Ruby:— Accept array as input and put it into hash

-3

User must enter the array as input. Hash has to accept the input array elements as values.

Create a Ruby program for this by using loops. If array completed print this statement “All array elements are assigned to keys in the hash”

A = [1, 6, 4, 5]

H = {“k1” => 1

“k2” => 6

“k3” => 4

“k4” => 5}

asked Nov 21 '18 at 10:37

Sagar

1):-User must enter the array as input.

– Sagar
Nov 21 '18 at 12:36

add a comment |

-3

User must enter the array as input. Hash has to accept the input array elements as values.

Create a Ruby program for this by using loops. If array completed print this statement “All array elements are assigned to keys in the hash”

A = [1, 6, 4, 5]

H = {“k1” => 1

“k2” => 6

“k3” => 4

“k4” => 5}

asked Nov 21 '18 at 10:37

Sagar

1):-User must enter the array as input.

– Sagar
Nov 21 '18 at 12:36

add a comment |

-3

User must enter the array as input. Hash has to accept the input array elements as values.

Create a Ruby program for this by using loops. If array completed print this statement “All array elements are assigned to keys in the hash”

A = [1, 6, 4, 5]

H = {“k1” => 1

“k2” => 6

“k3” => 4

“k4” => 5}

asked Nov 21 '18 at 10:37

Sagar

User must enter the array as input. Hash has to accept the input array elements as values.

Create a Ruby program for this by using loops. If array completed print this statement “All array elements are assigned to keys in the hash”

A = [1, 6, 4, 5]

H = {“k1” => 1

“k2” => 6

“k3” => 4

“k4” => 5}

ruby-on-rails ruby

asked Nov 21 '18 at 10:37

Sagar

asked Nov 21 '18 at 10:37

Sagar

asked Nov 21 '18 at 10:37

Sagar

asked Nov 21 '18 at 10:37

Sagar

asked Nov 21 '18 at 10:37

Sagar

1):-User must enter the array as input.

– Sagar
Nov 21 '18 at 12:36

add a comment |

1):-User must enter the array as input.

– Sagar
Nov 21 '18 at 12:36

1):-User must enter the array as input.

– Sagar
Nov 21 '18 at 12:36

add a comment |

2 Answers
2

active

oldest

votes

Another solution can be,

a.each_with_index.inject({}) { |m,(a,i)| m["k#{i+1}"] = a; m }

Update: answering to your question

puts 'Enter number of hash elements'

n = gets.to_i

n.times |i|

    hash["k#{i+1}"] = gets.to_i

end

puts 'All array elements are assigned to keys in the hash'

edited Nov 21 '18 at 12:28

answered Nov 21 '18 at 10:59

ray

2,6991725

1

You can use each_with_object and get rid of returning m a.each_with_object({}).with_index { |(a,m), i| m["k#{i+1}"] = a }.

– Sebastian Palma
Nov 21 '18 at 11:09

@SebastianPalma Yeah, it is better than I gave, thanks :)

– ray
Nov 21 '18 at 11:15

case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values.

– Sagar
Nov 21 '18 at 12:06

Using for loop only

– Sagar
Nov 21 '18 at 12:08

@Sagar check update

– ray
Nov 21 '18 at 12:28

add a comment |

A = [1, 6, 4, 5]

Hash[[*'k1'.."k#{A.length}"].zip(A)]

answered Nov 21 '18 at 10:51

Dyaniyal Nadar

37728

Case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values

– Sagar
Nov 21 '18 at 12:07

Using for loop only

– Sagar
Nov 21 '18 at 12:07

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Another solution can be,

a.each_with_index.inject({}) { |m,(a,i)| m["k#{i+1}"] = a; m }

Update: answering to your question

puts 'Enter number of hash elements'

n = gets.to_i

n.times |i|

    hash["k#{i+1}"] = gets.to_i

end

puts 'All array elements are assigned to keys in the hash'

edited Nov 21 '18 at 12:28

answered Nov 21 '18 at 10:59

ray

2,6991725

1

You can use each_with_object and get rid of returning m a.each_with_object({}).with_index { |(a,m), i| m["k#{i+1}"] = a }.

– Sebastian Palma
Nov 21 '18 at 11:09

@SebastianPalma Yeah, it is better than I gave, thanks :)

– ray
Nov 21 '18 at 11:15

case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values.

– Sagar
Nov 21 '18 at 12:06

Using for loop only

– Sagar
Nov 21 '18 at 12:08

@Sagar check update

– ray
Nov 21 '18 at 12:28

add a comment |

Another solution can be,

a.each_with_index.inject({}) { |m,(a,i)| m["k#{i+1}"] = a; m }

Update: answering to your question

puts 'Enter number of hash elements'

n = gets.to_i

n.times |i|

    hash["k#{i+1}"] = gets.to_i

end

puts 'All array elements are assigned to keys in the hash'

edited Nov 21 '18 at 12:28

answered Nov 21 '18 at 10:59

ray

2,6991725

1

You can use each_with_object and get rid of returning m a.each_with_object({}).with_index { |(a,m), i| m["k#{i+1}"] = a }.

