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Combine arrays with conditions in Ruby
2019 Community Moderator ElectionCreate ArrayList from arrayHow do I check if an array includes an object in JavaScript?How to append something to an array?PHP: Delete an element from an arrayHow to write a switch statement in RubyCheck if a value exists in an array in RubyLoop through an array in JavaScriptHow to check if an object is an array?How do I remove a particular element from an array in JavaScript?For-each over an array in JavaScript?
I have a class People with three properties
class People
attr_accessor :first_name, :last_name, :age
end
And I have two arrays:
a = [p1, p2]
b = [p3, p4]
Is there any easy way to combine these two arrays in a new array and remove the item with a condition like:
p1.first_name + p1.last_name == p3.first_name + p3.last_name
And after that all the item should be belong to array a
For example
p1.first_name = "Ada"
p1.last_name = "Wang"
p1.age = 28
p2.first_name = "Leon"
p2.last_name = "S"
p2.age = 28
p3.first_name = "Ada"
p3.last_name = "Wang"
p3.age = 18
p4.first_name = "Mario"
p4.last_name = "M"
p4.age = 80
the result should be [p1] the 28 years old Ada.Wang
arrays ruby
asked Mar 7 at 13:43
ShaggonShaggon
84
add a comment |
I have a class People with three properties
class People
attr_accessor :first_name, :last_name, :age
end
And I have two arrays:
a = [p1, p2]
b = [p3, p4]
Is there any easy way to combine these two arrays in a new array and remove the item with a condition like:
p1.first_name + p1.last_name == p3.first_name + p3.last_name
And after that all the item should be belong to array a
For example
p1.first_name = "Ada"
p1.last_name = "Wang"
p1.age = 28
p2.first_name = "Leon"
p2.last_name = "S"
p2.age = 28
p3.first_name = "Ada"
p3.last_name = "Wang"
p3.age = 18
p4.first_name = "Mario"
p4.last_name = "M"
p4.age = 80
the result should be [p1] the 28 years old Ada.Wang
arrays ruby
asked Mar 7 at 13:43
ShaggonShaggon
84
2
In your condition you comparep1's name top3's name. What aboutp2andp4? Could you give a more complete example with actual data and show the expected output?
– Stefan
Mar 7 at 13:52
@Shaggon do you want the result to be an array of persons with no duplicate first_name last_name ?
– steenslag
Mar 7 at 17:29
Yes. I want a result with no duplicate first_name last_name and all the item should be belong to array a. The result should be like[p1]because p2 have duplicate first_name last_name with p3 and p4 is inside array b, so only p1 is what I want
– Shaggon
Mar 8 at 2:26
Thanks stenfan and steenslag. I found the result I want is(b + a) - (b + a).uniq e
– Shaggon
Mar 8 at 5:15
add a comment |
I have a class People with three properties
class People
attr_accessor :first_name, :last_name, :age
end
And I have two arrays:
a = [p1, p2]
b = [p3, p4]
Is there any easy way to combine these two arrays in a new array and remove the item with a condition like:
p1.first_name + p1.last_name == p3.first_name + p3.last_name
And after that all the item should be belong to array a
For example
p1.first_name = "Ada"
p1.last_name = "Wang"
p1.age = 28
p2.first_name = "Leon"
p2.last_name = "S"
p2.age = 28
p3.first_name = "Ada"
p3.last_name = "Wang"
p3.age = 18
p4.first_name = "Mario"
p4.last_name = "M"
p4.age = 80
the result should be [p1] the 28 years old Ada.Wang
arrays ruby
asked Mar 7 at 13:43
ShaggonShaggon
84
I have a class People with three properties
class People
attr_accessor :first_name, :last_name, :age
end
And I have two arrays:
a = [p1, p2]
b = [p3, p4]
Is there any easy way to combine these two arrays in a new array and remove the item with a condition like:
p1.first_name + p1.last_name == p3.first_name + p3.last_name
And after that all the item should be belong to array a
For example
p1.first_name = "Ada"
p1.last_name = "Wang"
p1.age = 28
p2.first_name = "Leon"
p2.last_name = "S"
p2.age = 28
p3.first_name = "Ada"
p3.last_name = "Wang"
p3.age = 18
p4.first_name = "Mario"
p4.last_name = "M"
p4.age = 80
the result should be [p1] the 28 years old Ada.Wang
arrays ruby
arrays ruby
asked Mar 7 at 13:43
ShaggonShaggon
84
asked Mar 7 at 13:43
ShaggonShaggon
84
edited Mar 8 at 2:49
Shaggon
asked Mar 7 at 13:43
ShaggonShaggon
84
asked Mar 7 at 13:43
ShaggonShaggon
84
asked Mar 7 at 13:43
ShaggonShaggon
84
84
2
In your condition you comparep1's name top3's name. What aboutp2andp4? Could you give a more complete example with actual data and show the expected output?
