How do I build a Java `Map` out of `List<Map.Entry>`? [duplicate]

-1

This question already has an answer here:

Convert Set<Map.Entry<K, V>> to HashMap<K, V>

6 answers

I have a List<Map.Entry<Long, String>>.

How do I convert this into a Map?

Currently I am doing the following, but it seems a bit verbose (and most importantly, it is not "fluent", i.e. a single expression, but requires a code block).

Map<Long, String> result = new HashMap<>();
entries.forEach(e -> result.put(e.getKey(), e.getValue()));
return result;

Java 10 is fine.

asked Nov 11 at 11:25

Thilo

192k77413572

marked as duplicate by buræquete, Ferrybig, nullpointer java
Users with the java badge can single-handedly close java questions as duplicates and reopen them as needed.

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Nov 11 at 11:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1

I would go with Java 11 as 10 is eol.
– Peter Lawrey
Nov 11 at 11:42

1

@PeterLawrey already? Oh my, they are moving fast...
– Thilo
Nov 11 at 11:44

Java 11 is EOL in 4 months :P At that time many might shift to OpenJDK 11, instead of skipping to 12. e.g. adoptopenjdk.net
– Peter Lawrey
Nov 11 at 12:20

Anyway, I meant this as in "use of Java 10 API are permissible" (in the hopes that we have something less boilerplatey than the Java 8 streams collectors now). Does Java 11 bring something new to the table here (such as Collection literals)?
– Thilo
Nov 11 at 12:24

1

Not really imho, You can write entries.forEach((var e) -> result.put(e.getKey(), e.getValue())); note the var e, the real advantage is Long Term Support-ability.
– Peter Lawrey
Nov 11 at 12:30

add a comment |

-1

This question already has an answer here:

Convert Set<Map.Entry<K, V>> to HashMap<K, V>

6 answers

I have a List<Map.Entry<Long, String>>.

How do I convert this into a Map?

Currently I am doing the following, but it seems a bit verbose (and most importantly, it is not "fluent", i.e. a single expression, but requires a code block).

Map<Long, String> result = new HashMap<>();
entries.forEach(e -> result.put(e.getKey(), e.getValue()));
return result;

Java 10 is fine.

asked Nov 11 at 11:25

Thilo

192k77413572

marked as duplicate by buræquete, Ferrybig, nullpointer java
Users with the java badge can single-handedly close java questions as duplicates and reopen them as needed.

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Nov 11 at 11:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1

I would go with Java 11 as 10 is eol.
– Peter Lawrey
Nov 11 at 11:42

1

@PeterLawrey already? Oh my, they are moving fast...
– Thilo
Nov 11 at 11:44

Java 11 is EOL in 4 months :P At that time many might shift to OpenJDK 11, instead of skipping to 12. e.g. adoptopenjdk.net
– Peter Lawrey
Nov 11 at 12:20

Anyway, I meant this as in "use of Java 10 API are permissible" (in the hopes that we have something less boilerplatey than the Java 8 streams collectors now). Does Java 11 bring something new to the table here (such as Collection literals)?
– Thilo
Nov 11 at 12:24

1

Not really imho, You can write entries.forEach((var e) -> result.put(e.getKey(), e.getValue())); note the var e, the real advantage is Long Term Support-ability.
– Peter Lawrey
Nov 11 at 12:30

add a comment |

-1

This question already has an answer here:

Convert Set<Map.Entry<K, V>> to HashMap<K, V>

6 answers

I have a List<Map.Entry<Long, String>>.

How do I convert this into a Map?

Currently I am doing the following, but it seems a bit verbose (and most importantly, it is not "fluent", i.e. a single expression, but requires a code block).

Map<Long, String> result = new HashMap<>();
entries.forEach(e -> result.put(e.getKey(), e.getValue()));
return result;

Java 10 is fine.

asked Nov 11 at 11:25

Thilo

192k77413572

This question already has an answer here:

Convert Set<Map.Entry<K, V>> to HashMap<K, V>

6 answers

I have a List<Map.Entry<Long, String>>.

How do I convert this into a Map?

Currently I am doing the following, but it seems a bit verbose (and most importantly, it is not "fluent", i.e. a single expression, but requires a code block).

Map<Long, String> result = new HashMap<>();
entries.forEach(e -> result.put(e.getKey(), e.getValue()));
return result;

Java 10 is fine.

This question already has an answer here:

Convert Set<Map.Entry<K, V>> to HashMap<K, V>

6 answers

java

asked Nov 11 at 11:25

Thilo

192k77413572

asked Nov 11 at 11:25

Thilo

192k77413572

asked Nov 11 at 11:25

Thilo

192k77413572

asked Nov 11 at 11:25

Thilo

192k77413572

asked Nov 11 at 11:25

Thilo

192k77413572

marked as duplicate by buræquete, Ferrybig, nullpointer java
Users with the java badge can single-handedly close java questions as duplicates and reopen them as needed.

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Nov 11 at 11:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by buræquete, Ferrybig, nullpointer java
Users with the java badge can single-handedly close java questions as duplicates and reopen them as needed.

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Nov 11 at 11:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1

I would go with Java 11 as 10 is eol.
– Peter Lawrey
Nov 11 at 11:42

1

@PeterLawrey already? Oh my, they are moving fast...
– Thilo
Nov 11 at 11:44

Java 11 is EOL in 4 months :P At that time many might shift to OpenJDK 11, instead of skipping to 12. e.g. adoptopenjdk.net
– Peter Lawrey
Nov 11 at 12:20

Anyway, I meant this as in "use of Java 10 API are permissible" (in the hopes that we have something less boilerplatey than the Java 8 streams collectors now). Does Java 11 bring something new to the table here (such as Collection literals)?
– Thilo
Nov 11 at 12:24

1

Not really imho, You can write entries.forEach((var e) -> result.put(e.getKey(), e.getValue())); note the var e, the real advantage is Long Term Support-ability.
– Peter Lawrey
Nov 11 at 12:30

add a comment |

1

I would go with Java 11 as 10 is eol.
– Peter Lawrey
Nov 11 at 11:42

1

@PeterLawrey already? Oh my, they are moving fast...
– Thilo
Nov 11 at 11:44

Java 11 is EOL in 4 months :P At that time many might shift to OpenJDK 11, instead of skipping to 12. e.g. adoptopenjdk.net
– Peter Lawrey
Nov 11 at 12:20

Anyway, I meant this as in "use of Java 10 API are permissible" (in the hopes that we have something less boilerplatey than the Java 8 streams collectors now). Does Java 11 bring something new to the table here (such as Collection literals)?
– Thilo
Nov 11 at 12:24

1

Not really imho, You can write entries.forEach((var e) -> result.put(e.getKey(), e.getValue())); note the var e, the real advantage is Long Term Support-ability.
– Peter Lawrey
Nov 11 at 12:30

I would go with Java 11 as 10 is eol.
– Peter Lawrey
Nov 11 at 11:42

@PeterLawrey already? Oh my, they are moving fast...
– Thilo
Nov 11 at 11:44

Java 11 is EOL in 4 months :P At that time many might shift to OpenJDK 11, instead of skipping to 12. e.g. adoptopenjdk.net
– Peter Lawrey
Nov 11 at 12:20

Anyway, I meant this as in "use of Java 10 API are permissible" (in the hopes that we have something less boilerplatey than the Java 8 streams collectors now). Does Java 11 bring something new to the table here (such as Collection literals)?
– Thilo
Nov 11 at 12:24

Not really imho, You can write entries.forEach((var e) -> result.put(e.getKey(), e.getValue())); note the var e, the real advantage is Long Term Support-ability.
– Peter Lawrey
Nov 11 at 12:30

add a comment |

2 Answers
2

active

oldest

votes

If you know for sure there are no duplicate keys, this is sufficient:

Map<Long, String> result = entries.stream().collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue));

If there may be duplicates, you'll have to add a merge function to handle them.

