I have Spark 2.3 very big dataframe like this:
-------------------------
| col_key | col1 | col2 |
-------------------------
| AA | 1 | 2 |
| AB | 2 | 1 |
| AA | 2 | 3 |
| AC | 1 | 2 |
| AA | 3 | 2 |
| AC | 5 | 3 |
-------------------------
I need to "split" this dataframe by values in col_key column and save each splitted part in separate csv file, so I have to get smaller dataframes like
-------------------------
| col_key | col1 | col2 |
-------------------------
| AA | 1 | 2 |
| AA | 2 | 3 |
| AA | 3 | 2 |
-------------------------
and
-------------------------
| col_key | col1 | col2 |
-------------------------
| AC | 1 | 2 |
| AC | 5 | 3 |
-------------------------
and so far. Every result dataframe I need to save as different csv file.
Count of keys is not big (20-30) but total count of data is (~200 millions records).
I have the solution where in the loop is selected every part of data and then saved to file:
val keysList = df.select("col_key").distinct().map(r => r.getString(0)).collect.toList
keysList.foreach(k => {
val dfi = df.where($"col_key" === lit(k))
SaveDataByKey(dfi, path_to_save)
})
It works correct, but bad issue of this solution is that every selection of data by every key couse full passing through whole dataframe, and it get too many time. I think must be faster solution, where we pass through dataframe only once and during this put every record to "rigth" result dataframe (or directly to separate file). But I don't know how can to do it :) May be, someone have ideas about it?
Also I prefer to use Spark's DataFrame API because it provides fastest way of data processing (so using RDD's is not desirable, if possible).
SaveDataByKeydo? Is what you want to do simply to save the dataframe into different folders partitioned on thecol_keycolumn?SaveDataByKeydoes exactly this.