The following codes were tested under Julia v1.6.1 and DataFrames v1.2.0.

I saw a question of the same title for Python.

Let’s say we have `a = [1, 2, 3, 4, 5]`

and `b = ["a", "b", "c", "d", "e"]`

. What we want is `c = [(1, "a"), (2, "b"), (3, "c"), (4, "d"), (5, "e")]`

.

The first solution is hamza sadiqi’s own answer to his own question:

```
julia> using DataFrames
julia> a = 1:5
1:5
julia> b = 'a':'e'
'a':1:'e'
julia> c = [(a[i], b[i]) for i in range(1, length = length(a))]
5-element Vector{Tuple{Int64, Char}}:
(1, 'a')
(2, 'b')
(3, 'c')
(4, 'd')
(5, 'e')
```

from Cameron Bieganek’s answer
on Stack Overflow, I realized that I can use the `zip`

function
.

`zip`

will return an iterator. We’ll need the `collect`

function
which returns all items in the iterator created by `zip`

.

```
julia> z = zip(a,b)
zip(1:5, 'a':1:'e')
julia> typeof(z)
Base.Iterators.Zip{Tuple{UnitRange{Int64}, StepRange{Char, Int64}}}
julia> collect(z)
5-element Vector{Tuple{Int64, Char}}:
(1, 'a')
(2, 'b')
(3, 'c')
(4, 'd')
(5, 'e')
```

By the way, the equivalent way in Python is `list(zip())`

:

```
a = [1, 2, 3, 4, 5]
b = ["a", "b", "c", "d", "e"]
list(zip(a, b))
# returns a list: [(1, 'a'), (2, 'b'), (3, 'c'), (4, 'd'), (5, 'e')]
```

Last modified on 2021-07-16