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Posted to user@flink.apache.org by Michele Bertoni <mi...@mail.polimi.it> on 2015/06/26 12:12:26 UTC
open multiple file from list of uri
Hi everybody,
is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…) and open them as different files?
I know i may open the entire directory, but i want to be able to select a subset of files in the directory
thanks
Re: open multiple file from list of uri
Posted by Stephan Ewen <se...@apache.org>.
You are right, the implementation needs a place holder here.
The placeholder can probably be a "fake path", like
"file:///this/will/never/be/read/anyways", because you override the
"createSplits" method...
On Thu, Jul 16, 2015 at 12:03 AM, Michele Bertoni <
michele1.bertoni@mail.polimi.it> wrote:
> uhm, it doesn’t seem to work: it calls the configure() method that checks
> if filePath is null and throws an exception
> Actually i set that field only during the createInputSplits that is some
> steps later
>
>
>
> Il giorno 15/lug/2015, alle ore 13:16, Stephan Ewen <se...@apache.org>
> ha scritto:
>
> If you want to work without the placeholder, simply do: "env.createInput(new
> myDelimitedInputFormat(parser)(paths))
>
> The "createInputSplits()" method looks good.
>
> Greetings,
> Stephan
>
>
> On Tue, Jul 14, 2015 at 11:42 PM, Michele Bertoni <
> michele1.bertoni@mail.polimi.it> wrote:
>
>> Ok thank you, now I solved it!
>>
>>
>> The problem was in the env.readFile(myInputFormat, path)
>>
>> now that path is actually a list of paths what should I pass it?
>>
>>
>>
>> I solved in this way
>>
>> env.readFile(new myDelimitedInputFormat(parser)(paths), paths.head)
>>
>> where that paths.head gives to the read file a url that is just a
>> “placeholder” and seems to be never used, and the custom input format takes
>> care of creating the split out of the list of dir
>>
>> I tried and it works
>> is it correct way to do that? :)
>>
>>
>>
>> fyi the create input split is implemented in this way
>>
>> override def createInputSplits(minNumSplits : Int) = {
>> files.flatMap((f) => {
>> super.setFilePath(f)
>> super.createInputSplits(minNumSplits)
>> }).toArray
>> }
>>
>> where paths is a parameter of the input format constructor (as much as
>> the custom parser as shown above)
>>
>> do you think it is useful if a open a stack overflow post of it (maybe
>> with the custom parser too)?
>>
>>
>>
>>
>> cheers
>> michele
>>
>>
>> Il giorno 14/lug/2015, alle ore 18:50, Stephan Ewen <se...@apache.org>
>> ha scritto:
>>
>> For the approach that I outlined, you need to subclass of the file
>> input format.
>>
>> In that subclass, you store the list of URIs (in a new variable), and
>> override the "createInputSplits()" method.
>>
>> Stephan
>>
>> On Tue, Jul 14, 2015 at 6:42 PM, Michele Bertoni <
>> michele1.bertoni@mail.polimi.it> wrote:
>>
>>> Hi Stephan, I started working on this today, but I am having a problem
>>>
>>> Can you be a little more detailed in the procedure?
>>> actually I don’t understand how to give to the input format the list of
>>> URI since it will try putting it in a Path variable
>>>
>>> createinputsplit does not receive the path but takes a path from that
>>> variable
>>>
>>>
>>> Thanks,
>>> Michele
>>>
>>>
>>> Il giorno 26/giu/2015, alle ore 12:28, Michele Bertoni <
>>> michele1.bertoni@mail.polimi.it> ha scritto:
>>>
>>> Right!
>>> later I will do the question and quoting your answer with the solution :)
>>>
>>> Il giorno 26/giu/2015, alle ore 12:27, Stephan Ewen <se...@apache.org>
>>> ha scritto:
>>>
>>> Seems like a good idea to collect these questions.
>>>
>>> Stackoverflow is also a good place for "useful tricks"...
>>>
>>> On Fri, Jun 26, 2015 at 12:25 PM, Michele Bertoni <
>>> michele1.bertoni@mail.polimi.it> wrote:
>>>
>>>> Got it!
