How to convert a array of array to array of tuples

Question

I everyone,

I struggle to convert an array of array like this :

[
    ["foo","bar","foo","bar","foo"],
    ["foo","bar","foo","bar","foo"],
    ["foo","bar","foo","bar","foo"],
    ["foo","bar","foo","bar","foo"]
]

to an array of tuples (array, str) like this:

[
    (["foo","bar","foo","bar","foo"], "type1"), 
    (["foo","bar","foo","bar","foo"], "type1"), 
    (["foo","bar","foo","bar","foo"], "type1"), 
    (["foo","bar","foo","bar","foo"], "type1")
]

I did find a way to append the type to the array but it's not exactly what I want:

[
    ["foo","bar","foo","bar","foo", "type1"], 
    ["foo","bar","foo","bar","foo", "type1"], 
    ["foo","bar","foo","bar","foo", "type1"], 
    ["foo","bar","foo","bar","foo", "type1"]
]

Do you have something better ? Thanks :)

how are the types determined in association with the list of lists? what determined type1 type2 ect with the list — ListenSoftware Louise Ai Agent
– ListenSoftware Louise Ai Agent, Commented Jan 30, 2021 at 14:15

Martin Kleiven · Accepted Answer · 2021-01-29 10:07:00Z

2

Just initialize a tuple for every element in the list:

l = [
    ["foo","bar","foo","bar","foo"],
    ["foo","bar","foo","bar","foo"],
    ["foo","bar","foo","bar","foo"],
    ["foo","bar","foo","bar","foo"]]

new_l = [(i, "type1") for i in l]

print(new_l)

> [
    (['foo', 'bar', 'foo', 'bar', 'foo'], 'type1'),
    (['foo', 'bar', 'foo', 'bar', 'foo'], 'type1'),
    (['foo', 'bar', 'foo', 'bar', 'foo'], 'type1'),
    (['foo', 'bar', 'foo', 'bar', 'foo'], 'type1')
]

answered Jan 29, 2021 at 10:07

Martin Kleiven

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CypherX · Accepted Answer · 2021-01-29 11:47:02Z

1

Solution

Shortest solution: list(zip(vals, types)) 🔥🔥🔥

vals = [
    ["foo","bar","foo","bar","foo"], 
    ["foo","bar","foo","bar","foo"], 
    ["foo","bar","foo","bar","foo"], 
    ["foo","bar","foo","bar","foo"]
]

# If you must specify different types for each element
# uncomment the following line
# types = ['type1', 'type2', 'type3', 'type4']

# If all of them should have the same type
types = ['type1' for _ in range(len(vals))]

# Finally combine vals and types
list(zip(vals, types))

Output:

[(['foo', 'bar', 'foo', 'bar', 'foo'], 'type1'),
 (['foo', 'bar', 'foo', 'bar', 'foo'], 'type1'),
 (['foo', 'bar', 'foo', 'bar', 'foo'], 'type1'),
 (['foo', 'bar', 'foo', 'bar', 'foo'], 'type1')]

answered Jan 29, 2021 at 11:47

CypherX

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4 Comments

CypherX Over a year ago

@camile-saury Please try this and let me know if you have any questions.

Olvin Roght Over a year ago

1. You can create list with same elements much simpler: types = ['type1'] * len(vals). 2. You can use cycle() or repeat() which won't create long list in memory and will make your expression even shorter: list(zip(vals, cycle(('type1',)))) OR list(zip(vals, repeat('type1', len(vals)))) 3. Your solution is not shortest !!!

ListenSoftware Louise Ai Agent Over a year ago

type1 was the only type used. how did you determine to use type1

CypherX Over a year ago

@CamilleSAURY Please also consider voting-up, given you found the answer useful. Thank you.

azro · Accepted Answer · 2021-01-29 10:07:45Z

1

You may use a list comprehension, and for each build the tuple

vals = [["foo", "bar", "foo", "bar", "foo"], ["foo", "bar", "foo", "bar", "foo"],
        ["foo", "bar", "foo", "bar", "foo"], ["foo", "bar", "foo", "bar", "foo"]]

result = [(i, "type1") for i in vals]
print(result)

# [(["foo", "bar", "foo", "bar", "foo"], "type1"), (["foo", "bar", "foo", "bar", "foo"], "type1"),
   (["foo", "bar", "foo", "bar", "foo"], "type1"), (["foo", "bar", "foo", "bar", "foo"], "type1")]

This is equivalent, with a for-loop, to

result = []
for sublist in vals:
    result.append((sublist, "type1"))

answered Jan 29, 2021 at 10:07

azro

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