answered Nov 11 at 11:27

Eran

278k37447533

Hmm. I made it a List of Map.Entry specifically to avoid having to specify again what the ::getKey and ::getValue should be, but I guess I'll have to go with that.
– Thilo
Nov 11 at 11:38

1

I do like the Guava answer from the duplicate thread. stackoverflow.com/a/44904884/14955 And java.util.Map.ofEntries is close, too, but does not work with a list.
– Thilo
Nov 11 at 11:40

What happens if I have duplicates but no merge function?
– Thilo
Nov 11 at 12:27

1

@Thilo An exception is thrown.
– Eran
Nov 11 at 12:28

add a comment |

Flatten it:

Map<Long, String> result = 
 entries.stream()
 .collect(toMap(e -> e.getKey(), e -> e.getValue(), (a, b) -> b);

The (a, b) -> b means that the last value for duplicate keys will be taken, which matches the semantics of your current approach.

edited Nov 11 at 12:52

answered Nov 11 at 11:28

Andy Turner

80k879133

Map.Entry has no method called entrySet
– Ferrybig
Nov 11 at 11:29

@Ferrybig my mistake, I thought it was a list of maps.
– Andy Turner
Nov 11 at 11:29

add a comment |

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

If you know for sure there are no duplicate keys, this is sufficient:

Map<Long, String> result = entries.stream().collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue));

If there may be duplicates, you'll have to add a merge function to handle them.

answered Nov 11 at 11:27

Eran

278k37447533

Hmm. I made it a List of Map.Entry specifically to avoid having to specify again what the ::getKey and ::getValue should be, but I guess I'll have to go with that.
– Thilo
Nov 11 at 11:38

1

I do like the Guava answer from the duplicate thread. stackoverflow.com/a/44904884/14955 And java.util.Map.ofEntries is close, too, but does not work with a list.
– Thilo
Nov 11 at 11:40

What happens if I have duplicates but no merge function?
– Thilo
Nov 11 at 12:27

1

@Thilo An exception is thrown.
– Eran
Nov 11 at 12:28

add a comment |

If you know for sure there are no duplicate keys, this is sufficient:

Map<Long, String> result = entries.stream().collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue));

If there may be duplicates, you'll have to add a merge function to handle them.

answered Nov 11 at 11:27

Eran

278k37447533

Hmm. I made it a List of Map.Entry specifically to avoid having to specify again what the ::getKey and ::getValue should be, but I guess I'll have to go with that.
– Thilo
Nov 11 at 11:38

1

I do like the Guava answer from the duplicate thread. stackoverflow.com/a/44904884/14955 And java.util.Map.ofEntries is close, too, but does not work with a list.
– Thilo
Nov 11 at 11:40

What happens if I have duplicates but no merge function?
– Thilo
Nov 11 at 12:27

1

@Thilo An exception is thrown.
– Eran
Nov 11 at 12:28

add a comment |

If you know for sure there are no duplicate keys, this is sufficient:

Map<Long, String> result = entries.stream().collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue));

If there may be duplicates, you'll have to add a merge function to handle them.

answered Nov 11 at 11:27

Eran

278k37447533

If you know for sure there are no duplicate keys, this is sufficient:

Map<Long, String> result = entries.stream().collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue));

If there may be duplicates, you'll have to add a merge function to handle them.

answered Nov 11 at 11:27

Eran

278k37447533

answered Nov 11 at 11:27

Eran

278k37447533

answered Nov 11 at 11:27

Eran

278k37447533

answered Nov 11 at 11:27

Eran

278k37447533

Hmm. I made it a List of Map.Entry specifically to avoid having to specify again what the ::getKey and ::getValue should be, but I guess I'll have to go with that.
– Thilo
Nov 11 at 11:38

1

I do like the Guava answer from the duplicate thread. stackoverflow.com/a/44904884/14955 And java.util.Map.ofEntries is close, too, but does not work with a list.
– Thilo
Nov 11 at 11:40