>>>> i will try thanks! :)
>>>>
>>>> What about writing a section of it in the programming guide?
>>>> I found a couple of topic about the readers in the mailing list, it
>>>> seems it may be helpful
>>>>
>>>>
>>>>
>>>> Il giorno 26/giu/2015, alle ore 12:21, Stephan Ewen <se...@apache.org>
>>>> ha scritto:
>>>>
>>>> Sure, just override the "createInputSplits()" method. Call for each
>>>> of your file paths "super.createInputSplits()" and then combine the results
>>>> into one array that you return.
>>>>
>>>> That should do it...
>>>>
>>>> On Fri, Jun 26, 2015 at 12:19 PM, Michele Bertoni <
>>>> michele1.bertoni@mail.polimi.it> wrote:
>>>>
>>>>> Hi Stephan, thanks for answering,
>>>>> right now I am using an extension of the DelimitedInputFormat, is
>>>>> there a way to merge it with the option 2?
>>>>>
>>>>>
>>>>>
>>>>> Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <se...@apache.org>
>>>>> ha scritto:
>>>>>
>>>>> There are two ways you can realize that:
>>>>>
>>>>> 1) Create multiple sources and union them. This is easy, but
>>>>> probably a bit less efficient.
>>>>>
>>>>> 2) Override the FileInputFormat's createInputSplits method to take a
>>>>> union of the paths to create a list of all files and fils splits that will
>>>>> be read.
>>>>>
>>>>> Stephan
>>>>>
>>>>>
>>>>> On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <
>>>>> michele1.bertoni@mail.polimi.it> wrote:
>>>>>
>>>>>> Hi everybody,
>>>>>> is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…)
>>>>>> and open them as different files?
>>>>>> I know i may open the entire directory, but i want to be able to
>>>>>> select a subset of files in the directory
>>>>>>
>>>>>> thanks
>>>>>
>>>>>
>>>>>
>>>>>
>>>>
>>>>
>>>
>>>
>>>
>>
>>
>
>
Re: open multiple file from list of uri
Posted by Michele Bertoni <mi...@mail.polimi.it>.
uhm, it doesn’t seem to work: it calls the configure() method that checks if filePath is null and throws an exception
Actually i set that field only during the createInputSplits that is some steps later
Il giorno 15/lug/2015, alle ore 13:16, Stephan Ewen <se...@apache.org>> ha scritto:
If you want to work without the placeholder, simply do: "env.createInput(new myDelimitedInputFormat(parser)(paths))
The "createInputSplits()" method looks good.
Greetings,
Stephan
On Tue, Jul 14, 2015 at 11:42 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Ok thank you, now I solved it!
The problem was in the env.readFile(myInputFormat, path)
now that path is actually a list of paths what should I pass it?
I solved in this way
env.readFile(new myDelimitedInputFormat(parser)(paths), paths.head)
where that paths.head gives to the read file a url that is just a “placeholder” and seems to be never used, and the custom input format takes care of creating the split out of the list of dir
I tried and it works
is it correct way to do that? :)
fyi the create input split is implemented in this way
override def createInputSplits(minNumSplits : Int) = {
files.flatMap((f) => {
super.setFilePath(f)
super.createInputSplits(minNumSplits)
}).toArray
}
where paths is a parameter of the input format constructor (as much as the custom parser as shown above)
do you think it is useful if a open a stack overflow post of it (maybe with the custom parser too)?
cheers
michele
Il giorno 14/lug/2015, alle ore 18:50, Stephan Ewen <se...@apache.org>> ha scritto:
For the approach that I outlined, you need to subclass of the file input format.
In that subclass, you store the list of URIs (in a new variable), and override the "createInputSplits()" method.
Stephan
On Tue, Jul 14, 2015 at 6:42 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi Stephan, I started working on this today, but I am having a problem
Can you be a little more detailed in the procedure?
actually I don’t understand how to give to the input format the list of URI since it will try putting it in a Path variable
createinputsplit does not receive the path but takes a path from that variable
Thanks,
Michele
Il giorno 26/giu/2015, alle ore 12:28, Michele Bertoni <mi...@mail.polimi.it>> ha scritto:
Right!
later I will do the question and quoting your answer with the solution :)
Il giorno 26/giu/2015, alle ore 12:27, Stephan Ewen <se...@apache.org>> ha scritto:
Seems like a good idea to collect these questions.
