I have the following list:
columns = [('url','string'),('count','bigint'),('isindex','boolean')]
I want to add this columns to my df with empty values:
for column in columns:
df = df.withColumn(column[0], f.lit(?).cast(?))
I am not sure what I need to put in the lit function and in the cast in order to have the suitable empty value for each type
Thank you!
Related
In pyspark , i tried to do this
df = df.select(F.col("id"),
F.col("mp_code"),
F.col("mp_def"),
F.col("mp_desc"),
F.col("mp_code_desc"),
F.col("zdmtrt06_zstation").alias("station"),
F.to_timestamp(F.col("date_time"), "yyyyMMddHHmmss").alias("date_time_utc"))
df = df.groupBy("id", "mp_code", "mp_def", "mp_desc", "mp_code_desc", "station").min(F.col("date_time_utc"))
But, i have an issue
raise TypeError("Column is not iterable")
TypeError: Column is not iterable
Here is an extract of the pyspark documentation
GroupedData.min(*cols)[source]
Computes the min value for each numeric column for each group.
New in version 1.3.0.
Parameters: cols : str
In other words, the min function does not support column arguments. It only works with column names (strings) like this:
df.groupBy("x").min("date_time_utc")
# you can also specify several column names
df.groupBy("x").min("y", "z")
Note that if you want to use a column object, you have to use agg:
df.groupBy("x").agg(F.min(F.col("date_time_utc")))
As in the title. I have a list of columns and need to replace a certain string with 0 in these columns. I can do that using select statement with nested when function but I want to preserve my original dataframe and only change the columns in question. df.replace(string, 0, list_of_columns) doesn't work as there is a data type mismatch.
So I ended up with something like this which worked for me:
for column in column_list:
df = df.withColumn(column, F.when((F.col(column) == "string"), "0").otherwise(F.col(column)))
I have a dataframe which contains months and will change quite frequently. I am saving this dataframe values as list e.g. months = ['202111', '202112', '202201']. Using a for loop to to iterate through all list elements and trying to provide dynamic column values with following code:
for i in months:
df = (
adjustment_1_prepared_df.select("product", "mnth", "col1", "col2")
.groupBy("product")
.agg(
f.min(f.when(condition, f.col("col1")).otherwise(9999999)).alias(
concat("col3_"), f.lit(i.col)
)
)
)
So basically in alias I am trying to give column name as a combination of constant (minInv_) and a variable (e.g. 202111) but I am getting error. How can I give a column name as combination of fixed string and a variable.
Thanks in advance!
.alias("col3_"+str(i.col))
I have a config defined which contains a list of column for each table to be used as a dedup key
for ex:
config 1 :
val lst = List(section_xid, learner_xid)
these are the column that needs to be used as a dedup keys. This list is dynamic some table will have 1 value some will have 2 or 3 values in it
what I am trying to do is build a single key column from this list
df.
.withColumn( "dedup_key_sk", uuid(md5(concat($"lst(0)",$"lst(1)"))) )
how do I make this dynamic which will work for any number of columns in list .
I tried doing this
df.withColumn("dedup_key_sk", concat(Seq($"col1", $"col2"):_*))
For this to work I had to convert list to Df and each value in list needs to be in separate columns I was not able to figure that out.
tried doing this but didn't work
val res = sc.parallelize(List((lst))).toDF
ANy input here will be appreciated . Thank you
The list of strings can be mapped to a list of columns (using functions.col). This list of columns can then be used with concat:
val lst: List[String] = List("section_xid", "learner_xid")
df.withColumn("dedup_key_sk", concat(lst.map(col):_*)).show()
Hello guys i have this function that gets the row Values from a DataFrame, converts them into a list and the makes a Dataframe from it.
//Gets the row content from the "content column"
val dfList = df.select("content").rdd.map(r => r(0).toString).collect.toList
val dataSet = sparkSession.createDataset(dfList)
//Makes a new DataFrame
sparkSession.read.json(dataSet)
What i need to do to make a list with other column values so i can have another DataFrame with the other columns values
val dfList = df.select("content","collection", "h").rdd.map(r => {
println("******ROW********")
println(r(0).toString)
println(r(1).toString)
println(r(2).toString) //These have the row values from the other
//columns in the select
}).collect.toList
thanks
Approach doesn't look right, you don't need to collect dataframe to just add new columns. Try adding columns to directly to dataframe using withColumn() withColumnRenamed() https://docs.azuredatabricks.net/spark/1.6/sparkr/functions/withColumn.html.
If you want to bring columns from another dataframe try joining. In any case it's not good idea to use collect as it will bring all your data to driver.