Java 8 Strem filter map in map — Map<String,Map>
How to filter a Map<String,Map<String,Employee>>
using Java 8 Filter?
I have to filter only when any of employee in the list having a field value Gender = "M".
Input:
Map<String,Map<String,Employee>>
Output:
Map<String,Map<String,Employee>>
Filter criteria:
Employee.genter = "M"
Also i have to return empty map if the filtered result is empty.
I tried the below, but it is not working as expected. It is returning the only if all the Employees are with gender "M".
tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().allMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
java java-8 hashmap java-stream
add a comment |
How to filter a Map<String,Map<String,Employee>>
using Java 8 Filter?
I have to filter only when any of employee in the list having a field value Gender = "M".
Input:
Map<String,Map<String,Employee>>
Output:
Map<String,Map<String,Employee>>
Filter criteria:
Employee.genter = "M"
Also i have to return empty map if the filtered result is empty.
I tried the below, but it is not working as expected. It is returning the only if all the Employees are with gender "M".
tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().allMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
java java-8 hashmap java-stream
Return all the employees whose gender is M.
– user1578872
4 hours ago
add a comment |
How to filter a Map<String,Map<String,Employee>>
using Java 8 Filter?
I have to filter only when any of employee in the list having a field value Gender = "M".
Input:
Map<String,Map<String,Employee>>
Output:
Map<String,Map<String,Employee>>
Filter criteria:
Employee.genter = "M"
Also i have to return empty map if the filtered result is empty.
I tried the below, but it is not working as expected. It is returning the only if all the Employees are with gender "M".
tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().allMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
java java-8 hashmap java-stream
How to filter a Map<String,Map<String,Employee>>
using Java 8 Filter?
I have to filter only when any of employee in the list having a field value Gender = "M".
Input:
Map<String,Map<String,Employee>>
Output:
Map<String,Map<String,Employee>>
Filter criteria:
Employee.genter = "M"
Also i have to return empty map if the filtered result is empty.
I tried the below, but it is not working as expected. It is returning the only if all the Employees are with gender "M".
tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().allMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
java java-8 hashmap java-stream
java java-8 hashmap java-stream
edited 4 hours ago
Aomine
38.6k73265
38.6k73265
asked 4 hours ago
user1578872
1,53852563
1,53852563
Return all the employees whose gender is M.
– user1578872
4 hours ago
add a comment |
Return all the employees whose gender is M.
– user1578872
4 hours ago
Return all the employees whose gender is M.
– user1578872
4 hours ago
Return all the employees whose gender is M.
– user1578872
4 hours ago
add a comment |
5 Answers
5
active
oldest
votes
You could simply iterate on the key-value pairs and filter as:
Map<String, Map<String, Employee>> output = new HashMap<>();
tempCollection.forEach((k, v) -> {
if (v.values().stream().anyMatch(i -> "M".equals(i.getGender()))) {
output.putIfAbsent(k, v.entrySet()
.stream()
.filter(i -> "M".equals(i.getValue().getGender()))
.collect(toMap(Map.Entry::getKey, Map.Entry::getValue)));
}
});
anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
– user1578872
4 hours ago
If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
– user1578872
4 hours ago
@user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
– nullpointer
4 hours ago
It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
– user1578872
4 hours ago
1
I thinkput
would be more ideal rather thanputIfAbsent
as thekey
will always be unique.
– Aomine
3 hours ago
|
show 3 more comments
The function allMatch
only matches if every element in the stream matches the predicate; you can use anyMatch
to match if any element matches the predicate:
tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().anyMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
add a comment |
Seems like what you're after is given a Entry<String,Map<String,Employee>>
if there's any employee who has a gender of "M" then filter the inner Map<String,Employee>
to contain only entries with a gender "M".
in which case you can filter
along with anyMatch
for the first criterion. Further, at the collecting phase, you can then apply the filtering on the inner map:
tempCollection.entrySet().stream()
.filter(i -> i.getValue().values().stream().anyMatch(e -> "M".equals(e.getGender())))
.collect(toMap(Map.Entry::getKey,
v -> v.getValue().entrySet().stream()
.filter(i -> "M".equals(i.getValue().getGender()))
.collect(toMap(Map.Entry::getKey, Map.Entry::getValue))));
anyMatch returns all the values in the map even there is only one employee with Gender M.
