Re: add an auto_increment column
I have got the answer from Mich's answer. Thank you both.
frakass
On 08/02/2022 16:36, Gourav Sengupta wrote:
Hi,
so do you want to rank apple and tomato both as 2? Not quite clear on
the use case here though.
Regards,
Gourav Sengupta
On Tue, Feb 8, 2022 at 7:10 AM wrote:
Hello Gourav
As you see here orderBy has already give the solution for "equal
amount":
df =
sc.parallelize([("orange",2),("apple",3),("tomato",3),("cherry",5)]).toDF(['fruit','amount'])
df.select("*").orderBy("amount",ascending=False).show()
+--+--+
| fruit|amount|
+--+--+
|cherry| 5|
| apple| 3|
|tomato| 3|
|orange| 2|
+--+--+
I want to add a column at the right whose name is "top" and the
value
auto_increment from 1 to N.
Thank you.
On 08/02/2022 13:52, Gourav Sengupta wrote:
Hi,
sorry once again, will try to understand the problem first :)
As we can clearly see that the initial responses were incorrectly
guessing the solution to be monotonically_increasing function
What if there are two fruits with equal amount? For any real life
application, can we understand what are trying to achieve by the
rankings?
Regards,
Gourav Sengupta
On Tue, Feb 8, 2022 at 4:22 AM ayan guha
wrote:
For this req you can rank or dense rank.
On Tue, 8 Feb 2022 at 1:12 pm, wrote:
Hello,
For this query:
df.select("*").orderBy("amount",ascending=False).show()
+--+--+
| fruit|amount|
+--+--+
|tomato| 9|
| apple| 6|
|cherry| 5|
|orange| 3|
+--+--+
I want to add a column "top", in which the value is: 1,2,3...
meaning
top1, top2, top3...
How can I do it?
Thanks.
On 07/02/2022 21:18, Gourav Sengupta wrote:
Hi,
can we understand the requirement first?
What is that you are trying to achieve by auto increment id? Do
you
just want different ID's for rows, or you may want to keep
track
of
the record count of a table as well, or do you want to do use
them for
surrogate keys?
If you are going to insert records multiple times in a table,
and
still have different values?
I think without knowing the requirements all the above
responses, like
everything else where solutions are reached before
understanding
the
problem, has high chances of being wrong.
Regards,
Gourav Sengupta
On Mon, Feb 7, 2022 at 2:21 AM Siva Samraj
wrote:
Monotonically_increasing_id() will give the same functionality
On Mon, 7 Feb, 2022, 6:57 am , wrote:
For a dataframe object, how to add a column who is
auto_increment
like
mysql's behavior?
Thank you.
-
To unsubscribe e-mail: user-unsubscr...@spark.apache.org
-
To unsubscribe e-mail: user-unsubscr...@spark.apache.org
--
Best Regards,
Ayan Guha
-
To unsubscribe e-mail: user-unsubscr...@spark.apache.org
Re: add an auto_increment column
Hi,
so do you want to rank apple and tomato both as 2? Not quite clear on the
use case here though.
Regards,
Gourav Sengupta
On Tue, Feb 8, 2022 at 7:10 AM wrote:
>
> Hello Gourav
>
>
> As you see here orderBy has already give the solution for "equal
> amount":
>
> >>> df =
> >>>
> sc.parallelize([("orange",2),("apple",3),("tomato",3),("cherry",5)]).toDF(['fruit','amount'])
>
> >>> df.select("*").orderBy("amount",ascending=False).show()
> +--+--+
> | fruit|amount|
> +--+--+
> |cherry| 5|
> | apple| 3|
> |tomato| 3|
> |orange| 2|
> +--+--+
>
>
> I want to add a column at the right whose name is "top" and the value
> auto_increment from 1 to N.
>
> Thank you.
>
>
>
> On 08/02/2022 13:52, Gourav Sengupta wrote:
> > Hi,
> >
> > sorry once again, will try to understand the problem first :)
> >
> > As we can clearly see that the initial responses were incorrectly
> > guessing the solution to be monotonically_increasing function
> >
> > What if there are two fruits with equal amount? For any real life
> > application, can we understand what are trying to achieve by the
> > rankings?
