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Ruby - Best way to map objects into one array with fixed key
Serializing to JSON in jQueryCan you use a trailing comma in a JSON object?How do I format a Microsoft JSON date?Convert form data to JavaScript object with jQueryHow to sum array of numbers in Ruby?Check if a value exists in an array in RubySort hash by key, return hash in RubyRuby objects and JSON serialization (without Rails)Map and Remove nil values in RubyDetermining type of an object in ruby
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0
I try to merge some object into one array
I did output by
q = [["99","99","99"],["9"]]
o = [["b","1"],["c","3"],["d","1"],["c","30"]]
puts q.zip(o).map .to_json
=> [["99",["b","1"]],["99",["c","3"]],["99",["d","1"]],["9",["c","30"]]]
i'm looking for best way to
["99"=>"b"=>"1", "c"=>"3", "d"=>"1",{"9"=>"c"=>"30"]
json ruby
asked Mar 22 at 9:21
Tho NguyenTho Nguyen
116
|
show 5 more comments
0
I try to merge some object into one array
I did output by
q = [["99","99","99"],["9"]]
o = [["b","1"],["c","3"],["d","1"],["c","30"]]
puts q.zip(o).map .to_json
=> [["99",["b","1"]],["99",["c","3"]],["99",["d","1"]],["9",["c","30"]]]
i'm looking for best way to
["99"=>"b"=>"1", "c"=>"3", "d"=>"1",{"9"=>"c"=>"30"]
json ruby
asked Mar 22 at 9:21
Tho NguyenTho Nguyen
116
1
Your "some array" is invalid.
– sawa
Mar 22 at 9:25
yes, I'm still fresher, so I don't know to call this, but I print with this output and I try to beautify this output :(
– Tho Nguyen
Mar 22 at 9:29
No, it is not reproducible. Your initial array is invalid.
– sawa
Mar 22 at 9:34
I have updated my question, for more my flow to do that output. Can you teach me how to make it smarter :(
– Tho Nguyen
Mar 22 at 9:39
3
Where does that"10"come from in the expected output?
– Stefan
Mar 22 at 9:42
|
show 5 more comments
0
0
0
I try to merge some object into one array
I did output by
q = [["99","99","99"],["9"]]
o = [["b","1"],["c","3"],["d","1"],["c","30"]]
puts q.zip(o).map .to_json
=> [["99",["b","1"]],["99",["c","3"]],["99",["d","1"]],["9",["c","30"]]]
i'm looking for best way to
["99"=>"b"=>"1", "c"=>"3", "d"=>"1",{"9"=>"c"=>"30"]
json ruby
asked Mar 22 at 9:21
Tho NguyenTho Nguyen
116
I try to merge some object into one array
I did output by
q = [["99","99","99"],["9"]]
o = [["b","1"],["c","3"],["d","1"],["c","30"]]
puts q.zip(o).map .to_json
=> [["99",["b","1"]],["99",["c","3"]],["99",["d","1"]],["9",["c","30"]]]
i'm looking for best way to
["99"=>"b"=>"1", "c"=>"3", "d"=>"1",{"9"=>"c"=>"30"]
json ruby
json ruby
asked Mar 22 at 9:21
Tho NguyenTho Nguyen
116
asked Mar 22 at 9:21
Tho NguyenTho Nguyen
116
edited Mar 23 at 11:19
Tho Nguyen
asked Mar 22 at 9:21
Tho NguyenTho Nguyen
116
asked Mar 22 at 9:21
Tho NguyenTho Nguyen
116
asked Mar 22 at 9:21
Tho NguyenTho Nguyen
116
116
1
Your "some array" is invalid.
– sawa
Mar 22 at 9:25
yes, I'm still fresher, so I don't know to call this, but I print with this output and I try to beautify this output :(
– Tho Nguyen
Mar 22 at 9:29
No, it is not reproducible. Your initial array is invalid.
– sawa
Mar 22 at 9:34
I have updated my question, for more my flow to do that output. Can you teach me how to make it smarter :(
– Tho Nguyen
Mar 22 at 9:39
3
Where does that"10"come from in the expected output?