Stackoverflow is also a good place for "useful tricks"...
On Fri, Jun 26, 2015 at 12:25 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Got it!
i will try thanks! :)
What about writing a section of it in the programming guide?
I found a couple of topic about the readers in the mailing list, it seems it may be helpful
Il giorno 26/giu/2015, alle ore 12:21, Stephan Ewen <se...@apache.org>> ha scritto:
Sure, just override the "createInputSplits()" method. Call for each of your file paths "super.createInputSplits()" and then combine the results into one array that you return.
That should do it...
On Fri, Jun 26, 2015 at 12:19 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi Stephan, thanks for answering,
right now I am using an extension of the DelimitedInputFormat, is there a way to merge it with the option 2?
Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <se...@apache.org>> ha scritto:
There are two ways you can realize that:
1) Create multiple sources and union them. This is easy, but probably a bit less efficient.
2) Override the FileInputFormat's createInputSplits method to take a union of the paths to create a list of all files and fils splits that will be read.
Stephan
On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi everybody,
is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…) and open them as different files?
I know i may open the entire directory, but i want to be able to select a subset of files in the directory
thanks
Re: open multiple file from list of uri
Posted by Stephan Ewen <se...@apache.org>.
If you want to work without the placeholder, simply do: "env.createInput(new
myDelimitedInputFormat(parser)(paths))
The "createInputSplits()" method looks good.
Greetings,
Stephan
On Tue, Jul 14, 2015 at 11:42 PM, Michele Bertoni <
michele1.bertoni@mail.polimi.it> wrote:
> Ok thank you, now I solved it!
>
>
> The problem was in the env.readFile(myInputFormat, path)
>
> now that path is actually a list of paths what should I pass it?
>
>
>
> I solved in this way
>
> env.readFile(new myDelimitedInputFormat(parser)(paths), paths.head)
>
> where that paths.head gives to the read file a url that is just a
> “placeholder” and seems to be never used, and the custom input format takes
> care of creating the split out of the list of dir
>
> I tried and it works
> is it correct way to do that? :)
>
>
>
> fyi the create input split is implemented in this way
>
> override def createInputSplits(minNumSplits : Int) = {
> files.flatMap((f) => {
> super.setFilePath(f)
> super.createInputSplits(minNumSplits)
> }).toArray
> }
>
> where paths is a parameter of the input format constructor (as much as
> the custom parser as shown above)
>
> do you think it is useful if a open a stack overflow post of it (maybe
> with the custom parser too)?
>
>
>
>
> cheers
> michele
>
>
> Il giorno 14/lug/2015, alle ore 18:50, Stephan Ewen <se...@apache.org>
> ha scritto:
>
> For the approach that I outlined, you need to subclass of the file input
> format.
>
> In that subclass, you store the list of URIs (in a new variable), and
> override the "createInputSplits()" method.
>
> Stephan
>
> On Tue, Jul 14, 2015 at 6:42 PM, Michele Bertoni <
> michele1.bertoni@mail.polimi.it> wrote:
>
>> Hi Stephan, I started working on this today, but I am having a problem
>>
>> Can you be a little more detailed in the procedure?
>> actually I don’t understand how to give to the input format the list of
>> URI since it will try putting it in a Path variable
>>
>> createinputsplit does not receive the path but takes a path from that
>> variable
>>
>>
>> Thanks,
>> Michele
>>
>>
>> Il giorno 26/giu/2015, alle ore 12:28, Michele Bertoni <
>> michele1.bertoni@mail.polimi.it> ha scritto:
>>
>> Right!
>> later I will do the question and quoting your answer with the solution :)
>>
>> Il giorno 26/giu/2015, alle ore 12:27, Stephan Ewen <se...@apache.org>
>> ha scritto:
>>
>> Seems like a good idea to collect these questions.
>>
>> Stackoverflow is also a good place for "useful tricks"...
>>
>> On Fri, Jun 26, 2015 at 12:25 PM, Michele Bertoni <
>> michele1.bertoni@mail.polimi.it> wrote:
>>
>>> Got it!
