Mixing
Accessing map values from neo4, apoc, Cypher
I am still rather new to Neo4j, Cypher and programming in general.
Is there a way to access the posted output below, i.e. access the "count" values for every "item“ (which has to be the pair), and also access the "item" values? I need the amount of how often a pair, i.e. specific neighboring nodes occur not only as information, but as values with which I can further work with in order to adjust my graph.
My last lines of code (in the preceding lines I just ordered the nodes sequentially):
...
WITH apoc.coll.pairs(a) as pairsOfa
WITH apoc.coll.frequencies(pairsOfa) AS giveBackFrequencyOfPairsOfa
UNWIND giveBackFrequencyOfPairsOfa AS x
WITH DISTINCT x
RETURN x
Output from the Neo4j Browser that I need to work with:
"x"
│{"count":1,"item":[{"aName“:"Rob","time":1},{"aName":"Edwin“,"time“:2}]},{„count“:4,“item":[{"aName":"Edwin","time":2},{"aName“:"Celesta","time":3}]}
...
neo4j cypher neo4j-apoc
edited Jan 2 at 23:13
cybersam
40.8k53353
asked Jan 2 at 22:38
CodeIslandCodeIsland
276
add a comment |
I am still rather new to Neo4j, Cypher and programming in general.
Is there a way to access the posted output below, i.e. access the "count" values for every "item“ (which has to be the pair), and also access the "item" values? I need the amount of how often a pair, i.e. specific neighboring nodes occur not only as information, but as values with which I can further work with in order to adjust my graph.
My last lines of code (in the preceding lines I just ordered the nodes sequentially):
...
WITH apoc.coll.pairs(a) as pairsOfa
WITH apoc.coll.frequencies(pairsOfa) AS giveBackFrequencyOfPairsOfa
UNWIND giveBackFrequencyOfPairsOfa AS x
WITH DISTINCT x
RETURN x
Output from the Neo4j Browser that I need to work with:
"x"
│{"count":1,"item":[{"aName“:"Rob","time":1},{"aName":"Edwin“,"time“:2}]},{„count“:4,“item":[{"aName":"Edwin","time":2},{"aName“:"Celesta","time":3}]}
...
neo4j cypher neo4j-apoc
edited Jan 2 at 23:13
cybersam
40.8k53353
asked Jan 2 at 22:38
CodeIslandCodeIsland
276
Your description is a little confusing. Can you provide an example of desired output with some additional context?
– InverseFalcon
Jan 2 at 23:01
Thank you, cybersam solved it below :)
– CodeIsland
Jan 3 at 9:08
add a comment |
I am still rather new to Neo4j, Cypher and programming in general.
Is there a way to access the posted output below, i.e. access the "count" values for every "item“ (which has to be the pair), and also access the "item" values? I need the amount of how often a pair, i.e. specific neighboring nodes occur not only as information, but as values with which I can further work with in order to adjust my graph.
My last lines of code (in the preceding lines I just ordered the nodes sequentially):
...
WITH apoc.coll.pairs(a) as pairsOfa
WITH apoc.coll.frequencies(pairsOfa) AS giveBackFrequencyOfPairsOfa
UNWIND giveBackFrequencyOfPairsOfa AS x
WITH DISTINCT x
RETURN x
Output from the Neo4j Browser that I need to work with:
"x"
│{"count":1,"item":[{"aName“:"Rob","time":1},{"aName":"Edwin“,"time“:2}]},{„count“:4,“item":[{"aName":"Edwin","time":2},{"aName“:"Celesta","time":3}]}
...
neo4j cypher neo4j-apoc
edited Jan 2 at 23:13
cybersam
40.8k53353
asked Jan 2 at 22:38
CodeIslandCodeIsland
276
I am still rather new to Neo4j, Cypher and programming in general.
Is there a way to access the posted output below, i.e. access the "count" values for every "item“ (which has to be the pair), and also access the "item" values? I need the amount of how often a pair, i.e. specific neighboring nodes occur not only as information, but as values with which I can further work with in order to adjust my graph.
My last lines of code (in the preceding lines I just ordered the nodes sequentially):
...
WITH apoc.coll.pairs(a) as pairsOfa
WITH apoc.coll.frequencies(pairsOfa) AS giveBackFrequencyOfPairsOfa
UNWIND giveBackFrequencyOfPairsOfa AS x
WITH DISTINCT x
RETURN x
Output from the Neo4j Browser that I need to work with:
"x"
│{"count":1,"item":[{"aName“:"Rob","time":1},{"aName":"Edwin“,"time“:2}]},{„count“:4,“item":[{"aName":"Edwin","time":2},{"aName“:"Celesta","time":3}]}
...
neo4j cypher neo4j-apoc
neo4j cypher neo4j-apoc
edited Jan 2 at 23:13
cybersam
40.8k53353
asked Jan 2 at 22:38
CodeIslandCodeIsland
276
edited Jan 2 at 23:13
cybersam
40.8k53353
asked Jan 2 at 22:38
CodeIslandCodeIsland
276
edited Jan 2 at 23:13
cybersam
40.8k53353
edited Jan 2 at 23:13
cybersam
40.8k53353
edited Jan 2 at 23:13
cybersam
40.8k53353
40.8k53353
asked Jan 2 at 22:38
CodeIslandCodeIsland
276
asked Jan 2 at 22:38
CodeIslandCodeIsland
276
asked Jan 2 at 22:38
CodeIslandCodeIsland
276
276
Your description is a little confusing. Can you provide an example of desired output with some additional context?
