Example
let a = [a, b, c]
let b = [x, y, z]
I need this response
let c = [[a, b, c],[x, y, z]]
There are two approaches as I see it, either I concatenate both arrays (don't know if it's possible), or to append it directly in a multidimensional array.
If you will try to concatenate it:
let a = [1, 2, 3]
let b = [4, 5]
let c = a + b // will receive [1, 2, 3, 4, 5]
let d = [a, b] // will receive multidimensional array [[1, 2, 3], [4, 5]]
var e: [[Int]] = []
e.append(a)
e.append(b)
print(e) // should be equal to d
Related
I have an array in three dimensions (a, b, c) and I need to modify the positions c indexed by an array in two dimensions (a, b).
I wrote a code that works as expected, but does anyone know if there is a way to solve this problem without the for loop?
import numpy as np
arr = np.random.randn(100, 5, 2)
mod = np.random.randn(100, 5)
ids = np.random.randint(0, 2, size=[100,5])
for i in range(100):
for j in range(5):
arr[i,j,ids[i,j]] = mod[i,j]
You can refer to and set each slice of the array directly. I think this code shows the behaviour you are asking about:
import numpy as np
arr = np.array([[[1, 2, 3], [4, 5, 6]], [[7, 8, 9], [10, 11, 12]]])
print(arr[:,:,2])
# Output:
#[[ 3 6]
# [ 9 12]]
new2DSlice = np.array([[23, 26], [29, 32]])
arr[:,:,2] = new2DSlice
print(arr[:,:,2])
# Outupt:
#[[23 26]
# [29 32]]
arr[:,:,2] refers to the third slice of the array and, in this example, sets it directly.
You can read about NumPy's array indexing and array slicing on W3Schools.
I got with this:
import numpy as np
arr = np.random.randn(100, 5, 2)
mod = np.random.randn(100, 5)
ids = np.random.randint(0, 2, size=[100,5])
x = np.arange(100)
y = np.arange(5)
arr[x[:,None],y,ids] = mod
I have two numpy arrays:
a = np.array([1, 2, 3])
b = np.array([4, 5, 6])
and I want to concatenate them into two columns like,
1 4
2 5
3 6
is there any way to do this without transposing or reshaping the arrays?
You can try:
a = np.array([1, 2, 3])
b = np.array([4, 5, 6])
c = np.concatenate((a[np.newaxis, :], b[np.newaxis, :]), axis = 0).T
And you get :
c = array([[1, 4],
[2, 5],
[3, 6]])
Best,
Is there a concise way in Swift of creating an array by applying a binary operation on the elements of two other arrays?
For example:
let a = [1, 2, 3]
let b = [4, 5, 6]
let c = (0..<3).map{a[$0]+b[$0]} // c = [5, 7, 9]
If you use zip to combine the elements, you can refer to + with just +:
let a = [1, 2, 3]
let b = [4, 5, 6]
let c = zip(a, b).map(+) // [5, 7, 9]
Update:
You can use indices like this:
for index in a.indices{
sum.append(a[index] + b[index])
}
print(sum)// [5, 7, 9]
(Thanks to Alexander's comment this is better, because we don't have to deal with the element itself and we just deal with the index)
Old answer:
you can enumerate to get the index:
var sum = [Int]()
for (index, _) in a.enumerated(){
sum.append(a[index] + b[index])
}
print(sum)// [5, 7, 9]
Is there a concise way in Swift of creating an array by applying a binary operation on the elements of two other arrays?
For example:
let a = [1, 2, 3]
let b = [4, 5, 6]
let c = (0..<3).map{a[$0]+b[$0]} // c = [5, 7, 9]
If you use zip to combine the elements, you can refer to + with just +:
let a = [1, 2, 3]
let b = [4, 5, 6]
let c = zip(a, b).map(+) // [5, 7, 9]
Update:
You can use indices like this:
for index in a.indices{
sum.append(a[index] + b[index])
}
print(sum)// [5, 7, 9]
(Thanks to Alexander's comment this is better, because we don't have to deal with the element itself and we just deal with the index)
Old answer:
you can enumerate to get the index:
var sum = [Int]()
for (index, _) in a.enumerated(){
sum.append(a[index] + b[index])
}
print(sum)// [5, 7, 9]
This question already has answers here:
Sum of arrays of different size [closed]
(4 answers)
Closed 6 years ago.
I want to add values of two different length array.
a =[1,2,3]
b= [1,2]
c = [1,2,3,4]
and so on..
I want result to be like [3,6,6,4]. How to do this in ruby on rails.
In order to make it dynamic, I would create arrays of array with your a, b, c =>
a = [1, 2, 3]
b = [1, 2]
c = [1, 2, 3, 4]
arrays = [a, b, c]
Then I would retrieve the max size :
max_size = arrays.map(&:size).max #=> 4
Then the following line would give you your answer :
max_size.times.map{ |i| arrays.reduce(0){|s, a| s + a.fetch(i, 0)}} #=> [3, 6, 6, 4]
You can build a new array that consists of all those array, and then can write the following code to get the array that has combined entries of each array:
a = [1,2,3]
b = [1,2,3]
c = [1,2,3,4]
Before you apply the rest of the code, you need to make sure that each array has the same length. For that, you can append 0 in all the arrays if need be, to ensure that each array has the same length as the rest of the arrays have.
a = [1,2,3,0]
b = [1,2,3,0]
c = [1,2,3,4]
combined_array = [a,b,c]
result = combined_array.transpose.map { |a| a.reduce :+ }
Extending the answer from #Arslan Ali
I added a way to make all the arrays the same size, so that his method of summing can be applied:
a = [1,2,3]
b = [1,2,3]
c = [1,2,3,4]
arrays = [a, b, c]
size = [a, b, c].map{|a| a.size}.max # Compute maximum size
combined_array = [a,b,c].map{|a| a.fill(a.size...size){0}} # Fill arrays to maximum size
result = combined_array.transpose.map { |a| a.reduce :+ } # Sum everything
Here's one way:
a = [1, 2, 3]
b = [1, 2, 3]
c = [1, 2, 3, 4]
[a,b,c].inject([]) do |totals, add|
add.each_with_index do |n, i|
totals[i] = (totals[i] || 0) + n
end
totals
end