– Stefan
Mar 22 at 9:42
|
show 5 more comments
1
Your "some array" is invalid.
– sawa
Mar 22 at 9:25
yes, I'm still fresher, so I don't know to call this, but I print with this output and I try to beautify this output :(
– Tho Nguyen
Mar 22 at 9:29
No, it is not reproducible. Your initial array is invalid.
– sawa
Mar 22 at 9:34
I have updated my question, for more my flow to do that output. Can you teach me how to make it smarter :(
– Tho Nguyen
Mar 22 at 9:39
3
Where does that"10"come from in the expected output?
– Stefan
Mar 22 at 9:42
1
1
Your "some array" is invalid.
– sawa
Mar 22 at 9:25
Your "some array" is invalid.
– sawa
Mar 22 at 9:25
yes, I'm still fresher, so I don't know to call this, but I print with this output and I try to beautify this output :(
– Tho Nguyen
Mar 22 at 9:29
yes, I'm still fresher, so I don't know to call this, but I print with this output and I try to beautify this output :(
– Tho Nguyen
Mar 22 at 9:29
No, it is not reproducible. Your initial array is invalid.
– sawa
Mar 22 at 9:34
No, it is not reproducible. Your initial array is invalid.
– sawa
Mar 22 at 9:34
I have updated my question, for more my flow to do that output. Can you teach me how to make it smarter :(
– Tho Nguyen
Mar 22 at 9:39
I have updated my question, for more my flow to do that output. Can you teach me how to make it smarter :(
– Tho Nguyen
Mar 22 at 9:39
3
3
Where does that
"10" come from in the expected output?– Stefan
Mar 22 at 9:42
Where does that
"10" come from in the expected output?– Stefan
Mar 22 at 9:42
|
show 5 more comments
3 Answers
3
active
oldest
votes
1
a = [["9",["b","8"]],["9",["c","2"]],["9",["d","6"]]]
a.group_by(&:first).transform_values
# => "9"=>"b"=>"8", "c"=>"2", "d"=>"6"
a.group_by(&:first).transform_values.map k => v
# => ["9"=>"b"=>"8", "c"=>"2", "d"=>"6"]
answered Mar 22 at 9:35
sawasawa
134k31210309
add a comment |
1
Looking for something like this?:
some_array = [["9",["b","8"]], ["9",["c","2"]], ["9",["d","6"]]]
some_hash = some_array.each_with_object(Hash.new ) do |(k, (sub_key, sub_val)), hash|
hash[k][sub_key] = sub_val
end
p some_hash
#=> "9"=>"b"=>"8", "c"=>"2", "d"=>"6"
answered Mar 22 at 9:40
SuryaSurya
13.2k23558
add a comment |
-1
After the question was updated:
q = [["99","99","99"],["9"]]
o = [["b","1"],["c","3"],["d","1"],["c","30"]]
q.flatten.map.with_index.with_object(Hash.new ) .transform_values(&:to_h)
#=> "99"=>"b"=>"1", "c"=>"3", "d"=>"1", "9"=>"c"=>"30"
answered Mar 22 at 14:43
iGianiGian
5,4902725
I have update my question, for full output. Thank @iGian :)
– Tho Nguyen
Mar 23 at 11:13
[["100",["b","5"]],["101",["d","5"]],["102",["b","5"]],["103",["b","5"]],["104",["b","5"]],["105",["c","5"]],["106",["a","4"]],["106",["b","1"]],["107",["a","5"]],["108",["a","1"]],["108",["b","3"]],["108",["c","1"]],["109",["a","2"]],["109",["b","2"]],["109",["d","1"]]] I have cut output for you, I think we need edit q to another difference with first
– Tho Nguyen
Mar 23 at 11:28
@ThoNguyen, please, check my edit. I started from the original arrays.
