Efficient way to add new rules to a sublist within a nested list of rules
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4
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I often build nested lists (as a way to keep data sets organised) with some form like:
nestru = {"a" -> {"x" -> 5, "y" -> 7}, "b" -> {"x" -> 4, "y" -> 6}};
Here "a" and "b" are first order keys that associate a series of second order keys and values with the first order entity. If I want to add a new set of second order rules I can do something like:
newru = {"z" -> "x"/"y", "xx" -> 2 "x"} /. nestru[[All, 2]]
{{"z" -> 5/7, "xx" -> 10},
{"z" -> 2/3, "xx" -> 8}}
Then I rebuild the whole data structure again:
MapThread[#1 -> #2 &, {nestru[[All, 1]],
Flatten[#] & /@ Transpose@{nestru[[All, 2]], newru}}]
{"a" -> {"x" -> 5, "y" -> 7, "z" -> 5/7, "xx" -> 10},
"b" -> {"x" -> 4, "y" -> 6, "z" -> 2/3, "xx" -> 8}}
This feels like a clunky approach. Is there an 'cleaner'/more efficient way to add new sub-rules to a nested list of rules?
list-manipulation replacement
asked Nov 26 at 9:07
geordie
1,9661530
add a comment |
up vote
4
down vote
favorite
I often build nested lists (as a way to keep data sets organised) with some form like:
nestru = {"a" -> {"x" -> 5, "y" -> 7}, "b" -> {"x" -> 4, "y" -> 6}};
Here "a" and "b" are first order keys that associate a series of second order keys and values with the first order entity. If I want to add a new set of second order rules I can do something like:
newru = {"z" -> "x"/"y", "xx" -> 2 "x"} /. nestru[[All, 2]]
{{"z" -> 5/7, "xx" -> 10},
{"z" -> 2/3, "xx" -> 8}}
Then I rebuild the whole data structure again:
MapThread[#1 -> #2 &, {nestru[[All, 1]],
Flatten[#] & /@ Transpose@{nestru[[All, 2]], newru}}]
{"a" -> {"x" -> 5, "y" -> 7, "z" -> 5/7, "xx" -> 10},
"b" -> {"x" -> 4, "y" -> 6, "z" -> 2/3, "xx" -> 8}}
This feels like a clunky approach. Is there an 'cleaner'/more efficient way to add new sub-rules to a nested list of rules?
list-manipulation replacement
asked Nov 26 at 9:07
geordie
1,9661530
2
Instead of using nested lists, I would suggest to use nestedAssociations along withAssociateTo.
– Henrik Schumacher
Nov 26 at 9:35
@HenrikSchumacher If I convert toAssociationhow do I then useAssociateToto add values to a 'sub'-Association? It's not clear from the documentation.
– geordie
Nov 26 at 11:10
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I often build nested lists (as a way to keep data sets organised) with some form like:
nestru = {"a" -> {"x" -> 5, "y" -> 7}, "b" -> {"x" -> 4, "y" -> 6}};
Here "a" and "b" are first order keys that associate a series of second order keys and values with the first order entity. If I want to add a new set of second order rules I can do something like:
newru = {"z" -> "x"/"y", "xx" -> 2 "x"} /. nestru[[All, 2]]
{{"z" -> 5/7, "xx" -> 10},
{"z" -> 2/3, "xx" -> 8}}
Then I rebuild the whole data structure again:
MapThread[#1 -> #2 &, {nestru[[All, 1]],
Flatten[#] & /@ Transpose@{nestru[[All, 2]], newru}}]
{"a" -> {"x" -> 5, "y" -> 7, "z" -> 5/7, "xx" -> 10},
"b" -> {"x" -> 4, "y" -> 6, "z" -> 2/3, "xx" -> 8}}
This feels like a clunky approach. Is there an 'cleaner'/more efficient way to add new sub-rules to a nested list of rules?