>>> i will try thanks! :)
>>>
>>> What about writing a section of it in the programming guide?
>>> I found a couple of topic about the readers in the mailing list, it
>>> seems it may be helpful
>>>
>>>
>>>
>>> Il giorno 26/giu/2015, alle ore 12:21, Stephan Ewen <se...@apache.org>
>>> ha scritto:
>>>
>>> Sure, just override the "createInputSplits()" method. Call for each of
>>> your file paths "super.createInputSplits()" and then combine the results
>>> into one array that you return.
>>>
>>> That should do it...
>>>
>>> On Fri, Jun 26, 2015 at 12:19 PM, Michele Bertoni <
>>> michele1.bertoni@mail.polimi.it> wrote:
>>>
>>>> Hi Stephan, thanks for answering,
>>>> right now I am using an extension of the DelimitedInputFormat, is there
>>>> a way to merge it with the option 2?
>>>>
>>>>
>>>>
>>>> Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <se...@apache.org>
>>>> ha scritto:
>>>>
>>>> There are two ways you can realize that:
>>>>
>>>> 1) Create multiple sources and union them. This is easy, but probably
>>>> a bit less efficient.
>>>>
>>>> 2) Override the FileInputFormat's createInputSplits method to take a
>>>> union of the paths to create a list of all files and fils splits that will
>>>> be read.
>>>>
>>>> Stephan
>>>>
>>>>
>>>> On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <
>>>> michele1.bertoni@mail.polimi.it> wrote:
>>>>
>>>>> Hi everybody,
>>>>> is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…)
>>>>> and open them as different files?
>>>>> I know i may open the entire directory, but i want to be able to
>>>>> select a subset of files in the directory
>>>>>
>>>>> thanks
>>>>
>>>>
>>>>
>>>>
>>>
>>>
>>
>>
>>
>
>
Re: open multiple file from list of uri
Posted by Michele Bertoni <mi...@mail.polimi.it>.
Ok thank you, now I solved it!
The problem was in the env.readFile(myInputFormat, path)
now that path is actually a list of paths what should I pass it?
I solved in this way
env.readFile(new myDelimitedInputFormat(parser)(paths), paths.head)
where that paths.head gives to the read file a url that is just a “placeholder” and seems to be never used, and the custom input format takes care of creating the split out of the list of dir
I tried and it works
is it correct way to do that? :)
fyi the create input split is implemented in this way
override def createInputSplits(minNumSplits : Int) = {
files.flatMap((f) => {
super.setFilePath(f)
super.createInputSplits(minNumSplits)
}).toArray
}
where paths is a parameter of the input format constructor (as much as the custom parser as shown above)
do you think it is useful if a open a stack overflow post of it (maybe with the custom parser too)?
cheers
michele
Il giorno 14/lug/2015, alle ore 18:50, Stephan Ewen <se...@apache.org>> ha scritto:
For the approach that I outlined, you need to subclass of the file input format.
In that subclass, you store the list of URIs (in a new variable), and override the "createInputSplits()" method.
Stephan
On Tue, Jul 14, 2015 at 6:42 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi Stephan, I started working on this today, but I am having a problem
Can you be a little more detailed in the procedure?
actually I don’t understand how to give to the input format the list of URI since it will try putting it in a Path variable
createinputsplit does not receive the path but takes a path from that variable
Thanks,
Michele
Il giorno 26/giu/2015, alle ore 12:28, Michele Bertoni <mi...@mail.polimi.it>> ha scritto:
Right!
later I will do the question and quoting your answer with the solution :)
Il giorno 26/giu/2015, alle ore 12:27, Stephan Ewen <se...@apache.org>> ha scritto:
Seems like a good idea to collect these questions.
Stackoverflow is also a good place for "useful tricks"...
On Fri, Jun 26, 2015 at 12:25 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Got it!
i will try thanks! :)
What about writing a section of it in the programming guide?
I found a couple of topic about the readers in the mailing list, it seems it may be helpful
Il giorno 26/giu/2015, alle ore 12:21, Stephan Ewen <se...@apache.org>> ha scritto:
Sure, just override the "createInputSplits()" method. Call for each of your file paths "super.createInputSplits()" and then combine the results into one array that you return.