> >
> > Regards,
> > Gourav Sengupta
> >
> > On Tue, Feb 8, 2022 at 4:22 AM ayan guha wrote:
> >
> >> For this req you can rank or dense rank.
> >>
> >> On Tue, 8 Feb 2022 at 1:12 pm, wrote:
> >>
> >>> Hello,
> >>>
> >>> For this query:
> >>>
> >> df.select("*").orderBy("amount",ascending=False).show()
> >>> +--+--+
> >>> | fruit|amount|
> >>> +--+--+
> >>> |tomato| 9|
> >>> | apple| 6|
> >>> |cherry| 5|
> >>> |orange| 3|
> >>> +--+--+
> >>>
> >>> I want to add a column "top", in which the value is: 1,2,3...
> >>> meaning
> >>> top1, top2, top3...
> >>>
> >>> How can I do it?
> >>>
> >>> Thanks.
> >>>
> >>> On 07/02/2022 21:18, Gourav Sengupta wrote:
> Hi,
>
> can we understand the requirement first?
>
> What is that you are trying to achieve by auto increment id? Do
> >>> you
> just want different ID's for rows, or you may want to keep track
> >>> of
> the record count of a table as well, or do you want to do use
> >>> them for
> surrogate keys?
>
> If you are going to insert records multiple times in a table,
> >>> and
> still have different values?
>
> I think without knowing the requirements all the above
> >>> responses, like
> everything else where solutions are reached before understanding
> >>> the
> problem, has high chances of being wrong.
>
> Regards,
> Gourav Sengupta
>
> On Mon, Feb 7, 2022 at 2:21 AM Siva Samraj
> >>>
> wrote:
>
> > Monotonically_increasing_id() will give the same functionality
> >
> > On Mon, 7 Feb, 2022, 6:57 am , wrote:
> >
> >> For a dataframe object, how to add a column who is
> >>> auto_increment
> >> like
> >> mysql's behavior?
> >>
> >> Thank you.
> >>
> >>
> >
>
> >>>
> >>
> > -
> >> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
> >>>
> >>>
> >>
> > -
> >>> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
> >> --
> >> Best Regards,
> >> Ayan Guha
>
Re: add an auto_increment column
Maybe col func is not even needed here. :)
>>> df.select(F.dense_rank().over(wOrder).alias("rank"),
"fruit","amount").show()
++--+--+
|rank| fruit|amount|
++--+--+
| 1|cherry| 5|
| 2| apple| 3|
| 2|tomato| 3|
| 3|orange| 2|
++--+--+
On Tue, Feb 8, 2022 at 3:50 PM Mich Talebzadeh
wrote:
> simple either rank() or desnse_rank()
>
> >>> from pyspark.sql import functions as F
> >>> from pyspark.sql.functions import col
> >>> from pyspark.sql.window import Window
> >>> wOrder = Window().orderBy(df['amount'].desc())
> >>> df.select(F.rank().over(wOrder).alias("rank"), col('fruit'),
> col('amount')).show()
> ++--+--+
> |rank| fruit|amount|
> ++--+--+
> | 1|cherry| 5|
> | 2| apple| 3|
> | 2|tomato| 3|
> | 4|orange| 2|
> ++--+--+
>
> >>> df.select(F.dense_rank().over(wOrder).alias("rank"), col('fruit'),
> col('amount')).show()
> ++--+--+
> |rank| fruit|amount|
> ++--+--+
> | 1|cherry| 5|
> | 2| apple| 3|
> | 2|tomato| 3|
> | 3|orange| 2|
> ++--+--+
>
> HTH
>
>
>view my Linkedin profile
>
Re: add an auto_increment column
simple either rank() or desnse_rank()
>>> from pyspark.sql import functions as F
>>> from pyspark.sql.functions import col
>>> from pyspark.sql.window import Window
>>> wOrder = Window().orderBy(df['amount'].desc())
>>> df.select(F.rank().over(wOrder).alias("rank"), col('fruit'),
col('amount')).show()
++--+--+
|rank| fruit|amount|
++--+--+
| 1|cherry| 5|
| 2| apple| 3|
| 2|tomato| 3|
| 4|orange| 2|
++--+--+
>>> df.select(F.dense_rank().over(wOrder).alias("rank"), col('fruit'),
col('amount')).show()
++--+--+
|rank| fruit|amount|
++--+--+
| 1|cherry| 5|
| 2| apple| 3|
| 2|tomato| 3|
| 3|orange| 2|
++--+--+
HTH
view my Linkedin profile
Re: add an auto_increment column
https://stackoverflow.com/a/51854022/299676
On Tue, 8 Feb 2022 at 09:25, Stelios Philippou wrote:
> This has the information that you require in order to add an extra column
> with a sequence to it.
>
>
> On Tue, 8 Feb 2022 at 09:11, wrote:
>
>>
>> Hello Gourav
>>
>>
>> As you see here orderBy has already give the solution for "equal
>> amount":
>>
>> >>> df =
>> >>>
>> sc.parallelize([("orange",2),("apple",3),("tomato",3),("cherry",5)]).toDF(['fruit','amount'])
>>
>> >>> df.select("*").orderBy("amount",ascending=False).show()
>> +--+--+
>> | fruit|amount|
>> +--+--+
>> |cherry| 5|
>> | apple| 3|
>> |tomato| 3|
>> |orange| 2|
>> +--+--+
>>
>>
>> I want to add a column at the right whose name is "top" and the value
>> auto_increment from 1 to N.
>>
>> Thank you.
>>
>>
>>
>> On 08/02/2022 13:52, Gourav Sengupta wrote:
>> > Hi,
>> >
>> > sorry once again, will try to understand the problem first :)
>> >
>> > As we can clearly see that the initial responses were incorrectly
>> > guessing the solution to be monotonically_increasing function
>> >
>> > What if there are two fruits with equal amount? For any real life
>> > application, can we understand what are trying to achieve by the
>> > rankings?
>> >
>> > Regards,
>> > Gourav Sengupta
>> >
>> > On Tue, Feb 8, 2022 at 4:22 AM ayan guha wrote:
>> >
>> >> For this req you can rank or dense rank.
>> >>
>> >> On Tue, 8 Feb 2022 at 1:12 pm, wrote:
>> >>
>> >>> Hello,
>> >>>
>> >>> For this query:
>> >>>
>> >> df.select("*").orderBy("amount",ascending=False).show()
>> >>> +--+--+
>> >>> | fruit|amount|
>> >>> +--+--+
>> >>> |tomato| 9|
>> >>> | apple| 6|
>> >>> |cherry| 5|
>> >>> |orange| 3|
>> >>> +--+--+
>> >>>
>> >>> I want to add a column "top", in which the value is: 1,2,3...
>> >>> meaning
>> >>> top1, top2, top3...
>> >>>
>> >>> How can I do it?
>> >>>
>> >>> Thanks.
>> >>>
>> >>> On 07/02/2022 21:18, Gourav Sengupta wrote:
>> Hi,
>>
>> can we understand the requirement first?
>>
>> What is that you are trying to achieve by auto increment id? Do
>> >>> you
>> just want different ID's for rows, or you may want to keep track
>> >>> of
>> the record count of a table as well, or do you want to do use
>> >>> them for
>> surrogate keys?
>>
>> If you are going to insert records multiple times in a table,
>> >>> and
>> still have different values?
>>
>> I think without knowing the requirements all the above
>> >>> responses, like
>> everything else where solutions are reached before understanding
>> >>> the
>> problem, has high chances of being wrong.
>>
>> Regards,
>> Gourav Sengupta
>>
>> On Mon, Feb 7, 2022 at 2:21 AM Siva Samraj
>> >>>
>> wrote:
>>
>> > Monotonically_increasing_id() will give the same functionality
>> >
>> > On Mon, 7 Feb, 2022, 6:57 am , wrote:
>> >
>> >> For a dataframe object, how to add a column who is
>> >>> auto_increment
>> >> like
>> >> mysql's behavior?
>> >>
>> >> Thank you.
>> >>
>> >>
>> >
>>
>> >>>
>> >>
>> > -
>> >> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
>> >>>
>> >>>
>> >>
>> > -
>> >>> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
>> >> --
>> >> Best Regards,
>> >> Ayan Guha
>>
>> -
>> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
>>
>>
Re: add an auto_increment column
This has the information that you require in order to add an extra column
with a sequence to it.
