Insert values of list of lists in a dictionary declared with keys

Question

I have this list of lists:

x = [['x1', 'x2', 'x3', 'x4', 'x5', 'x6', 'x7', 'x8', 'x9', 'x10', 'x11'], 
    ['x1', 'x2', 'x3', 'x4', 'x5', 'x6', 'x7', 'x8', 'x9', 'x10', 'x11'], 
    ['x1', 'x2', 'x3', 'x4', 'x5', 'x6', 'x7', 'x8', 'x9', 'x10', 'x11']]

And I have a declared dictionary like:

d = {"x": None, "y": None, "z": None, "t": None, 
"a": None, "s": None, "m": None, "n": None, 
"u": None, "v": None, "b": None}

What I want to get is a list or dictionry such as:

result = [{"x": x1,
"y": x2,
"z": x3,
"t": x4,
"a": x5,
"s": x6,
"m": x7,
"n": x8,
"u": x9,
"v": x10,
"b": x11}, {"x": x1,
"y": x2,
"z": x3,
"t": x4,
"a": x5,
"s": x6,
"m": x7,
"n": x8,
"u": x9,
"v": x10,
"b": x11}...]

And so on. One dictionary inside the list per each element inside x (list of lists).

Solutions in python3 are correct,however in python2 would be even better.

If you are in control of creating the dictionary d, note that it's of little use to initialize a dictionary like that as for the preferred solution (see Andrej Kesely's answer) you only need the keys. Long story short, just make d a list of keys d = ['x', 'y', ...]. — fsimonjetz
– fsimonjetz, Commented Aug 30, 2022 at 13:25

Andrej Kesely · Accepted Answer · 2022-08-30 20:40:26Z

4

Try:

result = [dict(zip(d, subl)) for subl in x]
print(result)

Prints:

[
    {
        "x": "x1",
        "y": "x2",
        "z": "x3",
        "t": "x4",
        "a": "x5",
        "s": "x6",
        "m": "x7",
        "n": "x8",
        "u": "x9",
        "v": "x10",
        "b": "x11",
    },
...

The dict(zip(d, subl)) will iterate over keys of dictionary d and values of sublists of x at the same time and creates new dictionary (with keys from d and values from sublist). This works for Python 3.7+ as the dictionary keeps insertion order.

EDIT: I'd recommend to change d from dict to list:

from collections import OrderedDict

d = ["x", "y", "z", "t", "a", "s", "m", "n", "u", "v", "b"]
result = [OrderedDict(zip(d, subl)) for subl in x]
print(result)

OR:

Use collections.OrderedDict:

from collections import OrderedDict

d = OrderedDict(
    [
        ("x", None),
        ("y", None),
        ("z", None),
        ("t", None),
        ("a", None),
        ("s", None),
        ("m", None),
        ("n", None),
        ("u", None),
        ("v", None),
        ("b", None),
    ]
)

result = [OrderedDict(zip(d, subl)) for subl in x]
print(result)

edited Aug 30, 2022 at 20:40

answered Aug 30, 2022 at 13:22

Andrej Kesely

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5 Comments

Riccardo Bucco Over a year ago

A comment on python version is quite important in this case ;) I think you should add it

Jordi Lazo Over a year ago

@RiccardoBucco totally agree. In this case if we apply this solution with python 2.7 it will work as well but it will return the dictionary unsortered, the keys of the variable d will change the order of appearance.

Andrej Kesely Over a year ago

@JordiLazo For Python2.7 use OrderedDict for d or better, make d a list with the keys as items of this list.

Jordi Lazo Over a year ago

@AndrejKesely can you share both solutions please? Thanks.

Andrej Kesely Over a year ago

@JordiLazo See my edit.

Riccardo Bucco · Accepted Answer · 2022-08-30 13:24:42Z

1

From Python 3.7, dictionary order is guaranteed to be insertion order. So my answer does only make sense if you're using Python >=3.7.

Here is how you can do it:

result = [dict(zip(d, lst)) for lst in x]

answered Aug 30, 2022 at 13:24

Riccardo Bucco

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