With pyspark dataframe, how do you do the equivalent of Pandas df['col'].unique()
.
I want to list out all the unique values in a pyspark dataframe column.
Not the SQL type way (registertemplate then SQL query for distinct values).
Also I don't need groupby
then countDistinct
, instead I want to check distinct VALUES in that column.
This should help to get distinct values of a column:
df.select('column1').distinct().collect()
Note that .collect()
doesn't have any built-in limit on how many values can return so this might be slow -- use .show()
instead or add .limit(20)
before .collect()
to manage this.
Let's assume we're working with the following representation of data (two columns, k
and v
, where k
contains three entries, two unique:
+---+---+
| k| v|
+---+---+
|foo| 1|
|bar| 2|
|foo| 3|
+---+---+
With a Pandas dataframe:
import pandas as pd
p_df = pd.DataFrame([("foo", 1), ("bar", 2), ("foo", 3)], columns=("k", "v"))
p_df['k'].unique()
This returns an ndarray
, i.e. array(['foo', 'bar'], dtype=object)
You asked for a "pyspark dataframe alternative for pandas df['col'].unique()". Now, given the following Spark dataframe:
s_df = sqlContext.createDataFrame([("foo", 1), ("bar", 2), ("foo", 3)], ('k', 'v'))
If you want the same result from Spark, i.e. an ndarray
, use toPandas()
:
s_df.toPandas()['k'].unique()
Alternatively, if you don't need an ndarray
specifically and just want a list of the unique values of column k
:
s_df.select('k').distinct().rdd.map(lambda r: r[0]).collect()
Finally, you can also use a list comprehension as follows:
[i.k for i in s_df.select('k').distinct().collect()]
.rdd
call after distinct()
. It worked without that in Spark 1.6.2, but I just confirmed that the edited answer works in Spark 2.0.0 as well.
You can use df.dropDuplicates(['col1','col2'])
to get only distinct rows based on colX in the array.
If you want to see the distinct values of a specific column in your dataframe, you would just need to write the following code. It would show the 100 distinct values (if 100 values are available) for the colname
column in the df
dataframe.
df.select('colname').distinct().show(100, False)
If you want to do something fancy on the distinct values, you can save the distinct values in a vector:
a = df.select('colname').distinct()
collect_set
can help to get unique values from a given column of pyspark.sql.DataFrame
:
df.select(F.collect_set("column").alias("column")).first()["column"]
collect_set
, but I think the following would be cleaner: df.agg(F.collect_set("column")).collect()[0][0]
you could do
distinct_column = 'somecol'
distinct_column_vals = df.select(distinct_column).distinct().collect()
distinct_column_vals = [v[distinct_column] for v in distinct_column_vals]
In addition to the dropDuplicates
option there is the method named as we know it in pandas
drop_duplicates
:
drop_duplicates() is an alias for dropDuplicates().
Example
s_df = sqlContext.createDataFrame([("foo", 1),
("foo", 1),
("bar", 2),
("foo", 3)], ('k', 'v'))
s_df.show()
+---+---+
| k| v|
+---+---+
|foo| 1|
|foo| 1|
|bar| 2|
|foo| 3|
+---+---+
Drop by subset
s_df.drop_duplicates(subset = ['k']).show()
+---+---+
| k| v|
+---+---+
|bar| 2|
|foo| 1|
+---+---+
s_df.drop_duplicates().show()
+---+---+
| k| v|
+---+---+
|bar| 2|
|foo| 3|
|foo| 1|
+---+---+
If you want to select ALL(columns) data as distinct frrom a DataFrame (df), then
df.select('*').distinct().show(10,truncate=False)
Run this first
df.createOrReplaceTempView('df')
Then run
spark.sql("""
SELECT distinct
column name
FROM
df
""").show()
Let us suppose that your original DataFrame is called df
. Then, you can use:
df1 = df.groupBy('column_1').agg(F.count('column_1').alias('trip_count'))
df2 = df1.sort(df1.trip_count.desc()).show()
Success story sharing
Row
objects, you need to use a list comprehension like in this answer: stackoverflow.com/a/60896261/7465462