I want to use lazy groupby instead of eager dataframe pivot, but my result need use the one of the index as column indeed, so is there a way in rust-polars like unstack in pandas? how to rename the unstack new column? which is the "right" way to replace inefficient pivot to lazy groupy in rust-polars?
Related
i am trying to find the row number of a dataframe having 55k rows. I am having multiple index and hence i tried to search using np.where. But this is very very slow. Is there a better solution to achieve in less than a millisecond. I tried loc, iloc but seems of not better options.
lprowcount = np.where(dfreglist['lineports'] == lineport)[0]
Also any best method to locate and update a row in a large dataframe other loc, iloc methods.
Using the Kiva Loan_Data from Kaggle I aggregated the Loan Amounts by country. Pandas allows them to be easily turned into a DataFrame, but indexes on the country data. The reset_index can be used to create a numerical/sequential index, but I'm guessing I am adding an unnecessary step. Is there a way to create an automatic default index when creating a DataFrame like this?
Use as_index=False
groupby
split-apply-combine
df.groupby('country', as_index=False)['loan_amount'].sum()
I have a DataFrame df with a column column and I would like to convert column into a vector (e.g. a DenseVector) so that I can use it in vector and matrix products.
Beware: I don't need a column of vectors; I need a vector object.
How to do this?
I found out the vectorAssembler function (link) but this doesn't help me, as it converts some DataFrame columns into a vector columns, which is still a DataFrame column; my desired output should instead be a vector.
About the goal of this question: why am I trying to convert a DF column into a vector? Assume I have a DF with a numerical column and I need to compute a product between a matrix and this column. How can I achieve this? (The same could hold for a DF numerical row.) Any alternative approach is welcome.
How:
DenseVector(df.select("column_name").rdd.map(lambda x: x[0]).collect())
but it doesn't make sense in any practical scenario.
Spark Vectors are not distributed, therefore are applicable only if data fits in memory of one (driver) node. If this is the case you wouldn't use Spark DataFrame for processing.
I have a pandas DataFrame filled with strings. I would like to apply a string operation to all entries, for example capitalize(). I know that for a series we can use series.str.capitlize(). I also know that I can loop over the column of the Dataframe and do this for each of the columns. But I want something more efficient and elegant, without looping. Thanks
use stack + unstack
stack makes a dataframe with a single level column index into a series. You can then perform your str.capitalize() and unstack to get back your original form.
df.stack().str.capitalize().unstack()
I am trying to get the top 5 values of a column of my dataframe.
A sample of the dataframe is given below. In fact the original dataframe has thousands of rows.
Row(item_id=u'2712821', similarity=5.0)
Row(item_id=u'1728166', similarity=6.0)
Row(item_id=u'1054467', similarity=9.0)
Row(item_id=u'2788825', similarity=5.0)
Row(item_id=u'1128169', similarity=1.0)
Row(item_id=u'1053461', similarity=3.0)
The solution I came up with is to sort all of the dataframe and then to take the first 5 values. (the code below does that)
items_of_common_users.sort(items_of_common_users.similarity.desc()).take(5)
I am wondering if there is a faster way of achieving this.
Thanks
You can use RDD.top method with key:
from operator import attrgetter
df.rdd.top(5, attrgetter("similarity"))
There is a significant overhead of DataFrame to RDD conversion but it should be worth it.