0.5| 0.001|
|  3|  1.0|   1.0|
|  4| 10.0|  10.0|
|  5| 25.0|  20.0|
|  6|100.0| 100.0|
|  7|300.0| 100.0|
|  8|  NaN|   NaN|
+---+-----+------+

Answer 1

这就是我的方法。

首先，我创建了数据框。

<Data ...>foo&#10;bar&#10;baz</Data>

然后，我将Bucketizer创建为一个单独的变量。

from pyspark.ml.feature import Bucketizer
from pyspark.sql.types import StringType

data = [(0, -1.0), (1, 0.0), (2, 0.5), (3, 1.0), (4, 10.0),(5, 25.0),(6, 100.0),(7, 300.0),(8,float("nan"))]
df = spark.createDataFrame(data, ["id", "value"])
splits = [-float("inf"),0,0.001, 1, 5,10, 20, 30, 40, 50, 60, 70, 80, 90, 100, float("inf")]
# here I created a dictionary with {index: name of split}
splits_dict = {i:splits[i] for i in range(len(splits))}

要获取标签，我只是使用我们之前定义的dict应用了replace函数。

# create bucketizer
bucketizer = Bucketizer(splits=splits, inputCol="value",outputCol="result")
# bucketed dataframe
bucketed = bucketizer.setHandleInvalid('skip').transform(df)

输出：

bucketed = bucketed.replace(to_replace=splits_dict, subset=['result'])
bucketed.show()

如何在pyspark.bucketizer

1 个答案: