Creating an ArrayList from a method which returns a String

-1

I have an custom class InfoAQ which has a method called public String getSeqInf(). Now I have an ArrayList<InfoAQ> infList and
I need an ArrayList<String>strList = new ArrayList<String>with the content from getSeqInf()for each element.

This is the way Im doing it right now ...

for(InfoAQ currentInf : infList)

  strList.add(currentInf.getSeqInf());

Is there an alternative way to make it ? Maybe a faster one or one liner ?

asked Nov 21 '18 at 13:11

Ahmet Kazaman

352324

1

Possible duplicate of Lambda expression to convert array/List of String to array/List of Integers

– Sven Hakvoort
Nov 21 '18 at 13:14

Try this --> List<String> outputList = list.stream().map(it -> it.getSeqInf()).collect(Collectors.toList());

– Suryavel TR
Nov 21 '18 at 13:15

1

What you wrote can be one line. Just delete the new line you added.

– matt
Nov 21 '18 at 13:21

add a comment |

-1

This is the way Im doing it right now ...

for(InfoAQ currentInf : infList)

  strList.add(currentInf.getSeqInf());

Is there an alternative way to make it ? Maybe a faster one or one liner ?

asked Nov 21 '18 at 13:11

Ahmet Kazaman

352324

1

Possible duplicate of Lambda expression to convert array/List of String to array/List of Integers

– Sven Hakvoort
Nov 21 '18 at 13:14

Try this --> List<String> outputList = list.stream().map(it -> it.getSeqInf()).collect(Collectors.toList());

– Suryavel TR
Nov 21 '18 at 13:15

1

What you wrote can be one line. Just delete the new line you added.

– matt
Nov 21 '18 at 13:21

add a comment |

-1

This is the way Im doing it right now ...

for(InfoAQ currentInf : infList)

  strList.add(currentInf.getSeqInf());

Is there an alternative way to make it ? Maybe a faster one or one liner ?

asked Nov 21 '18 at 13:11

Ahmet Kazaman

352324

This is the way Im doing it right now ...

for(InfoAQ currentInf : infList)

  strList.add(currentInf.getSeqInf());

Is there an alternative way to make it ? Maybe a faster one or one liner ?

java

asked Nov 21 '18 at 13:11

Ahmet Kazaman

352324

asked Nov 21 '18 at 13:11

Ahmet Kazaman

352324

asked Nov 21 '18 at 13:11

Ahmet Kazaman

352324

asked Nov 21 '18 at 13:11

Ahmet Kazaman

352324

asked Nov 21 '18 at 13:11

Ahmet Kazaman

352324

1

Possible duplicate of Lambda expression to convert array/List of String to array/List of Integers

– Sven Hakvoort
Nov 21 '18 at 13:14

Try this --> List<String> outputList = list.stream().map(it -> it.getSeqInf()).collect(Collectors.toList());

– Suryavel TR
Nov 21 '18 at 13:15

1

What you wrote can be one line. Just delete the new line you added.

– matt
Nov 21 '18 at 13:21

add a comment |

1

Possible duplicate of Lambda expression to convert array/List of String to array/List of Integers

– Sven Hakvoort
Nov 21 '18 at 13:14

Try this --> List<String> outputList = list.stream().map(it -> it.getSeqInf()).collect(Collectors.toList());

– Suryavel TR
Nov 21 '18 at 13:15

1

What you wrote can be one line. Just delete the new line you added.

– matt
Nov 21 '18 at 13:21

Possible duplicate of Lambda expression to convert array/List of String to array/List of Integers

– Sven Hakvoort
Nov 21 '18 at 13:14

Try this --> List<String> outputList = list.stream().map(it -> it.getSeqInf()).collect(Collectors.toList());

– Suryavel TR
Nov 21 '18 at 13:15

What you wrote can be one line. Just delete the new line you added.