list-manipulation replacement
asked Nov 26 at 9:07
geordie
1,9661530
I often build nested lists (as a way to keep data sets organised) with some form like:
nestru = {"a" -> {"x" -> 5, "y" -> 7}, "b" -> {"x" -> 4, "y" -> 6}};
Here "a" and "b" are first order keys that associate a series of second order keys and values with the first order entity. If I want to add a new set of second order rules I can do something like:
newru = {"z" -> "x"/"y", "xx" -> 2 "x"} /. nestru[[All, 2]]
{{"z" -> 5/7, "xx" -> 10},
{"z" -> 2/3, "xx" -> 8}}
Then I rebuild the whole data structure again:
MapThread[#1 -> #2 &, {nestru[[All, 1]],
Flatten[#] & /@ Transpose@{nestru[[All, 2]], newru}}]
{"a" -> {"x" -> 5, "y" -> 7, "z" -> 5/7, "xx" -> 10},
"b" -> {"x" -> 4, "y" -> 6, "z" -> 2/3, "xx" -> 8}}
This feels like a clunky approach. Is there an 'cleaner'/more efficient way to add new sub-rules to a nested list of rules?
list-manipulation replacement
list-manipulation replacement
asked Nov 26 at 9:07
geordie
1,9661530
asked Nov 26 at 9:07
geordie
1,9661530
edited Nov 26 at 11:11
asked Nov 26 at 9:07
geordie
1,9661530
asked Nov 26 at 9:07
geordie
1,9661530
asked Nov 26 at 9:07
geordie
1,9661530
1,9661530
2
Instead of using nested lists, I would suggest to use nestedAssociations along withAssociateTo.
– Henrik Schumacher
Nov 26 at 9:35
@HenrikSchumacher If I convert toAssociationhow do I then useAssociateToto add values to a 'sub'-Association? It's not clear from the documentation.
– geordie
Nov 26 at 11:10
add a comment |
2
Instead of using nested lists, I would suggest to use nestedAssociations along withAssociateTo.
– Henrik Schumacher
Nov 26 at 9:35
@HenrikSchumacher If I convert toAssociationhow do I then useAssociateToto add values to a 'sub'-Association? It's not clear from the documentation.
– geordie
Nov 26 at 11:10
2
2
Instead of using nested lists, I would suggest to use nested
Associations along with AssociateTo.– Henrik Schumacher
Nov 26 at 9:35
Instead of using nested lists, I would suggest to use nested
Associations along with AssociateTo.– Henrik Schumacher
Nov 26 at 9:35
@HenrikSchumacher If I convert to
Association how do I then use AssociateTo to add values to a 'sub'-Association? It's not clear from the documentation.– geordie
Nov 26 at 11:10
@HenrikSchumacher If I convert to
Association how do I then use AssociateTo to add values to a 'sub'-Association? It's not clear from the documentation.– geordie
Nov 26 at 11:10
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
Let me show you another way. I'd go with associations and I'd make newgru longer but more robust (here "z", or "xx" could be replaced by rules from nestgru if you are not careful).
data = {"a" -> {"x" -> 5, "y" -> 7}, "b" -> {"x" -> 4, "y" -> 6}} //
GeneralUtilities`ToAssociations
<|"a" -> <|"x" -> 5, "y" -> 7|>, "b" -> <|"x" -> 4, "y" -> 6|>|>
<|#, "z" -> #x/#y, "xx" -> 2 #x|> & /@ data
<|"a" -> <|"x" -> 5, "y" -> 7, "z" -> 5/7, "xx" -> 10|>,
"b" -> <|"x" -> 4, "y" -> 6, "z" -> 2/3, "xx" -> 8|>|>
answered Nov 26 at 9:35
Kuba♦
103k12200511
It's about time for me to read more about associations... The only problem I see is how to turn it back into a list of rules? If I useNormalonly the top level changes to aList. Again It feels like I have to rebuild the data structure if I want utilise lists of rules.
– geordie
Nov 26 at 10:43
3
@geordie don't go back! But if you really need to:data /. Association -> Normal@*Association
– Kuba♦
Nov 26 at 10:51
Moving forward, I'll try not to go back (!) but I currently work with large ecosystem built on lists of rules. It may be that I can use most of this will work withAssociation'ed structures but I'll need to do some testing. For the moment (until I can brush up) 'old school' seems like the safest way to proceed...
