Pyspark- Take total sum of a column and use the value to divide another column
I have a dataframe df
>>> df = spark.createDataFrame([[1,0], [2,1], [3,1], [4,0], [5,1]], ['a', 'b'])
>>> df.show()
+---+---+
| a| b|
+---+---+
| 1| 0|
| 2| 1|
| 3| 1|
| 4| 0|
| 5| 1|
+---+---+
and
>>> nrows = df.count()
Using df, I created a new dataframe a that is an aggregate of df.
>>> a = df.groupby('b').count()
>>> a.show()
+---+-----+
| b|count|
+---+-----+
| 0| 2|
| 1| 3|
+---+-----+
I need to create a new column in a called ev. The value of ev on the ith row is given by
This is the output I'm expecting
+---+-----+------------------+
| b|count| ev_norm|
+---+-----+------------------+
| 0| 2| 1.25|
| 1| 3|0.8333333333333334|
+---+-----+------------------+
But I reached here by first, creating a new column num for the numerator (nrows-count) which yields 3,2 on the two rows. I then proceeded to calculate the denominator (denom=0.48) which is the same for all rows. Lastly, I created a new column using .withColumn that divides the num column with denom.
Is there a way I can perform the calculation of denom on the fly without having to precalculate it and perform the above operations in a single step?
I tried something like
a = a.withColumn('ev_norm', (nrows - F.col('count'))/F.sum(F.col('count')*(nrows - F.col('count'))))`
and
a = a.withColumn('ev_norm', (sum([F.col('count')*(nrows-F.col('count'))]))
but it either gives me an error or the wrong answer.
python pyspark apache-spark-sql
asked Nov 27 '18 at 6:40
Clock SlaveClock Slave
2,26852360
add a comment |
I have a dataframe df
>>> df = spark.createDataFrame([[1,0], [2,1], [3,1], [4,0], [5,1]], ['a', 'b'])
>>> df.show()
+---+---+
| a| b|
+---+---+
| 1| 0|
| 2| 1|
| 3| 1|
| 4| 0|
| 5| 1|
+---+---+
and
>>> nrows = df.count()
Using df, I created a new dataframe a that is an aggregate of df.
>>> a = df.groupby('b').count()
>>> a.show()
+---+-----+
| b|count|
+---+-----+
| 0| 2|
| 1| 3|
+---+-----+
I need to create a new column in a called ev. The value of ev on the ith row is given by
This is the output I'm expecting
+---+-----+------------------+
| b|count| ev_norm|
+---+-----+------------------+
| 0| 2| 1.25|
| 1| 3|0.8333333333333334|
+---+-----+------------------+
But I reached here by first, creating a new column num for the numerator (nrows-count) which yields 3,2 on the two rows. I then proceeded to calculate the denominator (denom=0.48) which is the same for all rows. Lastly, I created a new column using .withColumn that divides the num column with denom.
Is there a way I can perform the calculation of denom on the fly without having to precalculate it and perform the above operations in a single step?
I tried something like
a = a.withColumn('ev_norm', (nrows - F.col('count'))/F.sum(F.col('count')*(nrows - F.col('count'))))`
and
a = a.withColumn('ev_norm', (sum([F.col('count')*(nrows-F.col('count'))]))
but it either gives me an error or the wrong answer.
python pyspark apache-spark-sql
asked Nov 27 '18 at 6:40
Clock SlaveClock Slave
2,26852360
You should notice that if you can do this computation on the fly, you will do it for each row and it has a high computational cost. Hence, it would be plausible to precompute this single value!
– OmG
Nov 27 '18 at 7:44
Don't post pictures of code/data/formulas.
– pault
Nov 27 '18 at 14:38
add a comment |
I have a dataframe df
>>> df = spark.createDataFrame([[1,0], [2,1], [3,1], [4,0], [5,1]], ['a', 'b'])
>>> df.show()
+---+---+
| a| b|
+---+---+
| 1| 0|
| 2| 1|
| 3| 1|
| 4| 0|
| 5| 1|
+---+---+
and
>>> nrows = df.count()
Using df, I created a new dataframe a that is an aggregate of df.