On Tue, 8 Feb 2022 at 09:11, wrote:
>
> Hello Gourav
>
>
> As you see here orderBy has already give the solution for "equal
> amount":
>
> >>> df =
> >>>
> sc.parallelize([("orange",2),("apple",3),("tomato",3),("cherry",5)]).toDF(['fruit','amount'])
>
> >>> df.select("*").orderBy("amount",ascending=False).show()
> +--+--+
> | fruit|amount|
> +--+--+
> |cherry| 5|
> | apple| 3|
> |tomato| 3|
> |orange| 2|
> +--+--+
>
>
> I want to add a column at the right whose name is "top" and the value
> auto_increment from 1 to N.
>
> Thank you.
>
>
>
> On 08/02/2022 13:52, Gourav Sengupta wrote:
> > Hi,
> >
> > sorry once again, will try to understand the problem first :)
> >
> > As we can clearly see that the initial responses were incorrectly
> > guessing the solution to be monotonically_increasing function
> >
> > What if there are two fruits with equal amount? For any real life
> > application, can we understand what are trying to achieve by the
> > rankings?
> >
> > Regards,
> > Gourav Sengupta
> >
> > On Tue, Feb 8, 2022 at 4:22 AM ayan guha wrote:
> >
> >> For this req you can rank or dense rank.
> >>
> >> On Tue, 8 Feb 2022 at 1:12 pm, wrote:
> >>
> >>> Hello,
> >>>
> >>> For this query:
> >>>
> >> df.select("*").orderBy("amount",ascending=False).show()
> >>> +--+--+
> >>> | fruit|amount|
> >>> +--+--+
> >>> |tomato| 9|
> >>> | apple| 6|
> >>> |cherry| 5|
> >>> |orange| 3|
> >>> +--+--+
> >>>
> >>> I want to add a column "top", in which the value is: 1,2,3...
> >>> meaning
> >>> top1, top2, top3...
> >>>
> >>> How can I do it?
> >>>
> >>> Thanks.
> >>>
> >>> On 07/02/2022 21:18, Gourav Sengupta wrote:
> Hi,
>
> can we understand the requirement first?
>
> What is that you are trying to achieve by auto increment id? Do
> >>> you
> just want different ID's for rows, or you may want to keep track
> >>> of
> the record count of a table as well, or do you want to do use
> >>> them for
> surrogate keys?
>
> If you are going to insert records multiple times in a table,
> >>> and
> still have different values?
>
> I think without knowing the requirements all the above
> >>> responses, like
> everything else where solutions are reached before understanding
> >>> the
> problem, has high chances of being wrong.
>
> Regards,
> Gourav Sengupta
>
> On Mon, Feb 7, 2022 at 2:21 AM Siva Samraj
> >>>
> wrote:
>
> > Monotonically_increasing_id() will give the same functionality
> >
> > On Mon, 7 Feb, 2022, 6:57 am , wrote:
> >
> >> For a dataframe object, how to add a column who is
> >>> auto_increment
> >> like
> >> mysql's behavior?
> >>
> >> Thank you.
> >>
> >>
> >
>
> >>>
> >>
> > -
> >> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
> >>>
> >>>
> >>
> > -
> >>> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
> >> --
> >> Best Regards,
> >> Ayan Guha
>
> -
> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
>
>
Re: add an auto_increment column
Hello Gourav
As you see here orderBy has already give the solution for "equal
amount":
df =
sc.parallelize([("orange",2),("apple",3),("tomato",3),("cherry",5)]).toDF(['fruit','amount'])
df.select("*").orderBy("amount",ascending=False).show()
+--+--+
| fruit|amount|
+--+--+
|cherry| 5|
| apple| 3|
|tomato| 3|
|orange| 2|
+--+--+
I want to add a column at the right whose name is "top" and the value
auto_increment from 1 to N.
Thank you.
On 08/02/2022 13:52, Gourav Sengupta wrote:
Hi,
sorry once again, will try to understand the problem first :)
As we can clearly see that the initial responses were incorrectly
guessing the solution to be monotonically_increasing function
What if there are two fruits with equal amount? For any real life
application, can we understand what are trying to achieve by the
rankings?
Regards,
Gourav Sengupta
On Tue, Feb 8, 2022 at 4:22 AM ayan guha wrote:
For this req you can rank or dense rank.