– geordie
Nov 26 at 10:58
2
@geordie Just a remark:Associations are essentially hash tables, so -- in contrast to lists -- operations like insert and delete do not need a full copy of the data structure. ThusAssociationshould not only be much more convenient but also much more efficient. However, they come with a certain extra memory cost. You might also findDatasetinteresting.
– Henrik Schumacher
Nov 26 at 11:23
@HenrikSchumacher, about your insert/delete remark, does it hold forAssociations not assigned to a symbol? I have always thought that if an expression isn't assigned to a symbol, the Kernel will make a copy on any modification of this expression to provide immutability...
– Anton.Sakovich
Nov 27 at 8:21
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Let me show you another way. I'd go with associations and I'd make newgru longer but more robust (here "z", or "xx" could be replaced by rules from nestgru if you are not careful).
data = {"a" -> {"x" -> 5, "y" -> 7}, "b" -> {"x" -> 4, "y" -> 6}} //
GeneralUtilities`ToAssociations
<|"a" -> <|"x" -> 5, "y" -> 7|>, "b" -> <|"x" -> 4, "y" -> 6|>|>
<|#, "z" -> #x/#y, "xx" -> 2 #x|> & /@ data
<|"a" -> <|"x" -> 5, "y" -> 7, "z" -> 5/7, "xx" -> 10|>,
"b" -> <|"x" -> 4, "y" -> 6, "z" -> 2/3, "xx" -> 8|>|>
answered Nov 26 at 9:35
Kuba♦
103k12200511
It's about time for me to read more about associations... The only problem I see is how to turn it back into a list of rules? If I useNormalonly the top level changes to aList. Again It feels like I have to rebuild the data structure if I want utilise lists of rules.
– geordie
Nov 26 at 10:43
3
@geordie don't go back! But if you really need to:data /. Association -> Normal@*Association
– Kuba♦
Nov 26 at 10:51
Moving forward, I'll try not to go back (!) but I currently work with large ecosystem built on lists of rules. It may be that I can use most of this will work withAssociation'ed structures but I'll need to do some testing. For the moment (until I can brush up) 'old school' seems like the safest way to proceed...
– geordie
Nov 26 at 10:58
2
@geordie Just a remark:Associations are essentially hash tables, so -- in contrast to lists -- operations like insert and delete do not need a full copy of the data structure. ThusAssociationshould not only be much more convenient but also much more efficient. However, they come with a certain extra memory cost. You might also findDatasetinteresting.
– Henrik Schumacher
Nov 26 at 11:23
@HenrikSchumacher, about your insert/delete remark, does it hold forAssociations not assigned to a symbol? I have always thought that if an expression isn't assigned to a symbol, the Kernel will make a copy on any modification of this expression to provide immutability...
– Anton.Sakovich
Nov 27 at 8:21
add a comment |
up vote
4
down vote
Let me show you another way. I'd go with associations and I'd make newgru longer but more robust (here "z", or "xx" could be replaced by rules from nestgru if you are not careful).
data = {"a" -> {"x" -> 5, "y" -> 7}, "b" -> {"x" -> 4, "y" -> 6}} //
GeneralUtilities`ToAssociations
<|"a" -> <|"x" -> 5, "y" -> 7|>, "b" -> <|"x" -> 4, "y" -> 6|>|>
<|#, "z" -> #x/#y, "xx" -> 2 #x|> & /@ data
<|"a" -> <|"x" -> 5, "y" -> 7, "z" -> 5/7, "xx" -> 10|>,
"b" -> <|"x" -> 4, "y" -> 6, "z" -> 2/3, "xx" -> 8|>|>
answered Nov 26 at 9:35
Kuba♦
103k12200511
It's about time for me to read more about associations... The only problem I see is how to turn it back into a list of rules? If I useNormalonly the top level changes to aList. Again It feels like I have to rebuild the data structure if I want utilise lists of rules.
– geordie
Nov 26 at 10:43
3
@geordie don't go back! But if you really need to:data /. Association -> Normal@*Association
– Kuba♦
Nov 26 at 10:51
Moving forward, I'll try not to go back (!) but I currently work with large ecosystem built on lists of rules. It may be that I can use most of this will work withAssociation'ed structures but I'll need to do some testing. For the moment (until I can brush up) 'old school' seems like the safest way to proceed...
