英文:
Take a spark dataframe and collect all rows into one single row
问题
Certainly! You can achieve that by using the concat_ws function in Spark. Here's the code to transform your DataFrame as you described:
from pyspark.sql import functions as F
# Assuming you already have 'df' DataFrame defined
# Concatenate all values into one column separated by space
new_df = df.withColumn("new_column", F.concat_ws(" ", "id", "label"))
# Show the new DataFrame
new_df.show()
This will result in a DataFrame with a single column named "new_column" containing values like '1 foo' and '2 bar', just as you wanted.
英文:
is there a way to take a relational spark dataframe like the data below:
df = spark.createDataFrame(
[
(1, "foo"),
(2, "bar"),
],
["id", "label"]
)
df.show()
And collect all of the values (I don't care about the column names) into one column so it looks like below
new_df = spark.createDataFrame(["1 foo 2 bar"], "string").toDF("new_column")
new_df.show()
I do need to keep the order, so it has to be a string of '1 foo 2 bar', and not '1 2 foo bar' for example.
Is there a way to do this?
Thanks
答案1
得分: 1
Sure, here is the translated code:
是的，请尝试使用 **`concat_ws()`** 和 **`collect_list() + array_join()`** 函数。
**`示例:`**
```python
from pyspark.sql.functions import *
df = spark.createDataFrame([(1, "foo"), (2, "bar")], ["id", "label"])
df.withColumn("temp", concat_ws(" ", *df.columns)).groupBy(lit(1)).agg(array_join(collect_list(col("temp")), " ").alias("new_column")).\
drop("1").\
show(10, False)
#+-----------+
#|new_column |
#+-----------+
#|1 foo 2 bar|
#+-----------+
Please note that the code and comments have been translated, and only the translated parts are provided.
英文:
Yes try with concat_ws() and collect_list() + array_join() functions.
Example:
from pyspark.sql.functions import *
df = spark.createDataFrame([(1, "foo"),(2, "bar"),],["id", "label"])
df.withColumn("temp", concat_ws(" ", *df.columns)).groupBy(lit(1)).agg(array_join(collect_list(col("temp"))," ").alias("new_column")).\
drop("1").\
show(10,False)
#+-----------+
#|new_column |
#+-----------+
#|1 foo 2 bar|
#+-----------+
答案2
得分: 0
尝试这个：
df2 = spark.sql("select id, label, lead(id) over (order by id) as id_1, lead(label) over (order by id) as label_2 from df")
df2.createOrReplaceTempView("df2")
df3 = spark.sql("select concat(CONCAT(id, ' ', label), ' ', concat(id_1, ' ', label_2)) as one_col from df2 where id_1 is not null")
df3.show()
+-----------+
| one_col|
+-----------+
|1 foo 2 bar|
+-----------+
英文:
try this:
df2=spark.sql("select id,label,lead(id)over(order by id) as id_1,lead(label)over(order by id) as label_2 from df ")
df2.createOrReplaceTempView("df2")
df3=spark.sql("select concat(CONCAT(id, ' ', label) , ' ',concat(id_1 ,' ', label_2)) as one_col from df2 where id_1 is not null")
df3.show()
+-----------+
| one_col|
+-----------+
|1 foo 2 bar|
+-----------+
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