>>> a = df.groupby('b').count()
>>> a.show()
+---+-----+
| b|count|
+---+-----+
| 0| 2|
| 1| 3|
+---+-----+
I need to create a new column in a called ev. The value of ev on the ith row is given by
This is the output I'm expecting
+---+-----+------------------+
| b|count| ev_norm|
+---+-----+------------------+
| 0| 2| 1.25|
| 1| 3|0.8333333333333334|
+---+-----+------------------+
But I reached here by first, creating a new column num for the numerator (nrows-count) which yields 3,2 on the two rows. I then proceeded to calculate the denominator (denom=0.48) which is the same for all rows. Lastly, I created a new column using .withColumn that divides the num column with denom.
Is there a way I can perform the calculation of denom on the fly without having to precalculate it and perform the above operations in a single step?
I tried something like
a = a.withColumn('ev_norm', (nrows - F.col('count'))/F.sum(F.col('count')*(nrows - F.col('count'))))`
and
a = a.withColumn('ev_norm', (sum([F.col('count')*(nrows-F.col('count'))]))
but it either gives me an error or the wrong answer.
python pyspark apache-spark-sql
asked Nov 27 '18 at 6:40
Clock SlaveClock Slave
2,26852360
I have a dataframe df
>>> df = spark.createDataFrame([[1,0], [2,1], [3,1], [4,0], [5,1]], ['a', 'b'])
>>> df.show()
+---+---+
| a| b|
+---+---+
| 1| 0|
| 2| 1|
| 3| 1|
| 4| 0|
| 5| 1|
+---+---+
and
>>> nrows = df.count()
Using df, I created a new dataframe a that is an aggregate of df.
>>> a = df.groupby('b').count()
>>> a.show()
+---+-----+
| b|count|
+---+-----+
| 0| 2|
| 1| 3|
+---+-----+
I need to create a new column in a called ev. The value of ev on the ith row is given by
This is the output I'm expecting
+---+-----+------------------+
| b|count| ev_norm|
+---+-----+------------------+
| 0| 2| 1.25|
| 1| 3|0.8333333333333334|
+---+-----+------------------+
But I reached here by first, creating a new column num for the numerator (nrows-count) which yields 3,2 on the two rows. I then proceeded to calculate the denominator (denom=0.48) which is the same for all rows. Lastly, I created a new column using .withColumn that divides the num column with denom.
Is there a way I can perform the calculation of denom on the fly without having to precalculate it and perform the above operations in a single step?
I tried something like
a = a.withColumn('ev_norm', (nrows - F.col('count'))/F.sum(F.col('count')*(nrows - F.col('count'))))`
and
a = a.withColumn('ev_norm', (sum([F.col('count')*(nrows-F.col('count'))]))
but it either gives me an error or the wrong answer.
python pyspark apache-spark-sql
python pyspark apache-spark-sql
asked Nov 27 '18 at 6:40
Clock SlaveClock Slave
2,26852360
asked Nov 27 '18 at 6:40
Clock SlaveClock Slave
2,26852360
asked Nov 27 '18 at 6:40
Clock SlaveClock Slave
2,26852360
asked Nov 27 '18 at 6:40
Clock SlaveClock Slave
2,26852360
asked Nov 27 '18 at 6:40
Clock SlaveClock Slave
2,26852360
2,26852360
You should notice that if you can do this computation on the fly, you will do it for each row and it has a high computational cost. Hence, it would be plausible to precompute this single value!
– OmG
Nov 27 '18 at 7:44
Don't post pictures of code/data/formulas.
– pault
Nov 27 '18 at 14:38
add a comment |
You should notice that if you can do this computation on the fly, you will do it for each row and it has a high computational cost. Hence, it would be plausible to precompute this single value!
– OmG
Nov 27 '18 at 7:44
Don't post pictures of code/data/formulas.
– pault
Nov 27 '18 at 14:38
You should notice that if you can do this computation on the fly, you will do it for each row and it has a high computational cost. Hence, it would be plausible to precompute this single value!
– OmG
Nov 27 '18 at 7:44
You should notice that if you can do this computation on the fly, you will do it for each row and it has a high computational cost. Hence, it would be plausible to precompute this single value!
– OmG
Nov 27 '18 at 7:44
Don't post pictures of code/data/formulas.
– pault
Nov 27 '18 at 14:38
Don't post pictures of code/data/formulas.
– pault
Nov 27 '18 at 14:38
add a comment |
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You should notice that if you can do this computation on the fly, you will do it for each row and it has a high computational cost. Hence, it would be plausible to precompute this single value!
– OmG
Nov 27 '18 at 7:44
Don't post pictures of code/data/formulas.
– pault
Nov 27 '18 at 14:38