– geordie
Nov 26 at 10:58
2
@geordie Just a remark:Associations are essentially hash tables, so -- in contrast to lists -- operations like insert and delete do not need a full copy of the data structure. ThusAssociationshould not only be much more convenient but also much more efficient. However, they come with a certain extra memory cost. You might also findDatasetinteresting.
– Henrik Schumacher
Nov 26 at 11:23
@HenrikSchumacher, about your insert/delete remark, does it hold forAssociations not assigned to a symbol? I have always thought that if an expression isn't assigned to a symbol, the Kernel will make a copy on any modification of this expression to provide immutability...
– Anton.Sakovich
Nov 27 at 8:21
add a comment |
up vote
4
down vote
up vote
4
down vote
Let me show you another way. I'd go with associations and I'd make newgru longer but more robust (here "z", or "xx" could be replaced by rules from nestgru if you are not careful).
data = {"a" -> {"x" -> 5, "y" -> 7}, "b" -> {"x" -> 4, "y" -> 6}} //
GeneralUtilities`ToAssociations
<|"a" -> <|"x" -> 5, "y" -> 7|>, "b" -> <|"x" -> 4, "y" -> 6|>|>
<|#, "z" -> #x/#y, "xx" -> 2 #x|> & /@ data
<|"a" -> <|"x" -> 5, "y" -> 7, "z" -> 5/7, "xx" -> 10|>,
"b" -> <|"x" -> 4, "y" -> 6, "z" -> 2/3, "xx" -> 8|>|>
answered Nov 26 at 9:35
Kuba♦
103k12200511
Let me show you another way. I'd go with associations and I'd make newgru longer but more robust (here "z", or "xx" could be replaced by rules from nestgru if you are not careful).
data = {"a" -> {"x" -> 5, "y" -> 7}, "b" -> {"x" -> 4, "y" -> 6}} //
GeneralUtilities`ToAssociations
<|"a" -> <|"x" -> 5, "y" -> 7|>, "b" -> <|"x" -> 4, "y" -> 6|>|>
<|#, "z" -> #x/#y, "xx" -> 2 #x|> & /@ data
<|"a" -> <|"x" -> 5, "y" -> 7, "z" -> 5/7, "xx" -> 10|>,
"b" -> <|"x" -> 4, "y" -> 6, "z" -> 2/3, "xx" -> 8|>|>
answered Nov 26 at 9:35
Kuba♦
103k12200511
answered Nov 26 at 9:35
Kuba♦
103k12200511
answered Nov 26 at 9:35
Kuba♦
103k12200511
answered Nov 26 at 9:35
Kuba♦
103k12200511
103k12200511
It's about time for me to read more about associations... The only problem I see is how to turn it back into a list of rules? If I useNormalonly the top level changes to aList. Again It feels like I have to rebuild the data structure if I want utilise lists of rules.
– geordie
Nov 26 at 10:43
3
@geordie don't go back! But if you really need to:data /. Association -> Normal@*Association
– Kuba♦
Nov 26 at 10:51
Moving forward, I'll try not to go back (!) but I currently work with large ecosystem built on lists of rules. It may be that I can use most of this will work withAssociation'ed structures but I'll need to do some testing. For the moment (until I can brush up) 'old school' seems like the safest way to proceed...
– geordie
Nov 26 at 10:58
2
@geordie Just a remark:Associations are essentially hash tables, so -- in contrast to lists -- operations like insert and delete do not need a full copy of the data structure. ThusAssociationshould not only be much more convenient but also much more efficient. However, they come with a certain extra memory cost. You might also findDatasetinteresting.
– Henrik Schumacher
Nov 26 at 11:23
@HenrikSchumacher, about your insert/delete remark, does it hold forAssociations not assigned to a symbol? I have always thought that if an expression isn't assigned to a symbol, the Kernel will make a copy on any modification of this expression to provide immutability...