– iGian
Mar 23 at 11:35
Thank you, I think we can make it to an object is good for reusable :)
– Tho Nguyen
Mar 23 at 11:35
Sure, your solution can solved my problem :)
– Tho Nguyen
Mar 23 at 11:37
|
show 2 more comments
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
a = [["9",["b","8"]],["9",["c","2"]],["9",["d","6"]]]
a.group_by(&:first).transform_values
# => "9"=>"b"=>"8", "c"=>"2", "d"=>"6"
a.group_by(&:first).transform_values.map k => v
# => ["9"=>"b"=>"8", "c"=>"2", "d"=>"6"]
answered Mar 22 at 9:35
sawasawa
134k31210309
add a comment |
1
a = [["9",["b","8"]],["9",["c","2"]],["9",["d","6"]]]
a.group_by(&:first).transform_values
# => "9"=>"b"=>"8", "c"=>"2", "d"=>"6"
a.group_by(&:first).transform_values.map k => v
# => ["9"=>"b"=>"8", "c"=>"2", "d"=>"6"]
answered Mar 22 at 9:35
sawasawa
134k31210309
add a comment |
1
1
1
a = [["9",["b","8"]],["9",["c","2"]],["9",["d","6"]]]
a.group_by(&:first).transform_values
# => "9"=>"b"=>"8", "c"=>"2", "d"=>"6"
a.group_by(&:first).transform_values.map k => v
# => ["9"=>"b"=>"8", "c"=>"2", "d"=>"6"]
answered Mar 22 at 9:35
sawasawa
134k31210309
a = [["9",["b","8"]],["9",["c","2"]],["9",["d","6"]]]
a.group_by(&:first).transform_values
# => "9"=>"b"=>"8", "c"=>"2", "d"=>"6"
a.group_by(&:first).transform_values.map k => v
# => ["9"=>"b"=>"8", "c"=>"2", "d"=>"6"]
answered Mar 22 at 9:35
sawasawa
134k31210309
edited Mar 22 at 9:45
answered Mar 22 at 9:35
sawasawa
134k31210309
answered Mar 22 at 9:35
sawasawa
134k31210309
answered Mar 22 at 9:35
sawasawa
134k31210309
134k31210309
add a comment |
add a comment |
1
Looking for something like this?:
some_array = [["9",["b","8"]], ["9",["c","2"]], ["9",["d","6"]]]
some_hash = some_array.each_with_object(Hash.new ) do |(k, (sub_key, sub_val)), hash|
hash[k][sub_key] = sub_val
end
p some_hash
#=> "9"=>"b"=>"8", "c"=>"2", "d"=>"6"
answered Mar 22 at 9:40
SuryaSurya
13.2k23558
add a comment |
1
Looking for something like this?:
some_array = [["9",["b","8"]], ["9",["c","2"]], ["9",["d","6"]]]
some_hash = some_array.each_with_object(Hash.new ) do |(k, (sub_key, sub_val)), hash|
hash[k][sub_key] = sub_val
end
p some_hash
#=> "9"=>"b"=>"8", "c"=>"2", "d"=>"6"
answered Mar 22 at 9:40
SuryaSurya
13.2k23558
add a comment |
1
1
1
Looking for something like this?:
some_array = [["9",["b","8"]], ["9",["c","2"]], ["9",["d","6"]]]
some_hash = some_array.each_with_object(Hash.new ) do |(k, (sub_key, sub_val)), hash|
hash[k][sub_key] = sub_val
end
p some_hash
#=> "9"=>"b"=>"8", "c"=>"2", "d"=>"6"
answered Mar 22 at 9:40
SuryaSurya
13.2k23558
Looking for something like this?:
some_array = [["9",["b","8"]], ["9",["c","2"]], ["9",["d","6"]]]
some_hash = some_array.each_with_object(Hash.new ) do |(k, (sub_key, sub_val)), hash|
hash[k][sub_key] = sub_val
end
p some_hash
#=> "9"=>"b"=>"8", "c"=>"2", "d"=>"6"
answered Mar 22 at 9:40
SuryaSurya
13.2k23558
answered Mar 22 at 9:40
SuryaSurya
13.2k23558
answered Mar 22 at 9:40
SuryaSurya
13.2k23558
answered Mar 22 at 9:40
SuryaSurya
13.2k23558
13.2k23558
add a comment |
add a comment |
-1
After the question was updated:
q = [["99","99","99"],["9"]]
o = [["b","1"],["c","3"],["d","1"],["c","30"]]
q.flatten.map.with_index.with_object(Hash.new ) .transform_values(&:to_h)
#=> "99"=>"b"=>"1", "c"=>"3", "d"=>"1", "9"=>"c"=>"30"
answered Mar 22 at 14:43
iGianiGian
5,4902725
I have update my question, for full output. Thank @iGian :)
– Tho Nguyen
Mar 23 at 11:13
[["100",["b","5"]],["101",["d","5"]],["102",["b","5"]],["103",["b","5"]],["104",["b","5"]],["105",["c","5"]],["106",["a","4"]],["106",["b","1"]],["107",["a","5"]],["108",["a","1"]],["108",["b","3"]],["108",["c","1"]],["109",["a","2"]],["109",["b","2"]],["109",["d","1"]]] I have cut output for you, I think we need edit q to another difference with first
– Tho Nguyen
Mar 23 at 11:28
@ThoNguyen, please, check my edit. I started from the original arrays.