– matt
Nov 21 '18 at 13:21

add a comment |

5 Answers
5

active

oldest

votes

Yes, there is:

strList = infList.stream().map(e -> g.getSeqInf()).collect(Collectors.toList());

The map step can be also written in another way:

strList = infList.stream().map(InfoAQ::getSeqInf).collect(Collectors.toList());

which is know as method reference passing. Those two solutions are equivalent.

edited Nov 21 '18 at 13:19

answered Nov 21 '18 at 13:14

Lorelorelore

2,02761527

4

InfoAQ::getSeqInf ?

– matt
Nov 21 '18 at 13:15

1

It is an equivalent way for map. Good suggestion.

– Lorelorelore
Nov 21 '18 at 13:18

add a comment |

Also maybe this one:

List<String> strList = new ArrayList<String>();

infList.forEach(e -> strList.add(e.getSeqInf()));

answered Nov 21 '18 at 13:18

Pulszar

44426

add a comment |

And there is another one (-liner, if you format it in a single line):

infList.forEach(currentInf -> {strList.add(currentInf.getSeqInf());});

while I would prefer a formatting in more lines:

infList.forEach(currentInf -> {

    strList.add(currentInf.getSeqInf());

});

answered Nov 21 '18 at 13:19

deHaar

2,45451628

add a comment |

Using streams

infList.stream()

   .map(InfoAQ::getSeqInf)

   .collect(Collectors.toCollection(ArrayList::new))

Using Collectors.toCollection here to create an ArrayList that will hold the results as you do in your case. (Important if you do care about the result list type as Collectors.toList() does not guarantee this)

May not be the fastest as using stream has some overhead. You need to measure/benchmark to find out its performance

edited Nov 21 '18 at 13:21

answered Nov 21 '18 at 13:15

user7

9,55932443

add a comment |

This code will iterate all the data in the list, as getSeqInf returns a String, the collect method will store all returns of the getSeqInf method in a list.





`List listString = infList.stream().map(InfoAQ::getSeqInf).collect(Collectors.toList());`



or 



`

ArrayList<String> listString = new ArrayList<>();

for(int i = 0; i < infoAq.size(); i++) {

     listString.add(infoAq.get(i).getSeqInf());

}`

answered Nov 21 '18 at 13:28

Marcelo Lubanco Thomé

111

add a comment |

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

Yes, there is:

strList = infList.stream().map(e -> g.getSeqInf()).collect(Collectors.toList());

The map step can be also written in another way:

strList = infList.stream().map(InfoAQ::getSeqInf).collect(Collectors.toList());

which is know as method reference passing. Those two solutions are equivalent.

edited Nov 21 '18 at 13:19

answered Nov 21 '18 at 13:14

Lorelorelore

2,02761527

4

InfoAQ::getSeqInf ?

– matt
Nov 21 '18 at 13:15

1

It is an equivalent way for map. Good suggestion.

– Lorelorelore
Nov 21 '18 at 13:18

add a comment |

Yes, there is:

strList = infList.stream().map(e -> g.getSeqInf()).collect(Collectors.toList());

The map step can be also written in another way:

strList = infList.stream().map(InfoAQ::getSeqInf).collect(Collectors.toList());

which is know as method reference passing. Those two solutions are equivalent.

edited Nov 21 '18 at 13:19

answered Nov 21 '18 at 13:14

Lorelorelore

2,02761527

4

InfoAQ::getSeqInf ?

– matt
Nov 21 '18 at 13:15

1

It is an equivalent way for map. Good suggestion.