– Stefan
Mar 7 at 13:52
@Shaggon do you want the result to be an array of persons with no duplicate first_name last_name ?
– steenslag
Mar 7 at 17:29
Yes. I want a result with no duplicate first_name last_name and all the item should be belong to array a. The result should be like[p1]because p2 have duplicate first_name last_name with p3 and p4 is inside array b, so only p1 is what I want
– Shaggon
Mar 8 at 2:26
Thanks stenfan and steenslag. I found the result I want is(b + a) - (b + a).uniq e
– Shaggon
Mar 8 at 5:15
add a comment |
2
In your condition you comparep1's name top3's name. What aboutp2andp4? Could you give a more complete example with actual data and show the expected output?
– Stefan
Mar 7 at 13:52
@Shaggon do you want the result to be an array of persons with no duplicate first_name last_name ?
– steenslag
Mar 7 at 17:29
Yes. I want a result with no duplicate first_name last_name and all the item should be belong to array a. The result should be like[p1]because p2 have duplicate first_name last_name with p3 and p4 is inside array b, so only p1 is what I want
– Shaggon
Mar 8 at 2:26
Thanks stenfan and steenslag. I found the result I want is(b + a) - (b + a).uniq e
– Shaggon
Mar 8 at 5:15
2
2
In your condition you compare
p1's name to p3's name. What about p2 and p4? Could you give a more complete example with actual data and show the expected output?– Stefan
Mar 7 at 13:52
In your condition you compare
p1's name to p3's name. What about p2 and p4? Could you give a more complete example with actual data and show the expected output?– Stefan
Mar 7 at 13:52
@Shaggon do you want the result to be an array of persons with no duplicate first_name last_name ?
– steenslag
Mar 7 at 17:29
@Shaggon do you want the result to be an array of persons with no duplicate first_name last_name ?
– steenslag
Mar 7 at 17:29
Yes. I want a result with no duplicate first_name last_name and all the item should be belong to array a. The result should be like
[p1] because p2 have duplicate first_name last_name with p3 and p4 is inside array b, so only p1 is what I want– Shaggon
Mar 8 at 2:26
Yes. I want a result with no duplicate first_name last_name and all the item should be belong to array a. The result should be like
[p1] because p2 have duplicate first_name last_name with p3 and p4 is inside array b, so only p1 is what I want– Shaggon
Mar 8 at 2:26
Thanks stenfan and steenslag. I found the result I want is
(b + a) - (b + a).uniq e– Shaggon
Mar 8 at 5:15
Thanks stenfan and steenslag. I found the result I want is
(b + a) - (b + a).uniq e– Shaggon
Mar 8 at 5:15
add a comment |
2 Answers
2
active
oldest
votes
I'm not sure I get your point, but maybe this is a possible option.
c = a + b
c.uniq! e
Call Array#uniq! with a block on c which is the concatenation of a and b.
answered Mar 7 at 14:36
iGianiGian
4,6842725
1
Thank you so much. I think I found what I want from your comment. It is(b + a) - (b + a).uniq e
– Shaggon
Mar 8 at 5:14
add a comment |
If arrays a and b themselves do not contain people with matching first and last names then this would work:
b.each_with_index do |p, i|
if !(b[i].first_name == a[i].first_name and b[i].last_name == a[i].last_name)
a.push(p) # as people p does not contain the same first/last names as a it can now be added to a
end
end
To check for other fields simply replace first_name / last_name with other variables.
answered Mar 7 at 14:20
Jack BranchJack Branch
373
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I'm not sure I get your point, but maybe this is a possible option.
c = a + b
c.uniq! e
Call Array#uniq! with a block on c which is the concatenation of a and b.
answered Mar 7 at 14:36
iGianiGian
4,6842725
1
Thank you so much. I think I found what I want from your comment. It is(b + a) - (b + a).uniq e
– Shaggon
Mar 8 at 5:14
add a comment |
I'm not sure I get your point, but maybe this is a possible option.