– InverseFalcon
Jan 2 at 23:01
Thank you, cybersam solved it below :)
– CodeIsland
Jan 3 at 9:08
add a comment |
Your description is a little confusing. Can you provide an example of desired output with some additional context?
– InverseFalcon
Jan 2 at 23:01
Thank you, cybersam solved it below :)
– CodeIsland
Jan 3 at 9:08
Your description is a little confusing. Can you provide an example of desired output with some additional context?
– InverseFalcon
Jan 2 at 23:01
Your description is a little confusing. Can you provide an example of desired output with some additional context?
– InverseFalcon
Jan 2 at 23:01
Thank you, cybersam solved it below :)
– CodeIsland
Jan 3 at 9:08
Thank you, cybersam solved it below :)
– CodeIsland
Jan 3 at 9:08
add a comment |
1 Answer
1
active
oldest
votes
Based on your code, your result should contain multiple x records (not a single record, as implied by the "output" provided in your question). Here is an example of what I would expect:
╒══════════════════════════════════════════════════════════════════════╕
│"x" │
╞══════════════════════════════════════════════════════════════════════╡
│{"count":1,"item":[{"aName":"Rob","time":1},{"aName":"Edwin","time":2}│
│]} │
├──────────────────────────────────────────────────────────────────────┤
│{"count":1,"item":[{"aName":"Edwin","time":2},{"aName":"Celesta","time│
│":3}]} │
└──────────────────────────────────────────────────────────────────────┘
If that is true, then you can just access the count and item properties of each x directly via x.count and x.item. To get each value within an item, you could use x.item[0] and x.item[1].
Asides: you probably want to use apoc.coll.pairsMin instead of apoc.coll.pairs, to avoid the generation of a spurious "pair" (whose second element is null) when the number of values to be paired is odd. Also, you probably do not need the DISTINCT step.
answered Jan 2 at 23:26
cybersamcybersam
40.8k53353
Thank you so very much! It is working. Such a simple solution that I just could not see yet. Now I can work from here :)
– CodeIsland
Jan 3 at 9:07
Any chance you could help improve the OP's question? It's difficult to see this Q/A as very helpful to many others.
– rickhg12hs
Jan 3 at 12:25
The question was really about the output - how to access "count" and "item" in the output; the output as was correctly given in cybersam's post.
– CodeIsland
Jan 3 at 14:21
add a comment |
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1 Answer
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1 Answer
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active
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oldest
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oldest
votes
Based on your code, your result should contain multiple x records (not a single record, as implied by the "output" provided in your question). Here is an example of what I would expect:
╒══════════════════════════════════════════════════════════════════════╕
│"x" │
╞══════════════════════════════════════════════════════════════════════╡
│{"count":1,"item":[{"aName":"Rob","time":1},{"aName":"Edwin","time":2}│
│]} │
├──────────────────────────────────────────────────────────────────────┤
│{"count":1,"item":[{"aName":"Edwin","time":2},{"aName":"Celesta","time│
│":3}]} │
└──────────────────────────────────────────────────────────────────────┘
If that is true, then you can just access the count and item properties of each x directly via x.count and x.item. To get each value within an item, you could use x.item[0] and x.item[1].
Asides: you probably want to use apoc.coll.pairsMin instead of apoc.coll.pairs, to avoid the generation of a spurious "pair" (whose second element is null) when the number of values to be paired is odd. Also, you probably do not need the DISTINCT step.
answered Jan 2 at 23:26
cybersamcybersam
40.8k53353
Thank you so very much! It is working. Such a simple solution that I just could not see yet. Now I can work from here :)
– CodeIsland
Jan 3 at 9:07
Any chance you could help improve the OP's question? It's difficult to see this Q/A as very helpful to many others.
– rickhg12hs
Jan 3 at 12:25
The question was really about the output - how to access "count" and "item" in the output; the output as was correctly given in cybersam's post.
– CodeIsland
Jan 3 at 14:21
add a comment |
Based on your code, your result should contain multiple x records (not a single record, as implied by the "output" provided in your question). Here is an example of what I would expect:
╒══════════════════════════════════════════════════════════════════════╕
│"x" │
╞══════════════════════════════════════════════════════════════════════╡
│{"count":1,"item":[{"aName":"Rob","time":1},{"aName":"Edwin","time":2}│
│]} │
├──────────────────────────────────────────────────────────────────────┤
│{"count":1,"item":[{"aName":"Edwin","time":2},{"aName":"Celesta","time│
│":3}]} │
└──────────────────────────────────────────────────────────────────────┘
If that is true, then you can just access the count and item properties of each x directly via x.count and x.item. To get each value within an item, you could use x.item[0] and x.item[1].
