How to create unique ID column in DASK_CUDF

user_11368603 0
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How to create unique id column in dsak cudf dataframe across all the partitions So far I am using following technique, but if I increase data to more than 10cr rows it is giving me memory error.

def unique_id(df):
    rag = cupy.arrange(len(df))
    df['unique_id']=rag
    return df
    
part = data.npartitions
data = data.repartitions(npartitions=1)
cols_meta={c:str(data[c].dtype) for c in data.columns}
data = data.map_partitions(lambda df:unique_id(df), meta={**cols_meta,'unique_id'})
data = data.repartitions(npartitions=part)

If there's any other way, or any modification in code, please suggest. Thank you for help

cudf dask python
2 Answers

The reason why you are running into memory error is this step:

data = data.repartitions(npartitions=1)

By having a single partition you are forcing all the data on a single worker, which will cause memory problems as the dataset increases in size. What you want to do instead is assign a unique identifier while maintaining each partition, see this answer.

user_106935960

I was doing that because wanted to create ids sequentially, till the length data.

The other suggestions will likely work. However, one of the easiest way to do this is to create a temporary column with value 1 and use cumsum, like the following:

import cudf
import dask_cudf
​
df = cudf.DataFrame({
    "a": ["dog"]*10
})
ddf = dask_cudf.from_cudf(df, 3)
​
ddf["temp"] = 1
ddf["monotonic_id"] = ddf["temp"].cumsum()
del ddf["temp"]
​
print(ddf.partitions[2].compute())
     a  monotonic_id
8  dog             9
9  dog            10

As expected, the two rows in the partition index 2 have IDs 9 and 10. If you need the indexes to start at 0, you can subtract 1.

user_60665280
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