我有以下数据框,其中包含2列:
第一列有列名
第二列有值列表。
+--------------------+--------------------+
| Column| Quantile|
+--------------------+--------------------+
| rent|[4000.0, 4500.0, ...|
| is_rent_changed|[0.0, 0.0, 0.0, 0...|
| phone|[7.022372888E9, 7...|
| Area_house|[1000.0, 1000.0, ...|
| bedroom_count|[1.0, 1.0, 1.0, 1...|
| bathroom_count|[1.0, 1.0, 1.0, 1...|
| maintenance_cost|[0.0, 0.0, 0.0, 0...|
| latitude|[12.8217605, 12.8...|
| Max_rent|[9000.0, 10000.0,...|
| Beds|[2.0, 2.0, 2.0, 2...|
| Area|[1000.0, 1000.0, ...|
| Avg_Rent|[3500.0, 4000.0, ...|
| deposit_amount|[0.0, 0.0, 0.0, 0...|
| commission|[0.0, 0.0, 0.0, 0...|
| monthly_rent|[0.0, 0.0, 0.0, 0...|
|is_min_rent_guara...|[0.0, 0.0, 0.0, 0...|
|min_guarantee_amount|[0.0, 0.0, 0.0, 0...|
|min_guarantee_dur...|[1.0, 1.0, 1.0, 1...|
| furnish_cost|[0.0, 0.0, 0.0, 0...|
| owner_furnish_part|[0.0, 0.0, 0.0, 0...|
+--------------------+--------------------+
如何将第二列拆分为多列保留相同的数据集。
我可以使用以下方式访问这些值:
univar_df10.select("Column", univar_df10.Quantile[0],univar_df10.Quantile[1],univar_df10.Quantile[2]).show()
+--------------------+-------------+-------------+------------+
| Column| Quantile[0]| Quantile[1]| Quantile[2]|
+--------------------+-------------+-------------+------------+
| rent| 4000.0| 4500.0| 5000.0|
| is_rent_changed| 0.0| 0.0| 0.0|
| phone|7.022372888E9|7.042022842E9|7.07333021E9|
| Area_house| 1000.0| 1000.0| 1000.0|
| bedroom_count| 1.0| 1.0| 1.0|
| bathroom_count| 1.0| 1.0| 1.0|
| maintenance_cost| 0.0| 0.0| 0.0|
| latitude| 12.8217605| 12.8490502| 12.863517|
| Max_rent| 9000.0| 10000.0| 11500.0|
| Beds| 2.0| 2.0| 2.0|
| Area| 1000.0| 1000.0| 1000.0|
| Avg_Rent| 3500.0| 4000.0| 4125.0|
| deposit_amount| 0.0| 0.0| 0.0|
| commission| 0.0| 0.0| 0.0|
| monthly_rent| 0.0| 0.0| 0.0|
|is_min_rent_guara...| 0.0| 0.0| 0.0|
|min_guarantee_amount| 0.0| 0.0| 0.0|
|min_guarantee_dur...| 1.0| 1.0| 1.0|
| furnish_cost| 0.0| 0.0| 0.0|
| owner_furnish_part| 0.0| 0.0| 0.0|
+--------------------+-------------+-------------+------------+
only showing top 20 rows
我希望我的新数据框能够将我的第二列列表拆分为多个列,就像上面的数据集一样。 提前谢谢。
答案 0 :(得分:4)
假设(您的问题被标记为关闭,因为不清楚您要问的是什么)您的问题是Quantile
列中的列表有一定长度,因此手动构建相应的命令是不方便的,这里是使用列表添加和理解作为select
的参数的解决方案:
spark.version
# u'2.2.1'
# make some toy data
from pyspark.sql import Row
df = spark.createDataFrame([Row([0,45,63,0,0,0,0]),
Row([0,0,0,85,0,69,0]),
Row([0,89,56,0,0,0,0])],
['features'])
df.show()
# result:
+-----------------------+
|features |
+-----------------------+
|[0, 45, 63, 0, 0, 0, 0]|
|[0, 0, 0, 85, 0, 69, 0]|
|[0, 89, 56, 0, 0, 0, 0]|
+-----------------------+
# get the length of your lists, if you don't know it already (here is 7):
length = len(df.select('features').take(1)[0][0])
length
# 7
df.select([df.features] + [df.features[i] for i in range(length)]).show()
# result:
+--------------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+
| features|features[0]|features[1]|features[2]|features[3]|features[4]|features[5]|features[6]|
+--------------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+
|[0, 45, 63, 0, 0,...| 0| 45| 63| 0| 0| 0| 0|
|[0, 0, 0, 85, 0, ...| 0| 0| 0| 85| 0| 69| 0|
|[0, 89, 56, 0, 0,...| 0| 89| 56| 0| 0| 0| 0|
+--------------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+
所以,在你的情况下,
univar_df10.select([univar_df10.Column] + [univar_df10.Quantile[i] for i in range(length)])
在将length
计算为之后,应该完成这项工作
length = len(univar_df10.select('Quantile').take(1)[0][0])
答案 1 :(得分:1)
这是在scala中执行此操作的伪代码:-
import org.apache.spark.sql.functions.split
import org.apache.spark.sql.functions.col
#Create column which you wanted to be .
val quantileColumn = Seq("quantile1","qunatile2","quantile3")
#Get the number of columns
val numberOfColums = quantileColumn.size
#Create a list of column
val columList = for (i <- 0 until numberOfColums ) yield split(col("Quantile"),",").getItem(i).alias(quantileColumn(i))
#Just perfom Select operation.
df.select(columList: _ *)
# If you want some columns to be added or dropped , use withColumn & dropp on df.