我对火种很陌生。我想应用StringIndexer将列的值更改为索引。我查看了这篇文章:Apply StringIndexer to several columns in a PySpark Dataframe
此解决方案将创建一个新列,而不是更新输入列。是否有更新当前列的方法?我试图在输入和输出中使用相同的名称,但它不起作用。
label_stringIdx = StringIndexer(inputCol ="WindGustDir", outputCol = "WindGustDir_index")发布于 2019-10-14 05:46:25
不能简单地更新该列。但你能做的是
使用StringIndexer
的名称重命名新列
您可以使用以下代码
from pyspark.ml.feature import StringIndexer
import pyspark.sql.functions as F
df = spark.createDataFrame([['a', 1], ['b', 1], ['c', 2], ['b', 5]], ['WindGustDir', 'value'])
df.show()
# +-----------+-----+
# |WindGustDir|value|
# +-----------+-----+
# | a| 1|
# | b| 1|
# | c| 2|
# | b| 5|
# +-----------+-----+
# 1. create new column
label_stringIdx = StringIndexer(inputCol ="WindGustDir", outputCol = "WindGustDir_index")
label_stringIdx_model = label_stringIdx.fit(df)
df = label_stringIdx_model.transform(df)
# 2. delete original column
df = df.drop("WindGustDir")
# 3. rename new column
to_rename = ['WindGustDir_index', 'value']
replace_with = ['WindGustDir', 'value']
mapping = dict(zip(to_rename, replace_with))
df = df.select([F.col(c).alias(mapping.get(c, c)) for c in to_rename])
df.show()
# +-----------+-----+
# |WindGustDir|value|
# +-----------+-----+
# | 1.0| 1|
# | 0.0| 1|
# | 2.0| 2|
# | 0.0| 5|
# +-----------+-----+https://stackoverflow.com/questions/58368568
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