How to implement SQL Groupby in RAPIDS

I'm seeking to translate an SQL query to use RAPIDS. Consider the simplified query below:

(SELECT min(a), max(b), c

FROM T

GROUP BY c) AS result

I have validated the code below, but is this the optimal solution? Is sorting on the group key necessary? Is there a cleaner / more idiomatic way to write it?

from pygdf import DataFrame as gdf



T = gdf(...)

df = gdf({'a':T.a, 'c':T.c}).groupby('c').min().sort_values(by='c')

df['max_b'] = gdf({'b':T.b, 'c':T.c}).groupby('c').max().sort_values(by='c').max_b

result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})

edited Nov 20 '18 at 16:59

asked Nov 20 '18 at 16:49

Thomas Ryan Stovall

405

add a comment |

I'm seeking to translate an SQL query to use RAPIDS. Consider the simplified query below:

(SELECT min(a), max(b), c

FROM T

GROUP BY c) AS result

I have validated the code below, but is this the optimal solution? Is sorting on the group key necessary? Is there a cleaner / more idiomatic way to write it?

from pygdf import DataFrame as gdf



T = gdf(...)

df = gdf({'a':T.a, 'c':T.c}).groupby('c').min().sort_values(by='c')

df['max_b'] = gdf({'b':T.b, 'c':T.c}).groupby('c').max().sort_values(by='c').max_b

result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})

edited Nov 20 '18 at 16:59

asked Nov 20 '18 at 16:49

Thomas Ryan Stovall

405

add a comment |

I'm seeking to translate an SQL query to use RAPIDS. Consider the simplified query below:

(SELECT min(a), max(b), c

FROM T

GROUP BY c) AS result

I have validated the code below, but is this the optimal solution? Is sorting on the group key necessary? Is there a cleaner / more idiomatic way to write it?

from pygdf import DataFrame as gdf



T = gdf(...)

df = gdf({'a':T.a, 'c':T.c}).groupby('c').min().sort_values(by='c')

df['max_b'] = gdf({'b':T.b, 'c':T.c}).groupby('c').max().sort_values(by='c').max_b

result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})

edited Nov 20 '18 at 16:59

asked Nov 20 '18 at 16:49

Thomas Ryan Stovall

405

I'm seeking to translate an SQL query to use RAPIDS. Consider the simplified query below:

(SELECT min(a), max(b), c

FROM T

GROUP BY c) AS result

I have validated the code below, but is this the optimal solution? Is sorting on the group key necessary? Is there a cleaner / more idiomatic way to write it?

from pygdf import DataFrame as gdf



T = gdf(...)

df = gdf({'a':T.a, 'c':T.c}).groupby('c').min().sort_values(by='c')

df['max_b'] = gdf({'b':T.b, 'c':T.c}).groupby('c').max().sort_values(by='c').max_b

result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})

python rapids

edited Nov 20 '18 at 16:59

asked Nov 20 '18 at 16:49

Thomas Ryan Stovall

405

edited Nov 20 '18 at 16:59

asked Nov 20 '18 at 16:49

Thomas Ryan Stovall

405

edited Nov 20 '18 at 16:59

asked Nov 20 '18 at 16:49

Thomas Ryan Stovall

405

asked Nov 20 '18 at 16:49

Thomas Ryan Stovall

405

asked Nov 20 '18 at 16:49

Thomas Ryan Stovall

405

add a comment |

1 Answer
1

active

oldest

votes

You can rewrite your aggregation using the .agg function to make it more straightforward:

from pygdf import DataFrame as gdf



T = gdf(...)

df = gdf({'a':T.a, 'b': T.b, 'c':T.c}).groupby('c').agg({'a': 'min', 'b': 'max'})

result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})

answered Nov 23 '18 at 20:35

Keith Kraus

361

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

You can rewrite your aggregation using the .agg function to make it more straightforward:

from pygdf import DataFrame as gdf



T = gdf(...)

df = gdf({'a':T.a, 'b': T.b, 'c':T.c}).groupby('c').agg({'a': 'min', 'b': 'max'})

result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})

answered Nov 23 '18 at 20:35

Keith Kraus

361

add a comment |

You can rewrite your aggregation using the .agg function to make it more straightforward:

from pygdf import DataFrame as gdf



T = gdf(...)

df = gdf({'a':T.a, 'b': T.b, 'c':T.c}).groupby('c').agg({'a': 'min', 'b': 'max'})

result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})

answered Nov 23 '18 at 20:35

Keith Kraus

361

add a comment |

You can rewrite your aggregation using the .agg function to make it more straightforward:

from pygdf import DataFrame as gdf



T = gdf(...)

df = gdf({'a':T.a, 'b': T.b, 'c':T.c}).groupby('c').agg({'a': 'min', 'b': 'max'})

result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})

answered Nov 23 '18 at 20:35

Keith Kraus

361

You can rewrite your aggregation using the .agg function to make it more straightforward:

from pygdf import DataFrame as gdf



T = gdf(...)

df = gdf({'a':T.a, 'b': T.b, 'c':T.c}).groupby('c').agg({'a': 'min', 'b': 'max'})

result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})

answered Nov 23 '18 at 20:35

Keith Kraus

361

answered Nov 23 '18 at 20:35

Keith Kraus

361

answered Nov 23 '18 at 20:35

Keith Kraus

361

answered Nov 23 '18 at 20:35

Keith Kraus

361

add a comment |

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