Asides: you probably want to use apoc.coll.pairsMin instead of apoc.coll.pairs, to avoid the generation of a spurious "pair" (whose second element is null) when the number of values to be paired is odd. Also, you probably do not need the DISTINCT step.
answered Jan 2 at 23:26
cybersamcybersam
40.8k53353
Thank you so very much! It is working. Such a simple solution that I just could not see yet. Now I can work from here :)
– CodeIsland
Jan 3 at 9:07
Any chance you could help improve the OP's question? It's difficult to see this Q/A as very helpful to many others.
– rickhg12hs
Jan 3 at 12:25
The question was really about the output - how to access "count" and "item" in the output; the output as was correctly given in cybersam's post.
– CodeIsland
Jan 3 at 14:21
add a comment |
Based on your code, your result should contain multiple x records (not a single record, as implied by the "output" provided in your question). Here is an example of what I would expect:
╒══════════════════════════════════════════════════════════════════════╕
│"x" │
╞══════════════════════════════════════════════════════════════════════╡
│{"count":1,"item":[{"aName":"Rob","time":1},{"aName":"Edwin","time":2}│
│]} │
├──────────────────────────────────────────────────────────────────────┤
│{"count":1,"item":[{"aName":"Edwin","time":2},{"aName":"Celesta","time│
│":3}]} │
└──────────────────────────────────────────────────────────────────────┘
If that is true, then you can just access the count and item properties of each x directly via x.count and x.item. To get each value within an item, you could use x.item[0] and x.item[1].
Asides: you probably want to use apoc.coll.pairsMin instead of apoc.coll.pairs, to avoid the generation of a spurious "pair" (whose second element is null) when the number of values to be paired is odd. Also, you probably do not need the DISTINCT step.
answered Jan 2 at 23:26
cybersamcybersam
40.8k53353
Based on your code, your result should contain multiple x records (not a single record, as implied by the "output" provided in your question). Here is an example of what I would expect:
╒══════════════════════════════════════════════════════════════════════╕
│"x" │
╞══════════════════════════════════════════════════════════════════════╡
│{"count":1,"item":[{"aName":"Rob","time":1},{"aName":"Edwin","time":2}│
│]} │
├──────────────────────────────────────────────────────────────────────┤
│{"count":1,"item":[{"aName":"Edwin","time":2},{"aName":"Celesta","time│
│":3}]} │
└──────────────────────────────────────────────────────────────────────┘
If that is true, then you can just access the count and item properties of each x directly via x.count and x.item. To get each value within an item, you could use x.item[0] and x.item[1].
Asides: you probably want to use apoc.coll.pairsMin instead of apoc.coll.pairs, to avoid the generation of a spurious "pair" (whose second element is null) when the number of values to be paired is odd. Also, you probably do not need the DISTINCT step.
answered Jan 2 at 23:26
cybersamcybersam
40.8k53353
answered Jan 2 at 23:26
cybersamcybersam
40.8k53353
answered Jan 2 at 23:26
cybersamcybersam
40.8k53353
answered Jan 2 at 23:26
cybersamcybersam
40.8k53353
40.8k53353
Thank you so very much! It is working. Such a simple solution that I just could not see yet. Now I can work from here :)
– CodeIsland
Jan 3 at 9:07
Any chance you could help improve the OP's question? It's difficult to see this Q/A as very helpful to many others.
– rickhg12hs
Jan 3 at 12:25
The question was really about the output - how to access "count" and "item" in the output; the output as was correctly given in cybersam's post.
– CodeIsland
Jan 3 at 14:21
add a comment |
Thank you so very much! It is working. Such a simple solution that I just could not see yet. Now I can work from here :)
– CodeIsland
Jan 3 at 9:07
Any chance you could help improve the OP's question? It's difficult to see this Q/A as very helpful to many others.
– rickhg12hs
Jan 3 at 12:25
The question was really about the output - how to access "count" and "item" in the output; the output as was correctly given in cybersam's post.
– CodeIsland
Jan 3 at 14:21
Thank you so very much! It is working. Such a simple solution that I just could not see yet. Now I can work from here :)
– CodeIsland
Jan 3 at 9:07
Thank you so very much! It is working. Such a simple solution that I just could not see yet. Now I can work from here :)
– CodeIsland
Jan 3 at 9:07
Any chance you could help improve the OP's question? It's difficult to see this Q/A as very helpful to many others.
– rickhg12hs
Jan 3 at 12:25
Any chance you could help improve the OP's question? It's difficult to see this Q/A as very helpful to many others.
– rickhg12hs
Jan 3 at 12:25
The question was really about the output - how to access "count" and "item" in the output; the output as was correctly given in cybersam's post.
– CodeIsland
Jan 3 at 14:21
The question was really about the output - how to access "count" and "item" in the output; the output as was correctly given in cybersam's post.
– CodeIsland
Jan 3 at 14:21
add a comment |
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Your description is a little confusing. Can you provide an example of desired output with some additional context?
– InverseFalcon
Jan 2 at 23:01
Thank you, cybersam solved it below :)
– CodeIsland
Jan 3 at 9:08