– user1578872
4 hours ago
@user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
– Aomine
4 hours ago
add a comment |
Other way would be like this:
map.values()
.removeIf(entry->entry.values().stream().anyMatch(e -> !"M".equals(e.getGender())));
map.entrySet()
.removeIf(entry->entry.getValue().size() == 0);
add a comment |
You may do it like so,
Map<String, Map<String, Employee>> maleEmpMap = tempCollection.entrySet().stream()
.map(e -> new AbstractMap.SimpleEntry<>(e.getKey(),
e.getValue().entrySet().stream().filter(emp -> "M".equals(emp.getValue().getGender()))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue))))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
Take each map entry from the outer map and create a new map entry using the same key and the newly constructed nested map with the same key and which has only male employees as it's value. Finally collect all those matching entries into a new outer map. This will gracefully handle the empty scenario which you mentioned in the problem statement above.
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could simply iterate on the key-value pairs and filter as:
Map<String, Map<String, Employee>> output = new HashMap<>();
tempCollection.forEach((k, v) -> {
if (v.values().stream().anyMatch(i -> "M".equals(i.getGender()))) {
output.putIfAbsent(k, v.entrySet()
.stream()
.filter(i -> "M".equals(i.getValue().getGender()))
.collect(toMap(Map.Entry::getKey, Map.Entry::getValue)));
}
});
anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
– user1578872
4 hours ago
If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
– user1578872
4 hours ago
@user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
– nullpointer
4 hours ago
It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
– user1578872
4 hours ago
1
I thinkput
would be more ideal rather thanputIfAbsent
as thekey
will always be unique.
– Aomine
3 hours ago
|
show 3 more comments
You could simply iterate on the key-value pairs and filter as:
Map<String, Map<String, Employee>> output = new HashMap<>();
tempCollection.forEach((k, v) -> {
if (v.values().stream().anyMatch(i -> "M".equals(i.getGender()))) {
output.putIfAbsent(k, v.entrySet()
.stream()
.filter(i -> "M".equals(i.getValue().getGender()))
.collect(toMap(Map.Entry::getKey, Map.Entry::getValue)));
}
});
anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
– user1578872
4 hours ago
If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
– user1578872
4 hours ago
@user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
– nullpointer
4 hours ago
It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
– user1578872
4 hours ago
1
I thinkput
would be more ideal rather thanputIfAbsent
as thekey
will always be unique.
– Aomine
3 hours ago
|
show 3 more comments
You could simply iterate on the key-value pairs and filter as:
Map<String, Map<String, Employee>> output = new HashMap<>();
tempCollection.forEach((k, v) -> {
if (v.values().stream().anyMatch(i -> "M".equals(i.getGender()))) {
output.putIfAbsent(k, v.entrySet()
.stream()
.filter(i -> "M".equals(i.getValue().getGender()))
.collect(toMap(Map.Entry::getKey, Map.Entry::getValue)));
}
});
You could simply iterate on the key-value pairs and filter as:
Map<String, Map<String, Employee>> output = new HashMap<>();
tempCollection.forEach((k, v) -> {
if (v.values().stream().anyMatch(i -> "M".equals(i.getGender()))) {
output.putIfAbsent(k, v.entrySet()
.stream()
.filter(i -> "M".equals(i.getValue().getGender()))
.collect(toMap(Map.Entry::getKey, Map.Entry::getValue)));
}
});
edited 4 hours ago
answered 4 hours ago
nullpointer
40.9k1085168
40.9k1085168
anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
– user1578872
4 hours ago
If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
– user1578872
4 hours ago
@user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
– nullpointer
4 hours ago
It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
– user1578872
4 hours ago
1
I thinkput
would be more ideal rather thanputIfAbsent
as thekey
will always be unique.
– Aomine
3 hours ago
|
show 3 more comments
anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
– user1578872
4 hours ago
If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
– user1578872
4 hours ago
@user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
– nullpointer
4 hours ago
It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
– user1578872
4 hours ago
1
I thinkput
would be more ideal rather thanputIfAbsent
as thekey
will always be unique.
– Aomine
3 hours ago
anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
– user1578872
4 hours ago
anyMatch returns all the result even there is a single match, I want the key value to be removed if there is no match,
– user1578872
4 hours ago
If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
– user1578872
4 hours ago
If none of the employee is with the Gender "M", it still returns Map<String,emptyMap()>
– user1578872
4 hours ago
@user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
– nullpointer
4 hours ago
@user1578872 i have to return empty map if the filtered result is empty... that's what you meant, right?
