CODEWITHSUNDEEP
rubyarrays
gweg
gweg

Reputation: 2882

How do I replace an array's element?

How can I substitue an element in an array?

a = [1,2,3,4,5]

I need to replace 5 with [11,22,33,44].flatten!

so that a now becomes

a = [1,2,3,4,11,22,33,44]

Upvotes: 27

Views: 50737

Answers (8)

Ryan McGeary
Ryan McGeary

Reputation: 239914

Not sure if you're looking to substitute a particular value or not, but this works:

a = [1, 2, 3, 4, 5]
b = [11, 22, 33, 44]
a.map! { |x| x == 5 ? b : x }.flatten!

This iterates over the values of a, and when it finds a value of 5, it replaces that value with array b, then flattens the arrays into one array.

Upvotes: 45

user456814
user456814

Reputation:

Here is another simple way to replace the value 5 in the array:

a[-1, 1] = [11, 22, 33, 44]

This uses the Array#[]= method. I'm not exactly sure why it works though.

Upvotes: 2

chrissrogers
chrissrogers

Reputation: 226

Array#delete will return the item or nil. You may use this to know whether or not to push your new values

a.push 11,22,33,44 if a.delete 5

Upvotes: 7

DigitalRoss
DigitalRoss

Reputation: 146043

Perhaps you mean:

a[4] = [11,22,33,44] # or a[-1] = ...
a.flatten!

A functional solution might be nicer, how about just:

a[0..-2] + [11, 22, 33, 44]

which yields...

=> [1, 2, 3, 4, 11, 22, 33, 44]

Upvotes: 19

Guillaume
Guillaume

Reputation: 705

The version of bta using a.index(5) is the fastest one:

a[a.index(5)] = b if a.index(5) # 5.133327 sec/10^6

At least 10% faster than Ryan McGeary's one:

a.map!{ |x| x == 5 ? b : x } # 5.647182 sec/10^6

However, note that a.index(5) only return the first index where 5 is found. So, given an array where 5 appears more than once, results will be different:

a = [1, 2, 3, 4, 5, 5]
b = [11,22,33,44]

a[a.index(5)] = b if a.index(5)
a.flatten # => [1, 2, 3, 4, 11, 22, 33, 44, 5]

a.map!{ |x| x == 5 ? b : x }
a.flatten # => [1, 2, 3, 4, 11, 22, 33, 44, 11, 22, 33, 44]

Upvotes: 14

bta
bta

Reputation: 45057

This variant will find the 5 no matter where in the array it is.

a = [1, 2, 3, 4, 5]
a[a.index(5)]=[11, 22, 33, 44] if a.index(5)
a.flatten!

Upvotes: 5

Benjamin Cox
Benjamin Cox

Reputation: 6120

You really don't have to flatten if you just concatenate. So trim the last element off the first array and concatenate them:

a = [ 1, 2, 3, 4, 5 ]           #=> [1, 2, 3, 4, 5]
t = [11, 22, 33, 44]            #=> [11, 22, 33, 44]
result = a[0..-2] + t           #=> [1, 2, 3, 4, 11, 22, 33, 44]

a[0..-2] is a slice operation that takes all but the last element of the array.

Hope it helps!

Upvotes: 5

Mike Seplowitz
Mike Seplowitz

Reputation: 10375

gweg, not sure what you're trying to do here, but are you looking for something like this?

a = [1, 2, 3, 4, 5]
a.delete_at(4)
a = a.concat([11,22,33,44])

There are a number of ways of doing this -- I don't think the code above is especially nice looking. It all depends on the significance of '5' in your original array.

Upvotes: 0

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