That should do it...
On Fri, Jun 26, 2015 at 12:19 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi Stephan, thanks for answering,
right now I am using an extension of the DelimitedInputFormat, is there a way to merge it with the option 2?
Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <se...@apache.org>> ha scritto:
There are two ways you can realize that:
1) Create multiple sources and union them. This is easy, but probably a bit less efficient.
2) Override the FileInputFormat's createInputSplits method to take a union of the paths to create a list of all files and fils splits that will be read.
Stephan
On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi everybody,
is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…) and open them as different files?
I know i may open the entire directory, but i want to be able to select a subset of files in the directory
thanks
Re: open multiple file from list of uri
Posted by Stephan Ewen <se...@apache.org>.
For the approach that I outlined, you need to subclass of the file input
format.
In that subclass, you store the list of URIs (in a new variable), and
override the "createInputSplits()" method.
Stephan
On Tue, Jul 14, 2015 at 6:42 PM, Michele Bertoni <
michele1.bertoni@mail.polimi.it> wrote:
> Hi Stephan, I started working on this today, but I am having a problem
>
> Can you be a little more detailed in the procedure?
> actually I don’t understand how to give to the input format the list of
> URI since it will try putting it in a Path variable
>
> createinputsplit does not receive the path but takes a path from that
> variable
>
>
> Thanks,
> Michele
>
>
> Il giorno 26/giu/2015, alle ore 12:28, Michele Bertoni <
> michele1.bertoni@mail.polimi.it> ha scritto:
>
> Right!
> later I will do the question and quoting your answer with the solution :)
>
> Il giorno 26/giu/2015, alle ore 12:27, Stephan Ewen <se...@apache.org>
> ha scritto:
>
> Seems like a good idea to collect these questions.
>
> Stackoverflow is also a good place for "useful tricks"...
>
> On Fri, Jun 26, 2015 at 12:25 PM, Michele Bertoni <
> michele1.bertoni@mail.polimi.it> wrote:
>
>> Got it!
>> i will try thanks! :)
>>
>> What about writing a section of it in the programming guide?
>> I found a couple of topic about the readers in the mailing list, it seems
>> it may be helpful
>>
>>
>>
>> Il giorno 26/giu/2015, alle ore 12:21, Stephan Ewen <se...@apache.org>
>> ha scritto:
>>
>> Sure, just override the "createInputSplits()" method. Call for each of
>> your file paths "super.createInputSplits()" and then combine the results
>> into one array that you return.
>>
>> That should do it...
>>
>> On Fri, Jun 26, 2015 at 12:19 PM, Michele Bertoni <
>> michele1.bertoni@mail.polimi.it> wrote:
>>
>>> Hi Stephan, thanks for answering,
>>> right now I am using an extension of the DelimitedInputFormat, is there
>>> a way to merge it with the option 2?
>>>
>>>
>>>
>>> Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <se...@apache.org>
>>> ha scritto:
>>>
>>> There are two ways you can realize that:
>>>
>>> 1) Create multiple sources and union them. This is easy, but probably
>>> a bit less efficient.
>>>
>>> 2) Override the FileInputFormat's createInputSplits method to take a
>>> union of the paths to create a list of all files and fils splits that will
>>> be read.
>>>
>>> Stephan
>>>
>>>
>>> On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <
>>> michele1.bertoni@mail.polimi.it> wrote:
>>>
>>>> Hi everybody,
>>>> is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…)
>>>> and open them as different files?
>>>> I know i may open the entire directory, but i want to be able to select
>>>> a subset of files in the directory
>>>>
>>>> thanks
>>>
>>>
>>>
>>>
>>
>>
>
>
>
Re: open multiple file from list of uri
Posted by Michele Bertoni <mi...@mail.polimi.it>.