– Anton.Sakovich
Nov 27 at 8:21
add a comment |
It's about time for me to read more about associations... The only problem I see is how to turn it back into a list of rules? If I useNormalonly the top level changes to aList. Again It feels like I have to rebuild the data structure if I want utilise lists of rules.
– geordie
Nov 26 at 10:43
3
@geordie don't go back! But if you really need to:data /. Association -> Normal@*Association
– Kuba♦
Nov 26 at 10:51
Moving forward, I'll try not to go back (!) but I currently work with large ecosystem built on lists of rules. It may be that I can use most of this will work withAssociation'ed structures but I'll need to do some testing. For the moment (until I can brush up) 'old school' seems like the safest way to proceed...
– geordie
Nov 26 at 10:58
2
@geordie Just a remark:Associations are essentially hash tables, so -- in contrast to lists -- operations like insert and delete do not need a full copy of the data structure. ThusAssociationshould not only be much more convenient but also much more efficient. However, they come with a certain extra memory cost. You might also findDatasetinteresting.
– Henrik Schumacher
Nov 26 at 11:23
@HenrikSchumacher, about your insert/delete remark, does it hold forAssociations not assigned to a symbol? I have always thought that if an expression isn't assigned to a symbol, the Kernel will make a copy on any modification of this expression to provide immutability...
– Anton.Sakovich
Nov 27 at 8:21
It's about time for me to read more about associations... The only problem I see is how to turn it back into a list of rules? If I use
Normal only the top level changes to a List. Again It feels like I have to rebuild the data structure if I want utilise lists of rules.– geordie
Nov 26 at 10:43
It's about time for me to read more about associations... The only problem I see is how to turn it back into a list of rules? If I use
Normal only the top level changes to a List. Again It feels like I have to rebuild the data structure if I want utilise lists of rules.– geordie
Nov 26 at 10:43
3
3
@geordie don't go back! But if you really need to:
data /. Association -> Normal@*Association– Kuba♦
Nov 26 at 10:51
@geordie don't go back! But if you really need to:
data /. Association -> Normal@*Association– Kuba♦
Nov 26 at 10:51
Moving forward, I'll try not to go back (!) but I currently work with large ecosystem built on lists of rules. It may be that I can use most of this will work with
Association'ed structures but I'll need to do some testing. For the moment (until I can brush up) 'old school' seems like the safest way to proceed...– geordie
Nov 26 at 10:58
Moving forward, I'll try not to go back (!) but I currently work with large ecosystem built on lists of rules. It may be that I can use most of this will work with
Association'ed structures but I'll need to do some testing. For the moment (until I can brush up) 'old school' seems like the safest way to proceed...– geordie
Nov 26 at 10:58
2
2
@geordie Just a remark:
Associations are essentially hash tables, so -- in contrast to lists -- operations like insert and delete do not need a full copy of the data structure. Thus Association should not only be much more convenient but also much more efficient. However, they come with a certain extra memory cost. You might also find Dataset interesting.– Henrik Schumacher
Nov 26 at 11:23
@geordie Just a remark:
Associations are essentially hash tables, so -- in contrast to lists -- operations like insert and delete do not need a full copy of the data structure. Thus Association should not only be much more convenient but also much more efficient. However, they come with a certain extra memory cost. You might also find Dataset interesting.– Henrik Schumacher
Nov 26 at 11:23
@HenrikSchumacher, about your insert/delete remark, does it hold for
Associations not assigned to a symbol? I have always thought that if an expression isn't assigned to a symbol, the Kernel will make a copy on any modification of this expression to provide immutability...– Anton.Sakovich
Nov 27 at 8:21
@HenrikSchumacher, about your insert/delete remark, does it hold for
Associations not assigned to a symbol? I have always thought that if an expression isn't assigned to a symbol, the Kernel will make a copy on any modification of this expression to provide immutability...– Anton.Sakovich
Nov 27 at 8:21
add a comment |
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2
Instead of using nested lists, I would suggest to use nested
Associations along withAssociateTo.– Henrik Schumacher
Nov 26 at 9:35
@HenrikSchumacher If I convert to
Associationhow do I then useAssociateToto add values to a 'sub'-Association? It's not clear from the documentation.– geordie
Nov 26 at 11:10