– Lorelorelore
Nov 21 '18 at 13:18

add a comment |

Yes, there is:

strList = infList.stream().map(e -> g.getSeqInf()).collect(Collectors.toList());

The map step can be also written in another way:

strList = infList.stream().map(InfoAQ::getSeqInf).collect(Collectors.toList());

which is know as method reference passing. Those two solutions are equivalent.

edited Nov 21 '18 at 13:19

answered Nov 21 '18 at 13:14

Lorelorelore

2,02761527

Yes, there is:

strList = infList.stream().map(e -> g.getSeqInf()).collect(Collectors.toList());

The map step can be also written in another way:

strList = infList.stream().map(InfoAQ::getSeqInf).collect(Collectors.toList());

which is know as method reference passing. Those two solutions are equivalent.

edited Nov 21 '18 at 13:19

answered Nov 21 '18 at 13:14

Lorelorelore

2,02761527

edited Nov 21 '18 at 13:19

answered Nov 21 '18 at 13:14

Lorelorelore

2,02761527

answered Nov 21 '18 at 13:14

Lorelorelore

2,02761527

answered Nov 21 '18 at 13:14

Lorelorelore

2,02761527

4

InfoAQ::getSeqInf ?

– matt
Nov 21 '18 at 13:15

1

It is an equivalent way for map. Good suggestion.

– Lorelorelore
Nov 21 '18 at 13:18

add a comment |

4

InfoAQ::getSeqInf ?

– matt
Nov 21 '18 at 13:15

1

It is an equivalent way for map. Good suggestion.

– Lorelorelore
Nov 21 '18 at 13:18

InfoAQ::getSeqInf ?

– matt
Nov 21 '18 at 13:15

It is an equivalent way for map. Good suggestion.

– Lorelorelore
Nov 21 '18 at 13:18

add a comment |

Also maybe this one:

List<String> strList = new ArrayList<String>();

infList.forEach(e -> strList.add(e.getSeqInf()));

answered Nov 21 '18 at 13:18

Pulszar

44426

add a comment |

Also maybe this one:

List<String> strList = new ArrayList<String>();

infList.forEach(e -> strList.add(e.getSeqInf()));

answered Nov 21 '18 at 13:18

Pulszar

44426

add a comment |

Also maybe this one:

List<String> strList = new ArrayList<String>();

infList.forEach(e -> strList.add(e.getSeqInf()));

answered Nov 21 '18 at 13:18

Pulszar

44426

Also maybe this one:

List<String> strList = new ArrayList<String>();

infList.forEach(e -> strList.add(e.getSeqInf()));

answered Nov 21 '18 at 13:18

Pulszar

44426

answered Nov 21 '18 at 13:18

Pulszar

44426

answered Nov 21 '18 at 13:18

Pulszar

44426

answered Nov 21 '18 at 13:18

Pulszar

44426

add a comment |

And there is another one (-liner, if you format it in a single line):

infList.forEach(currentInf -> {strList.add(currentInf.getSeqInf());});

while I would prefer a formatting in more lines:

infList.forEach(currentInf -> {

    strList.add(currentInf.getSeqInf());

});

answered Nov 21 '18 at 13:19

deHaar

2,45451628

add a comment |

And there is another one (-liner, if you format it in a single line):

infList.forEach(currentInf -> {strList.add(currentInf.getSeqInf());});

while I would prefer a formatting in more lines:

infList.forEach(currentInf -> {

    strList.add(currentInf.getSeqInf());

});

answered Nov 21 '18 at 13:19

deHaar

2,45451628

add a comment |

And there is another one (-liner, if you format it in a single line):

infList.forEach(currentInf -> {strList.add(currentInf.getSeqInf());});

while I would prefer a formatting in more lines:

infList.forEach(currentInf -> {

    strList.add(currentInf.getSeqInf());

});

answered Nov 21 '18 at 13:19

deHaar

2,45451628

And there is another one (-liner, if you format it in a single line):

infList.forEach(currentInf -> {strList.add(currentInf.getSeqInf());});

while I would prefer a formatting in more lines:

infList.forEach(currentInf -> {

    strList.add(currentInf.getSeqInf());

});

answered Nov 21 '18 at 13:19

deHaar

2,45451628

answered Nov 21 '18 at 13:19

deHaar

2,45451628

answered Nov 21 '18 at 13:19

deHaar

2,45451628

answered Nov 21 '18 at 13:19

deHaar

2,45451628

add a comment |

Using streams

infList.stream()