Hi Stephan, I started working on this today, but I am having a problem
Can you be a little more detailed in the procedure?
actually I don’t understand how to give to the input format the list of URI since it will try putting it in a Path variable
createinputsplit does not receive the path but takes a path from that variable
Thanks,
Michele
Il giorno 26/giu/2015, alle ore 12:28, Michele Bertoni <mi...@mail.polimi.it>> ha scritto:
Right!
later I will do the question and quoting your answer with the solution :)
Il giorno 26/giu/2015, alle ore 12:27, Stephan Ewen <se...@apache.org>> ha scritto:
Seems like a good idea to collect these questions.
Stackoverflow is also a good place for "useful tricks"...
On Fri, Jun 26, 2015 at 12:25 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Got it!
i will try thanks! :)
What about writing a section of it in the programming guide?
I found a couple of topic about the readers in the mailing list, it seems it may be helpful
Il giorno 26/giu/2015, alle ore 12:21, Stephan Ewen <se...@apache.org>> ha scritto:
Sure, just override the "createInputSplits()" method. Call for each of your file paths "super.createInputSplits()" and then combine the results into one array that you return.
That should do it...
On Fri, Jun 26, 2015 at 12:19 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi Stephan, thanks for answering,
right now I am using an extension of the DelimitedInputFormat, is there a way to merge it with the option 2?
Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <se...@apache.org>> ha scritto:
There are two ways you can realize that:
1) Create multiple sources and union them. This is easy, but probably a bit less efficient.
2) Override the FileInputFormat's createInputSplits method to take a union of the paths to create a list of all files and fils splits that will be read.
Stephan
On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi everybody,
is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…) and open them as different files?
I know i may open the entire directory, but i want to be able to select a subset of files in the directory
thanks
Re: open multiple file from list of uri
Posted by Michele Bertoni <mi...@mail.polimi.it>.
Right!
later I will do the question and quoting your answer with the solution :)
Il giorno 26/giu/2015, alle ore 12:27, Stephan Ewen <se...@apache.org>> ha scritto:
Seems like a good idea to collect these questions.
Stackoverflow is also a good place for "useful tricks"...
On Fri, Jun 26, 2015 at 12:25 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Got it!
i will try thanks! :)
What about writing a section of it in the programming guide?
I found a couple of topic about the readers in the mailing list, it seems it may be helpful
Il giorno 26/giu/2015, alle ore 12:21, Stephan Ewen <se...@apache.org>> ha scritto:
Sure, just override the "createInputSplits()" method. Call for each of your file paths "super.createInputSplits()" and then combine the results into one array that you return.
That should do it...
On Fri, Jun 26, 2015 at 12:19 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi Stephan, thanks for answering,
right now I am using an extension of the DelimitedInputFormat, is there a way to merge it with the option 2?
Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <se...@apache.org>> ha scritto:
There are two ways you can realize that:
1) Create multiple sources and union them. This is easy, but probably a bit less efficient.
2) Override the FileInputFormat's createInputSplits method to take a union of the paths to create a list of all files and fils splits that will be read.
Stephan
On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi everybody,
is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…) and open them as different files?
I know i may open the entire directory, but i want to be able to select a subset of files in the directory
thanks
Re: open multiple file from list of uri
Posted by Stephan Ewen <se...@apache.org>.
Seems like a good idea to collect these questions.
Stackoverflow is also a good place for "useful tricks"...
On Fri, Jun 26, 2015 at 12:25 PM, Michele Bertoni <
michele1.bertoni@mail.polimi.it> wrote:
> Got it!
> i will try thanks! :)
>
> What about writing a section of it in the programming guide?
> I found a couple of topic about the readers in the mailing list, it seems
> it may be helpful
>
>
>
> Il giorno 26/giu/2015, alle ore 12:21, Stephan Ewen <se...@apache.org>
> ha scritto:
>
> Sure, just override the "createInputSplits()" method. Call for each of
> your file paths "super.createInputSplits()" and then combine the results
> into one array that you return.
>
> That should do it...
>
> On Fri, Jun 26, 2015 at 12:19 PM, Michele Bertoni <
> michele1.bertoni@mail.polimi.it> wrote:
>
>> Hi Stephan, thanks for answering,
>> right now I am using an extension of the DelimitedInputFormat, is there a
>> way to merge it with the option 2?
>>
>>
>>
>> Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <se...@apache.org>
>> ha scritto:
>>
>> There are two ways you can realize that:
>>
>> 1) Create multiple sources and union them. This is easy, but probably a
>> bit less efficient.