– nullpointer
4 hours ago
It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
– user1578872
4 hours ago
It should be completeEmpty map in this case. Like, return emptyMap(), not Map<String,emptyMap()>
– user1578872
4 hours ago
1
1
I think
put
would be more ideal rather than putIfAbsent
as the key
will always be unique.– Aomine
3 hours ago
I think
put
would be more ideal rather than putIfAbsent
as the key
will always be unique.– Aomine
3 hours ago
|
show 3 more comments
The function allMatch
only matches if every element in the stream matches the predicate; you can use anyMatch
to match if any element matches the predicate:
tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().anyMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
add a comment |
The function allMatch
only matches if every element in the stream matches the predicate; you can use anyMatch
to match if any element matches the predicate:
tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().anyMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
add a comment |
The function allMatch
only matches if every element in the stream matches the predicate; you can use anyMatch
to match if any element matches the predicate:
tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().anyMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
The function allMatch
only matches if every element in the stream matches the predicate; you can use anyMatch
to match if any element matches the predicate:
tempCollection.entrySet().stream()
.filter(i -> i.getValue().entrySet().stream().anyMatch(e-> "M".equals(e.getValue().getGender())))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
answered 4 hours ago
Tordek
6,96622861
6,96622861
add a comment |
add a comment |
Seems like what you're after is given a Entry<String,Map<String,Employee>>
if there's any employee who has a gender of "M" then filter the inner Map<String,Employee>
to contain only entries with a gender "M".
in which case you can filter
along with anyMatch
for the first criterion. Further, at the collecting phase, you can then apply the filtering on the inner map:
tempCollection.entrySet().stream()
.filter(i -> i.getValue().values().stream().anyMatch(e -> "M".equals(e.getGender())))
.collect(toMap(Map.Entry::getKey,
v -> v.getValue().entrySet().stream()
.filter(i -> "M".equals(i.getValue().getGender()))
.collect(toMap(Map.Entry::getKey, Map.Entry::getValue))));
anyMatch returns all the values in the map even there is only one employee with Gender M.
– user1578872
4 hours ago
@user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
– Aomine
4 hours ago
add a comment |
Seems like what you're after is given a Entry<String,Map<String,Employee>>
if there's any employee who has a gender of "M" then filter the inner Map<String,Employee>
to contain only entries with a gender "M".
in which case you can filter
along with anyMatch
for the first criterion. Further, at the collecting phase, you can then apply the filtering on the inner map:
tempCollection.entrySet().stream()
.filter(i -> i.getValue().values().stream().anyMatch(e -> "M".equals(e.getGender())))
.collect(toMap(Map.Entry::getKey,
v -> v.getValue().entrySet().stream()
.filter(i -> "M".equals(i.getValue().getGender()))
.collect(toMap(Map.Entry::getKey, Map.Entry::getValue))));
anyMatch returns all the values in the map even there is only one employee with Gender M.
– user1578872
4 hours ago
@user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
– Aomine
4 hours ago
add a comment |
Seems like what you're after is given a Entry<String,Map<String,Employee>>
if there's any employee who has a gender of "M" then filter the inner Map<String,Employee>
to contain only entries with a gender "M".
in which case you can filter
along with anyMatch
for the first criterion. Further, at the collecting phase, you can then apply the filtering on the inner map:
tempCollection.entrySet().stream()
.filter(i -> i.getValue().values().stream().anyMatch(e -> "M".equals(e.getGender())))
.collect(toMap(Map.Entry::getKey,
v -> v.getValue().entrySet().stream()
.filter(i -> "M".equals(i.getValue().getGender()))
.collect(toMap(Map.Entry::getKey, Map.Entry::getValue))));
Seems like what you're after is given a Entry<String,Map<String,Employee>>
if there's any employee who has a gender of "M" then filter the inner Map<String,Employee>
to contain only entries with a gender "M".
in which case you can filter
along with anyMatch
for the first criterion. Further, at the collecting phase, you can then apply the filtering on the inner map:
tempCollection.entrySet().stream()
.filter(i -> i.getValue().values().stream().anyMatch(e -> "M".equals(e.getGender())))
.collect(toMap(Map.Entry::getKey,
v -> v.getValue().entrySet().stream()
.filter(i -> "M".equals(i.getValue().getGender()))
.collect(toMap(Map.Entry::getKey, Map.Entry::getValue))));
edited 3 hours ago
answered 4 hours ago
Aomine
38.6k73265
38.6k73265
anyMatch returns all the values in the map even there is only one employee with Gender M.
