我以为这很容易,但找不到答案:-)
如何将中的“名称”列转换为列表。我希望我可以得到isin的工作,而不是一个连接对另一个datframe列。但isin似乎需要一个列表(如果我理解正确的话)。
创建datframe:
from pyspark import SparkContext, SparkConf, SQLContext
from datetime import datetime
sc = SparkContext().getOrCreate()
sqlContext = SQLContext(sc)
data2 = [
('George', datetime(2010, 3, 24, 3, 19, 58), 3),
('Sally', datetime(2009, 12, 12, 17, 21, 30), 5),
('Frank', datetime(2010, 11, 22, 13, 29, 40), 2),
('Paul', datetime(2010, 2, 8, 3, 31, 23), 8),
('Jesus', datetime(2009, 1, 1, 4, 19, 47), 2),
('Lou', datetime(2010, 3, 2, 4, 33, 51), 3),
]
df2 = sqlContext.createDataFrame(data2, ['name', 'trial_start_time', 'purchase_time'])
df2.show(truncate=False)
应该是这样的:
+------+-------------------+-------------+
|name |trial_start_time |purchase_time|
+------+-------------------+-------------+
|George|2010-03-24 07:19:58|3 |
|Sally |2009-12-12 22:21:30|5 |
|Frank |2010-11-22 18:29:40|2 |
|Paul |2010-02-08 08:31:23|8 |
|Jesus |2009-01-01 09:19:47|2 |
|Lou |2010-03-02 09:33:51|3 |
+------+-------------------+-------------+
我不确定是否收集是最接近我能做到这一点。
df2.select(“name”).collect()
[Row(name='George'),
Row(name='Sally'),
Row(name='Frank'),
Row(name='Paul'),
Row(name='Jesus'),
Row(name='Lou')]
关于如何将name列输出到列表有什么建议吗?
它可能需要如下所示:
[乔治,萨利,弗兰克,保罗,耶稣,卢]
1条答案
按热度按时间aemubtdh1#
使用
collect_list
函数,然后collect获取列表变量。Example:
```from pyspark.sql.functions import *
df2.agg(collect_list(col("name")).alias("name")).show(10,False)
+----------------------------------------+
|name |
+----------------------------------------+
|[George, Sally, Frank, Paul, Jesus, Lou]|
+----------------------------------------+
lst=df2.agg(collect_list(col("name"))).collect()[0][0]
lst
['George', 'Sally', 'Frank', 'Paul', 'Jesus', 'Lou']