>>
>> 2) Override the FileInputFormat's createInputSplits method to take a
>> union of the paths to create a list of all files and fils splits that will
>> be read.
>>
>> Stephan
>>
>>
>> On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <
>> michele1.bertoni@mail.polimi.it> wrote:
>>
>>> Hi everybody,
>>> is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…)
>>> and open them as different files?
>>> I know i may open the entire directory, but i want to be able to select
>>> a subset of files in the directory
>>>
>>> thanks
>>
>>
>>
>>
>
>
Re: open multiple file from list of uri
Posted by Michele Bertoni <mi...@mail.polimi.it>.
Got it!
i will try thanks! :)
What about writing a section of it in the programming guide?
I found a couple of topic about the readers in the mailing list, it seems it may be helpful
Il giorno 26/giu/2015, alle ore 12:21, Stephan Ewen <se...@apache.org>> ha scritto:
Sure, just override the "createInputSplits()" method. Call for each of your file paths "super.createInputSplits()" and then combine the results into one array that you return.
That should do it...
On Fri, Jun 26, 2015 at 12:19 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi Stephan, thanks for answering,
right now I am using an extension of the DelimitedInputFormat, is there a way to merge it with the option 2?
Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <se...@apache.org>> ha scritto:
There are two ways you can realize that:
1) Create multiple sources and union them. This is easy, but probably a bit less efficient.
2) Override the FileInputFormat's createInputSplits method to take a union of the paths to create a list of all files and fils splits that will be read.
Stephan
On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi everybody,
is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…) and open them as different files?
I know i may open the entire directory, but i want to be able to select a subset of files in the directory
thanks
Re: open multiple file from list of uri
Posted by Stephan Ewen <se...@apache.org>.
Sure, just override the "createInputSplits()" method. Call for each of your
file paths "super.createInputSplits()" and then combine the results into
one array that you return.
That should do it...
On Fri, Jun 26, 2015 at 12:19 PM, Michele Bertoni <
michele1.bertoni@mail.polimi.it> wrote:
> Hi Stephan, thanks for answering,
> right now I am using an extension of the DelimitedInputFormat, is there a
> way to merge it with the option 2?
>
>
>
> Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <se...@apache.org>
> ha scritto:
>
> There are two ways you can realize that:
>
> 1) Create multiple sources and union them. This is easy, but probably a
> bit less efficient.
>
> 2) Override the FileInputFormat's createInputSplits method to take a
> union of the paths to create a list of all files and fils splits that will
> be read.
>
> Stephan
>
>
> On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <
> michele1.bertoni@mail.polimi.it> wrote:
>
>> Hi everybody,
>> is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…)
>> and open them as different files?
>> I know i may open the entire directory, but i want to be able to select a
>> subset of files in the directory
>>
>> thanks
>
>
>
>
Re: open multiple file from list of uri
Posted by Michele Bertoni <mi...@mail.polimi.it>.
Hi Stephan, thanks for answering,
right now I am using an extension of the DelimitedInputFormat, is there a way to merge it with the option 2?
Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <se...@apache.org>> ha scritto:
There are two ways you can realize that:
1) Create multiple sources and union them. This is easy, but probably a bit less efficient.
2) Override the FileInputFormat's createInputSplits method to take a union of the paths to create a list of all files and fils splits that will be read.
Stephan
On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <mi...@mail.polimi.it>> wrote:
Hi everybody,
is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…) and open them as different files?
I know i may open the entire directory, but i want to be able to select a subset of files in the directory
thanks
Re: open multiple file from list of uri
Posted by Stephan Ewen <se...@apache.org>.
There are two ways you can realize that:
1) Create multiple sources and union them. This is easy, but probably a bit
less efficient.
2) Override the FileInputFormat's createInputSplits method to take a union
of the paths to create a list of all files and fils splits that will be
read.
Stephan
On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <
michele1.bertoni@mail.polimi.it> wrote:
> Hi everybody,
> is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…)
> and open them as different files?
> I know i may open the entire directory, but i want to be able to select a
> subset of files in the directory
>
> thanks