– iGian
Mar 23 at 11:35
Thank you, I think we can make it to an object is good for reusable :)
– Tho Nguyen
Mar 23 at 11:35
Sure, your solution can solved my problem :)
– Tho Nguyen
Mar 23 at 11:37
|
show 2 more comments
-1
After the question was updated:
q = [["99","99","99"],["9"]]
o = [["b","1"],["c","3"],["d","1"],["c","30"]]
q.flatten.map.with_index.with_object(Hash.new ) .transform_values(&:to_h)
#=> "99"=>"b"=>"1", "c"=>"3", "d"=>"1", "9"=>"c"=>"30"
answered Mar 22 at 14:43
iGianiGian
5,4902725
I have update my question, for full output. Thank @iGian :)
– Tho Nguyen
Mar 23 at 11:13
[["100",["b","5"]],["101",["d","5"]],["102",["b","5"]],["103",["b","5"]],["104",["b","5"]],["105",["c","5"]],["106",["a","4"]],["106",["b","1"]],["107",["a","5"]],["108",["a","1"]],["108",["b","3"]],["108",["c","1"]],["109",["a","2"]],["109",["b","2"]],["109",["d","1"]]] I have cut output for you, I think we need edit q to another difference with first
– Tho Nguyen
Mar 23 at 11:28
@ThoNguyen, please, check my edit. I started from the original arrays.
– iGian
Mar 23 at 11:35
Thank you, I think we can make it to an object is good for reusable :)
– Tho Nguyen
Mar 23 at 11:35
Sure, your solution can solved my problem :)
– Tho Nguyen
Mar 23 at 11:37
|
show 2 more comments
-1
-1
-1
After the question was updated:
q = [["99","99","99"],["9"]]
o = [["b","1"],["c","3"],["d","1"],["c","30"]]
q.flatten.map.with_index.with_object(Hash.new ) .transform_values(&:to_h)
#=> "99"=>"b"=>"1", "c"=>"3", "d"=>"1", "9"=>"c"=>"30"
answered Mar 22 at 14:43
iGianiGian
5,4902725
After the question was updated:
q = [["99","99","99"],["9"]]
o = [["b","1"],["c","3"],["d","1"],["c","30"]]
q.flatten.map.with_index.with_object(Hash.new ) .transform_values(&:to_h)
#=> "99"=>"b"=>"1", "c"=>"3", "d"=>"1", "9"=>"c"=>"30"
answered Mar 22 at 14:43
iGianiGian
5,4902725
edited Mar 23 at 11:38
answered Mar 22 at 14:43
iGianiGian
5,4902725
answered Mar 22 at 14:43
iGianiGian
5,4902725
answered Mar 22 at 14:43
iGianiGian
5,4902725
5,4902725
I have update my question, for full output. Thank @iGian :)
– Tho Nguyen
Mar 23 at 11:13
[["100",["b","5"]],["101",["d","5"]],["102",["b","5"]],["103",["b","5"]],["104",["b","5"]],["105",["c","5"]],["106",["a","4"]],["106",["b","1"]],["107",["a","5"]],["108",["a","1"]],["108",["b","3"]],["108",["c","1"]],["109",["a","2"]],["109",["b","2"]],["109",["d","1"]]] I have cut output for you, I think we need edit q to another difference with first
– Tho Nguyen
Mar 23 at 11:28
@ThoNguyen, please, check my edit. I started from the original arrays.