On Tue, 8 Feb 2022 at 1:12 pm, wrote:
Hello,
For this query:
df.select("*").orderBy("amount",ascending=False).show()
+--+--+
| fruit|amount|
+--+--+
|tomato| 9|
| apple| 6|
|cherry| 5|
|orange| 3|
+--+--+
I want to add a column "top", in which the value is: 1,2,3...
meaning
top1, top2, top3...
How can I do it?
Thanks.
On 07/02/2022 21:18, Gourav Sengupta wrote:
Hi,
can we understand the requirement first?
What is that you are trying to achieve by auto increment id? Do
you
just want different ID's for rows, or you may want to keep track
of
the record count of a table as well, or do you want to do use
them for
surrogate keys?
If you are going to insert records multiple times in a table,
and
still have different values?
I think without knowing the requirements all the above
responses, like
everything else where solutions are reached before understanding
the
problem, has high chances of being wrong.
Regards,
Gourav Sengupta
On Mon, Feb 7, 2022 at 2:21 AM Siva Samraj
wrote:
Monotonically_increasing_id() will give the same functionality
On Mon, 7 Feb, 2022, 6:57 am , wrote:
For a dataframe object, how to add a column who is
auto_increment
like
mysql's behavior?
Thank you.
-
To unsubscribe e-mail: user-unsubscr...@spark.apache.org
-
To unsubscribe e-mail: user-unsubscr...@spark.apache.org
--
Best Regards,
Ayan Guha
-
To unsubscribe e-mail: user-unsubscr...@spark.apache.org
Re: add an auto_increment column
Hi,
sorry once again, will try to understand the problem first :)
As we can clearly see that the initial responses were incorrectly guessing
the solution to be monotonically_increasing function
What if there are two fruits with equal amount? For any real life
application, can we understand what are trying to achieve by the rankings?
Regards,
Gourav Sengupta
On Tue, Feb 8, 2022 at 4:22 AM ayan guha wrote:
> For this req you can rank or dense rank.
>
> On Tue, 8 Feb 2022 at 1:12 pm, wrote:
>
>> Hello,
>>
>> For this query:
>>
>> >>> df.select("*").orderBy("amount",ascending=False).show()
>> +--+--+
>> | fruit|amount|
>> +--+--+
>> |tomato| 9|
>> | apple| 6|
>> |cherry| 5|
>> |orange| 3|
>> +--+--+
>>
>>
>> I want to add a column "top", in which the value is: 1,2,3... meaning
>> top1, top2, top3...
>>
>> How can I do it?
>>
>> Thanks.
>>
>>
>>
>>
>> On 07/02/2022 21:18, Gourav Sengupta wrote:
>> > Hi,
>> >
>> > can we understand the requirement first?
>> >
>> > What is that you are trying to achieve by auto increment id? Do you
>> > just want different ID's for rows, or you may want to keep track of
>> > the record count of a table as well, or do you want to do use them for
>> > surrogate keys?
>> >
>> > If you are going to insert records multiple times in a table, and
>> > still have different values?
>> >
>> > I think without knowing the requirements all the above responses, like
>> > everything else where solutions are reached before understanding the
>> > problem, has high chances of being wrong.
>> >
>> > Regards,
>> > Gourav Sengupta
>> >
>> > On Mon, Feb 7, 2022 at 2:21 AM Siva Samraj
>> > wrote:
>> >
>> >> Monotonically_increasing_id() will give the same functionality
>> >>
>> >> On Mon, 7 Feb, 2022, 6:57 am , wrote:
>> >>
>> >>> For a dataframe object, how to add a column who is auto_increment
>> >>> like
>> >>> mysql's behavior?
>> >>>
>> >>> Thank you.
>> >>>
>> >>>
>> >>
>> > -
>> >>> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
>>
>> -
>> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
>>
>> --
> Best Regards,
> Ayan Guha
>
Re: add an auto_increment column
For this req you can rank or dense rank.
On Tue, 8 Feb 2022 at 1:12 pm, wrote:
> Hello,
>
> For this query:
>
> >>> df.select("*").orderBy("amount",ascending=False).show()
> +--+--+
> | fruit|amount|
> +--+--+
> |tomato| 9|
> | apple| 6|
> |cherry| 5|
> |orange| 3|
> +--+--+
>
>
> I want to add a column "top", in which the value is: 1,2,3... meaning
> top1, top2, top3...