– Sebastian Palma
Nov 21 '18 at 11:09

@SebastianPalma Yeah, it is better than I gave, thanks :)

– ray
Nov 21 '18 at 11:15

case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values.

– Sagar
Nov 21 '18 at 12:06

Using for loop only

– Sagar
Nov 21 '18 at 12:08

@Sagar check update

– ray
Nov 21 '18 at 12:28

add a comment |

Another solution can be,

a.each_with_index.inject({}) { |m,(a,i)| m["k#{i+1}"] = a; m }

Update: answering to your question

puts 'Enter number of hash elements'

n = gets.to_i

n.times |i|

    hash["k#{i+1}"] = gets.to_i

end

puts 'All array elements are assigned to keys in the hash'

edited Nov 21 '18 at 12:28

answered Nov 21 '18 at 10:59

ray

2,6991725

Another solution can be,

a.each_with_index.inject({}) { |m,(a,i)| m["k#{i+1}"] = a; m }

Update: answering to your question

puts 'Enter number of hash elements'

n = gets.to_i

n.times |i|

    hash["k#{i+1}"] = gets.to_i

end

puts 'All array elements are assigned to keys in the hash'

edited Nov 21 '18 at 12:28

answered Nov 21 '18 at 10:59

ray

2,6991725

edited Nov 21 '18 at 12:28

answered Nov 21 '18 at 10:59

ray

2,6991725

answered Nov 21 '18 at 10:59

ray

2,6991725

answered Nov 21 '18 at 10:59

ray

2,6991725

1

You can use each_with_object and get rid of returning m a.each_with_object({}).with_index { |(a,m), i| m["k#{i+1}"] = a }.

– Sebastian Palma
Nov 21 '18 at 11:09

@SebastianPalma Yeah, it is better than I gave, thanks :)

– ray
Nov 21 '18 at 11:15

case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values.

– Sagar
Nov 21 '18 at 12:06

Using for loop only

– Sagar
Nov 21 '18 at 12:08

@Sagar check update

– ray
Nov 21 '18 at 12:28

add a comment |

1

You can use each_with_object and get rid of returning m a.each_with_object({}).with_index { |(a,m), i| m["k#{i+1}"] = a }.

– Sebastian Palma
Nov 21 '18 at 11:09

@SebastianPalma Yeah, it is better than I gave, thanks :)

– ray
Nov 21 '18 at 11:15

case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values.

– Sagar
Nov 21 '18 at 12:06

Using for loop only

– Sagar
Nov 21 '18 at 12:08

@Sagar check update

– ray
Nov 21 '18 at 12:28

You can use each_with_object and get rid of returning m a.each_with_object({}).with_index { |(a,m), i| m["k#{i+1}"] = a }.

– Sebastian Palma
Nov 21 '18 at 11:09

@SebastianPalma Yeah, it is better than I gave, thanks :)

– ray
Nov 21 '18 at 11:15

case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values.

– Sagar
Nov 21 '18 at 12:06

Using for loop only

– Sagar
Nov 21 '18 at 12:08

@Sagar check update

– ray
Nov 21 '18 at 12:28

add a comment |

A = [1, 6, 4, 5]

Hash[[*'k1'.."k#{A.length}"].zip(A)]

answered Nov 21 '18 at 10:51

Dyaniyal Nadar

37728

Case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values

– Sagar
Nov 21 '18 at 12:07

Using for loop only

– Sagar
Nov 21 '18 at 12:07

add a comment |

A = [1, 6, 4, 5]

Hash[[*'k1'.."k#{A.length}"].zip(A)]

answered Nov 21 '18 at 10:51

Dyaniyal Nadar

37728

Case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values

– Sagar
Nov 21 '18 at 12:07

Using for loop only

– Sagar
Nov 21 '18 at 12:07

add a comment |

A = [1, 6, 4, 5]

Hash[[*'k1'.."k#{A.length}"].zip(A)]

answered Nov 21 '18 at 10:51

Dyaniyal Nadar

37728

A = [1, 6, 4, 5]

Hash[[*'k1'.."k#{A.length}"].zip(A)]

answered Nov 21 '18 at 10:51

Dyaniyal Nadar

37728

answered Nov 21 '18 at 10:51

Dyaniyal Nadar

37728

answered Nov 21 '18 at 10:51

Dyaniyal Nadar

37728

answered Nov 21 '18 at 10:51

Dyaniyal Nadar

37728

Case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values

– Sagar
Nov 21 '18 at 12:07

Using for loop only

– Sagar
Nov 21 '18 at 12:07

add a comment |

Case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values

– Sagar
Nov 21 '18 at 12:07

Using for loop only

– Sagar
Nov 21 '18 at 12:07

Case is i am getting array from user input, Then, create an empty hash = H=Hash.new{}. After that i need to iterate through that array(user input) and put values into hash values

– Sagar
Nov 21 '18 at 12:07

Using for loop only

– Sagar
Nov 21 '18 at 12:07

add a comment |

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