Why do I get a TypeError when building a dict()?

Question

I have a list of dicts from which I need to extract, into a new list, some elements. It works fine for one element:

d = [
    {
        "a": 1,
        "b": {
            "x": 3
        },
        "blah": 4
    },
    {
        "a": 10,
        "b": {
            "x": 30
        },
        "blah": 40
    },
]

z = [k["b"] for k in d]
print(z)

This outputs [{'x': 3}, {'x': 30}] which is the extracted data.

I now realize that I also need to include in the extracted dict another element from the original one: a (and its value).

I thought I would just take the value if b and update() it with a pair:

y = [k["b"].update({"a": k["a"]}) for k in d]
print(y)

This will not work (returns [None, None]) as update() does not return the resulting dict. A previous question led me to build a dict() from the components:

x = [dict(a=k["a"], **k["b"]) for k in d]
print(x)

This crashes with

Traceback (most recent call last):
  File "C:/dev/config/scratches/scratch_21.py", line 24, in <module>
    x = [dict(a=k["a"], **k["b"]) for k in d]
  File "C:/dev/config/scratches/scratch_21.py", line 24, in <listcomp>
    x = [dict(a=k["a"], **k["b"]) for k in d]
TypeError: type object got multiple values for keyword argument 'a'

What does that mean? Specifically: what are the "multiple values" when there is only one for that key (an integer) ?

After you add the data ... what do you want each entry in the final list to look like? — kdopen, May 31 '18 at 16:15

score 7 · Accepted Answer · answered May 31 '18 at 16:05

You are passing in multiple a keys to the dict() call; one as a keyword argument, the other in the ** expanded mapping:

>>> dict(a=41, **{'a': 82})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: type object got multiple values for keyword argument 'a'

It doesn't really matter what object you are calling here, keyword arguments must be unique in any call.

You presumably are using the same dictionaries you called .update() on, so now all your k['b'] dictionaries have an a key as well. Re-build your d list and your code will work:

>>> [k["b"] for k in d]
[{'x': 3}, {'x': 30}]
>>> [dict(a=k["a"], **k["b"]) for k in d]
[{'a': 1, 'x': 3}, {'a': 10, 'x': 30}]

Ahhh, I understand. The successive trials have modified `k['b']` - it is true that I was testing all the above code one after the other. Indeed, commenting out the intermediates and leaving only the last one fixed the problem. Thanks! — WoJ, May 31 '18 at 19:33

Why do I get a TypeError when building a dict()?

1 Answers1