c = a + b
c.uniq! e
Call Array#uniq! with a block on c which is the concatenation of a and b.
answered Mar 7 at 14:36
iGianiGian
4,6842725
1
Thank you so much. I think I found what I want from your comment. It is(b + a) - (b + a).uniq e
– Shaggon
Mar 8 at 5:14
add a comment |
I'm not sure I get your point, but maybe this is a possible option.
c = a + b
c.uniq! e
Call Array#uniq! with a block on c which is the concatenation of a and b.
answered Mar 7 at 14:36
iGianiGian
4,6842725
I'm not sure I get your point, but maybe this is a possible option.
c = a + b
c.uniq! e
Call Array#uniq! with a block on c which is the concatenation of a and b.
answered Mar 7 at 14:36
iGianiGian
4,6842725
answered Mar 7 at 14:36
iGianiGian
4,6842725
answered Mar 7 at 14:36
iGianiGian
4,6842725
answered Mar 7 at 14:36
iGianiGian
4,6842725
4,6842725
1
Thank you so much. I think I found what I want from your comment. It is(b + a) - (b + a).uniq e
– Shaggon
Mar 8 at 5:14
add a comment |
1
Thank you so much. I think I found what I want from your comment. It is(b + a) - (b + a).uniq e
– Shaggon
Mar 8 at 5:14
1
1
Thank you so much. I think I found what I want from your comment. It is
(b + a) - (b + a).uniq e– Shaggon
Mar 8 at 5:14
Thank you so much. I think I found what I want from your comment. It is
(b + a) - (b + a).uniq e– Shaggon
Mar 8 at 5:14
add a comment |
If arrays a and b themselves do not contain people with matching first and last names then this would work:
b.each_with_index do |p, i|
if !(b[i].first_name == a[i].first_name and b[i].last_name == a[i].last_name)
a.push(p) # as people p does not contain the same first/last names as a it can now be added to a
end
end
To check for other fields simply replace first_name / last_name with other variables.
answered Mar 7 at 14:20
Jack BranchJack Branch
373
add a comment |
If arrays a and b themselves do not contain people with matching first and last names then this would work:
b.each_with_index do |p, i|
if !(b[i].first_name == a[i].first_name and b[i].last_name == a[i].last_name)
a.push(p) # as people p does not contain the same first/last names as a it can now be added to a
end
end
To check for other fields simply replace first_name / last_name with other variables.
answered Mar 7 at 14:20
Jack BranchJack Branch
373
add a comment |
If arrays a and b themselves do not contain people with matching first and last names then this would work:
b.each_with_index do |p, i|
if !(b[i].first_name == a[i].first_name and b[i].last_name == a[i].last_name)
a.push(p) # as people p does not contain the same first/last names as a it can now be added to a
end
end
To check for other fields simply replace first_name / last_name with other variables.
answered Mar 7 at 14:20
Jack BranchJack Branch
373
If arrays a and b themselves do not contain people with matching first and last names then this would work:
b.each_with_index do |p, i|
if !(b[i].first_name == a[i].first_name and b[i].last_name == a[i].last_name)
a.push(p) # as people p does not contain the same first/last names as a it can now be added to a
end
end
To check for other fields simply replace first_name / last_name with other variables.
answered Mar 7 at 14:20
Jack BranchJack Branch
373
answered Mar 7 at 14:20
Jack BranchJack Branch
373
answered Mar 7 at 14:20
Jack BranchJack Branch
373
answered Mar 7 at 14:20
Jack BranchJack Branch
373
373
add a comment |
add a comment |
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2
In your condition you compare
p1's name top3's name. What aboutp2andp4? Could you give a more complete example with actual data and show the expected output?– Stefan
Mar 7 at 13:52
@Shaggon do you want the result to be an array of persons with no duplicate first_name last_name ?
– steenslag
Mar 7 at 17:29
Yes. I want a result with no duplicate first_name last_name and all the item should be belong to array a. The result should be like
[p1]because p2 have duplicate first_name last_name with p3 and p4 is inside array b, so only p1 is what I want– Shaggon
Mar 8 at 2:26
Thanks stenfan and steenslag. I found the result I want is
(b + a) - (b + a).uniq e– Shaggon
Mar 8 at 5:15