What happens if I have duplicates but no merge function?
– Thilo
Nov 11 at 12:27

1

@Thilo An exception is thrown.
– Eran
Nov 11 at 12:28

add a comment |

Hmm. I made it a List of Map.Entry specifically to avoid having to specify again what the ::getKey and ::getValue should be, but I guess I'll have to go with that.
– Thilo
Nov 11 at 11:38

1

I do like the Guava answer from the duplicate thread. stackoverflow.com/a/44904884/14955 And java.util.Map.ofEntries is close, too, but does not work with a list.
– Thilo
Nov 11 at 11:40

What happens if I have duplicates but no merge function?
– Thilo
Nov 11 at 12:27

1

@Thilo An exception is thrown.
– Eran
Nov 11 at 12:28

Hmm. I made it a List of Map.Entry specifically to avoid having to specify again what the ::getKey and ::getValue should be, but I guess I'll have to go with that.
– Thilo
Nov 11 at 11:38

I do like the Guava answer from the duplicate thread. stackoverflow.com/a/44904884/14955 And java.util.Map.ofEntries is close, too, but does not work with a list.
– Thilo
Nov 11 at 11:40

What happens if I have duplicates but no merge function?
– Thilo
Nov 11 at 12:27

@Thilo An exception is thrown.
– Eran
Nov 11 at 12:28

add a comment |

Flatten it:

Map<Long, String> result = 
 entries.stream()
 .collect(toMap(e -> e.getKey(), e -> e.getValue(), (a, b) -> b);

The (a, b) -> b means that the last value for duplicate keys will be taken, which matches the semantics of your current approach.

edited Nov 11 at 12:52

answered Nov 11 at 11:28

Andy Turner

80k879133

Map.Entry has no method called entrySet
– Ferrybig
Nov 11 at 11:29

@Ferrybig my mistake, I thought it was a list of maps.
– Andy Turner
Nov 11 at 11:29

add a comment |

Flatten it:

Map<Long, String> result = 
 entries.stream()
 .collect(toMap(e -> e.getKey(), e -> e.getValue(), (a, b) -> b);

The (a, b) -> b means that the last value for duplicate keys will be taken, which matches the semantics of your current approach.

edited Nov 11 at 12:52

answered Nov 11 at 11:28

Andy Turner

80k879133

Map.Entry has no method called entrySet
– Ferrybig
Nov 11 at 11:29

@Ferrybig my mistake, I thought it was a list of maps.
– Andy Turner
Nov 11 at 11:29

add a comment |

Flatten it:

Map<Long, String> result = 
 entries.stream()
 .collect(toMap(e -> e.getKey(), e -> e.getValue(), (a, b) -> b);

The (a, b) -> b means that the last value for duplicate keys will be taken, which matches the semantics of your current approach.

edited Nov 11 at 12:52

answered Nov 11 at 11:28

Andy Turner

80k879133

Flatten it:

Map<Long, String> result = 
 entries.stream()
 .collect(toMap(e -> e.getKey(), e -> e.getValue(), (a, b) -> b);

The (a, b) -> b means that the last value for duplicate keys will be taken, which matches the semantics of your current approach.

edited Nov 11 at 12:52

answered Nov 11 at 11:28

Andy Turner

80k879133

edited Nov 11 at 12:52

answered Nov 11 at 11:28

Andy Turner

80k879133

answered Nov 11 at 11:28

Andy Turner

80k879133

answered Nov 11 at 11:28

Andy Turner

80k879133

Map.Entry has no method called entrySet
– Ferrybig
Nov 11 at 11:29

@Ferrybig my mistake, I thought it was a list of maps.
– Andy Turner
Nov 11 at 11:29

add a comment |

Map.Entry has no method called entrySet
– Ferrybig
Nov 11 at 11:29

@Ferrybig my mistake, I thought it was a list of maps.
– Andy Turner
Nov 11 at 11:29

Map.Entry has no method called entrySet
– Ferrybig
Nov 11 at 11:29

@Ferrybig my mistake, I thought it was a list of maps.
– Andy Turner
Nov 11 at 11:29

add a comment |

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