You can pass the keyword arguments to assign as a dictionary, like so: kwargs = {“ln(A)” : lambda x: np.log(x.A)} df.assign(**kwargs) A B ln(A) 0 1 0.500033 0.000000 1 2 -0.392229 0.693147 2 3 0.385512 1.098612 3 4 -0.029816 1.386294 4 5 -2.386748 1.609438 5 6 -1.828487 1.791759 6 7 0.096117 1.945910 7 8 -2.867469 … Read more
One option is to create a script file. Right click on the database -> Tasks -> Generate Scripts Then you can select all the stored procedures and generate the script with all the sps. So you can find the reference from there. Or — Search in All Objects SELECT OBJECT_NAME(OBJECT_ID), definition FROM sys.sql_modules WHERE definition … Read more
You can get the names from the schema by doing spark_df.schema.names Printing the schema can be useful to visualize it as well spark_df.printSchema()
You can just get/set the index via its name property In [7]: df.index.name Out[7]: ‘Index Title’ In [8]: df.index.name=”foo” In [9]: df.index.name Out[9]: ‘foo’ In [10]: df Out[10]: Column 1 foo Apples 1 Oranges 2 Puppies 3 Ducks 4