>
> How can I do it?
>
> Thanks.
>
>
>
>
> On 07/02/2022 21:18, Gourav Sengupta wrote:
> > Hi,
> >
> > can we understand the requirement first?
> >
> > What is that you are trying to achieve by auto increment id? Do you
> > just want different ID's for rows, or you may want to keep track of
> > the record count of a table as well, or do you want to do use them for
> > surrogate keys?
> >
> > If you are going to insert records multiple times in a table, and
> > still have different values?
> >
> > I think without knowing the requirements all the above responses, like
> > everything else where solutions are reached before understanding the
> > problem, has high chances of being wrong.
> >
> > Regards,
> > Gourav Sengupta
> >
> > On Mon, Feb 7, 2022 at 2:21 AM Siva Samraj
> > wrote:
> >
> >> Monotonically_increasing_id() will give the same functionality
> >>
> >> On Mon, 7 Feb, 2022, 6:57 am , wrote:
> >>
> >>> For a dataframe object, how to add a column who is auto_increment
> >>> like
> >>> mysql's behavior?
> >>>
> >>> Thank you.
> >>>
> >>>
> >>
> > -
> >>> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
>
> -
> To unsubscribe e-mail: user-unsubscr...@spark.apache.org
>
> --
Best Regards,
Ayan Guha
Re: add an auto_increment column
Hello,
For this query:
df.select("*").orderBy("amount",ascending=False).show()
+--+--+
| fruit|amount|
+--+--+
|tomato| 9|
| apple| 6|
|cherry| 5|
|orange| 3|
+--+--+
I want to add a column "top", in which the value is: 1,2,3... meaning
top1, top2, top3...
How can I do it?
Thanks.
On 07/02/2022 21:18, Gourav Sengupta wrote:
Hi,
can we understand the requirement first?
What is that you are trying to achieve by auto increment id? Do you
just want different ID's for rows, or you may want to keep track of
the record count of a table as well, or do you want to do use them for
surrogate keys?
If you are going to insert records multiple times in a table, and
still have different values?
I think without knowing the requirements all the above responses, like
everything else where solutions are reached before understanding the
problem, has high chances of being wrong.
Regards,
Gourav Sengupta
On Mon, Feb 7, 2022 at 2:21 AM Siva Samraj
wrote:
Monotonically_increasing_id() will give the same functionality
On Mon, 7 Feb, 2022, 6:57 am , wrote:
For a dataframe object, how to add a column who is auto_increment
like
mysql's behavior?
Thank you.
-
To unsubscribe e-mail: user-unsubscr...@spark.apache.org
-
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Re: add an auto_increment column
Hi, can we understand the requirement first? What is that you are trying to achieve by auto increment id? Do you just want different ID's for rows, or you may want to keep track of the record count of a table as well, or do you want to do use them for surrogate keys? If you are going to insert records multiple times in a table, and still have different values? I think without knowing the requirements all the above responses, like everything else where solutions are reached before understanding the problem, has high chances of being wrong. Regards, Gourav Sengupta On Mon, Feb 7, 2022 at 2:21 AM Siva Samraj wrote: > Monotonically_increasing_id() will give the same functionality > > On Mon, 7 Feb, 2022, 6:57 am , wrote: > >> For a dataframe object, how to add a column who is auto_increment like >> mysql's behavior? >> >> Thank you. >> >> - >> To unsubscribe e-mail: user-unsubscr...@spark.apache.org >> >>
Re: add an auto_increment column
Monotonically_increasing_id() will give the same functionality On Mon, 7 Feb, 2022, 6:57 am , wrote: > For a dataframe object, how to add a column who is auto_increment like > mysql's behavior? > > Thank you. > > - > To unsubscribe e-mail: user-unsubscr...@spark.apache.org > >
Re: add an auto_increment column
Try this: https://spark.apache.org/docs/latest/api/python/reference/api/pyspark.sql.functions.monotonically_increasing_id.html On Mon, 7 Feb 2022 at 12:27 pm, wrote: > For a dataframe object, how to add a column who is auto_increment like > mysql's behavior? > > Thank you. > > - > To unsubscribe e-mail: user-unsubscr...@spark.apache.org > > -- Best Regards, Ayan Guha