– iGian
Mar 23 at 11:35
Thank you, I think we can make it to an object is good for reusable :)
– Tho Nguyen
Mar 23 at 11:35
Sure, your solution can solved my problem :)
– Tho Nguyen
Mar 23 at 11:37
|
show 2 more comments
I have update my question, for full output. Thank @iGian :)
– Tho Nguyen
Mar 23 at 11:13
[["100",["b","5"]],["101",["d","5"]],["102",["b","5"]],["103",["b","5"]],["104",["b","5"]],["105",["c","5"]],["106",["a","4"]],["106",["b","1"]],["107",["a","5"]],["108",["a","1"]],["108",["b","3"]],["108",["c","1"]],["109",["a","2"]],["109",["b","2"]],["109",["d","1"]]] I have cut output for you, I think we need edit q to another difference with first
– Tho Nguyen
Mar 23 at 11:28
@ThoNguyen, please, check my edit. I started from the original arrays.
– iGian
Mar 23 at 11:35
Thank you, I think we can make it to an object is good for reusable :)
– Tho Nguyen
Mar 23 at 11:35
Sure, your solution can solved my problem :)
– Tho Nguyen
Mar 23 at 11:37
I have update my question, for full output. Thank @iGian :)
– Tho Nguyen
Mar 23 at 11:13
I have update my question, for full output. Thank @iGian :)
– Tho Nguyen
Mar 23 at 11:13
[["100",["b","5"]],["101",["d","5"]],["102",["b","5"]],["103",["b","5"]],["104",["b","5"]],["105",["c","5"]],["106",["a","4"]],["106",["b","1"]],["107",["a","5"]],["108",["a","1"]],["108",["b","3"]],["108",["c","1"]],["109",["a","2"]],["109",["b","2"]],["109",["d","1"]]] I have cut output for you, I think we need edit q to another difference with first
– Tho Nguyen
Mar 23 at 11:28
[["100",["b","5"]],["101",["d","5"]],["102",["b","5"]],["103",["b","5"]],["104",["b","5"]],["105",["c","5"]],["106",["a","4"]],["106",["b","1"]],["107",["a","5"]],["108",["a","1"]],["108",["b","3"]],["108",["c","1"]],["109",["a","2"]],["109",["b","2"]],["109",["d","1"]]] I have cut output for you, I think we need edit q to another difference with first
– Tho Nguyen
Mar 23 at 11:28
@ThoNguyen, please, check my edit. I started from the original arrays.
– iGian
Mar 23 at 11:35
@ThoNguyen, please, check my edit. I started from the original arrays.
– iGian
Mar 23 at 11:35
Thank you, I think we can make it to an object is good for reusable :)
– Tho Nguyen
Mar 23 at 11:35
Thank you, I think we can make it to an object is good for reusable :)
– Tho Nguyen
Mar 23 at 11:35
Sure, your solution can solved my problem :)
– Tho Nguyen
Mar 23 at 11:37
Sure, your solution can solved my problem :)
– Tho Nguyen
Mar 23 at 11:37
|
show 2 more comments
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1
Your "some array" is invalid.
– sawa
Mar 22 at 9:25
yes, I'm still fresher, so I don't know to call this, but I print with this output and I try to beautify this output :(
– Tho Nguyen
Mar 22 at 9:29
No, it is not reproducible. Your initial array is invalid.
– sawa
Mar 22 at 9:34
I have updated my question, for more my flow to do that output. Can you teach me how to make it smarter :(
– Tho Nguyen
Mar 22 at 9:39
3
Where does that
"10"come from in the expected output?– Stefan
Mar 22 at 9:42