– user1578872
4 hours ago
@user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
– Aomine
4 hours ago
add a comment |
anyMatch returns all the values in the map even there is only one employee with Gender M.
– user1578872
4 hours ago
@user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
– Aomine
4 hours ago
anyMatch returns all the values in the map even there is only one employee with Gender M.
– user1578872
4 hours ago
anyMatch returns all the values in the map even there is only one employee with Gender M.
– user1578872
4 hours ago
@user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
– Aomine
4 hours ago
@user1578872 right, thanks to the accepted answer, I can now understand your requirement. the description wasn't very clear IMO. anyway edited to accommodate.
– Aomine
4 hours ago
add a comment |
Other way would be like this:
map.values()
.removeIf(entry->entry.values().stream().anyMatch(e -> !"M".equals(e.getGender())));
map.entrySet()
.removeIf(entry->entry.getValue().size() == 0);
add a comment |
Other way would be like this:
map.values()
.removeIf(entry->entry.values().stream().anyMatch(e -> !"M".equals(e.getGender())));
map.entrySet()
.removeIf(entry->entry.getValue().size() == 0);
add a comment |
Other way would be like this:
map.values()
.removeIf(entry->entry.values().stream().anyMatch(e -> !"M".equals(e.getGender())));
map.entrySet()
.removeIf(entry->entry.getValue().size() == 0);
Other way would be like this:
map.values()
.removeIf(entry->entry.values().stream().anyMatch(e -> !"M".equals(e.getGender())));
map.entrySet()
.removeIf(entry->entry.getValue().size() == 0);
answered 3 hours ago
Hadi J
9,70731641
9,70731641
add a comment |
add a comment |
You may do it like so,
Map<String, Map<String, Employee>> maleEmpMap = tempCollection.entrySet().stream()
.map(e -> new AbstractMap.SimpleEntry<>(e.getKey(),
e.getValue().entrySet().stream().filter(emp -> "M".equals(emp.getValue().getGender()))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue))))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
Take each map entry from the outer map and create a new map entry using the same key and the newly constructed nested map with the same key and which has only male employees as it's value. Finally collect all those matching entries into a new outer map. This will gracefully handle the empty scenario which you mentioned in the problem statement above.
add a comment |
You may do it like so,
Map<String, Map<String, Employee>> maleEmpMap = tempCollection.entrySet().stream()
.map(e -> new AbstractMap.SimpleEntry<>(e.getKey(),
e.getValue().entrySet().stream().filter(emp -> "M".equals(emp.getValue().getGender()))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue))))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
Take each map entry from the outer map and create a new map entry using the same key and the newly constructed nested map with the same key and which has only male employees as it's value. Finally collect all those matching entries into a new outer map. This will gracefully handle the empty scenario which you mentioned in the problem statement above.
add a comment |
You may do it like so,
Map<String, Map<String, Employee>> maleEmpMap = tempCollection.entrySet().stream()
.map(e -> new AbstractMap.SimpleEntry<>(e.getKey(),
e.getValue().entrySet().stream().filter(emp -> "M".equals(emp.getValue().getGender()))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue))))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
Take each map entry from the outer map and create a new map entry using the same key and the newly constructed nested map with the same key and which has only male employees as it's value. Finally collect all those matching entries into a new outer map. This will gracefully handle the empty scenario which you mentioned in the problem statement above.
You may do it like so,
Map<String, Map<String, Employee>> maleEmpMap = tempCollection.entrySet().stream()
.map(e -> new AbstractMap.SimpleEntry<>(e.getKey(),
e.getValue().entrySet().stream().filter(emp -> "M".equals(emp.getValue().getGender()))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue))))
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
Take each map entry from the outer map and create a new map entry using the same key and the newly constructed nested map with the same key and which has only male employees as it's value. Finally collect all those matching entries into a new outer map. This will gracefully handle the empty scenario which you mentioned in the problem statement above.
edited 3 hours ago
answered 3 hours ago
Ravindra Ranwala
8,30731634
8,30731634
add a comment |
add a comment |
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Return all the employees whose gender is M.
– user1578872
4 hours ago