   .map(InfoAQ::getSeqInf)

   .collect(Collectors.toCollection(ArrayList::new))

May not be the fastest as using stream has some overhead. You need to measure/benchmark to find out its performance

edited Nov 21 '18 at 13:21

answered Nov 21 '18 at 13:15

user7

9,55932443

add a comment |

Using streams

infList.stream()

   .map(InfoAQ::getSeqInf)

   .collect(Collectors.toCollection(ArrayList::new))

May not be the fastest as using stream has some overhead. You need to measure/benchmark to find out its performance

edited Nov 21 '18 at 13:21

answered Nov 21 '18 at 13:15

user7

9,55932443

add a comment |

Using streams

infList.stream()

   .map(InfoAQ::getSeqInf)

   .collect(Collectors.toCollection(ArrayList::new))

May not be the fastest as using stream has some overhead. You need to measure/benchmark to find out its performance

edited Nov 21 '18 at 13:21

answered Nov 21 '18 at 13:15

user7

9,55932443

Using streams

infList.stream()

   .map(InfoAQ::getSeqInf)

   .collect(Collectors.toCollection(ArrayList::new))

May not be the fastest as using stream has some overhead. You need to measure/benchmark to find out its performance

edited Nov 21 '18 at 13:21

answered Nov 21 '18 at 13:15

user7

9,55932443

edited Nov 21 '18 at 13:21

answered Nov 21 '18 at 13:15

user7

9,55932443

answered Nov 21 '18 at 13:15

user7

9,55932443

answered Nov 21 '18 at 13:15

user7

9,55932443

add a comment |

This code will iterate all the data in the list, as getSeqInf returns a String, the collect method will store all returns of the getSeqInf method in a list.





`List listString = infList.stream().map(InfoAQ::getSeqInf).collect(Collectors.toList());`



or 



`

ArrayList<String> listString = new ArrayList<>();

for(int i = 0; i < infoAq.size(); i++) {

     listString.add(infoAq.get(i).getSeqInf());

}`

answered Nov 21 '18 at 13:28

Marcelo Lubanco Thomé

111

add a comment |

This code will iterate all the data in the list, as getSeqInf returns a String, the collect method will store all returns of the getSeqInf method in a list.





`List listString = infList.stream().map(InfoAQ::getSeqInf).collect(Collectors.toList());`



or 



`

ArrayList<String> listString = new ArrayList<>();

for(int i = 0; i < infoAq.size(); i++) {

     listString.add(infoAq.get(i).getSeqInf());

}`

answered Nov 21 '18 at 13:28

Marcelo Lubanco Thomé

111

add a comment |

This code will iterate all the data in the list, as getSeqInf returns a String, the collect method will store all returns of the getSeqInf method in a list.





`List listString = infList.stream().map(InfoAQ::getSeqInf).collect(Collectors.toList());`



or 



`

ArrayList<String> listString = new ArrayList<>();

for(int i = 0; i < infoAq.size(); i++) {

     listString.add(infoAq.get(i).getSeqInf());

}`

answered Nov 21 '18 at 13:28

Marcelo Lubanco Thomé

111

This code will iterate all the data in the list, as getSeqInf returns a String, the collect method will store all returns of the getSeqInf method in a list.





`List listString = infList.stream().map(InfoAQ::getSeqInf).collect(Collectors.toList());`



or 



`

ArrayList<String> listString = new ArrayList<>();

for(int i = 0; i < infoAq.size(); i++) {

     listString.add(infoAq.get(i).getSeqInf());

}`

answered Nov 21 '18 at 13:28

Marcelo Lubanco Thomé

111

answered Nov 21 '18 at 13:28

Marcelo Lubanco Thomé

111

answered Nov 21 '18 at 13:28

Marcelo Lubanco Thomé

111

answered Nov 21 '18 at 13:28

Marcelo Lubanco Thomé

111

add a comment |

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