Dict comprehension with numpy array?

Question

I have 4 NumPy arrays: a, b, c, d

I am running: {i: i.shape for i in [a,b,c,d]}

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
    {i:i.shape for i in [a,b,c,d]}
  File "<stdin>", line 1, in <dictcomp>
    {i:i.shape for i in [a,b,c,d]}
TypeError: unhashable type: 'numpy.ndarray'

What am I doing wrong?

You're trying to make a numpy array as your dict key...you can't do that — David
– David, Commented Mar 28, 2018 at 18:02
@David a dict like you've produced in your answer that I have accepted! — Dino Alessi
– Dino Alessi, Commented Mar 28, 2018 at 18:39
Dictionary keys have to be immutable, unlike values which don't need to be. — gxpr
– gxpr, Commented Mar 28, 2018 at 18:51

David · Accepted Answer · 2018-03-28 18:03:50Z

4

Your code tries to make the numpy array as the key, rather than the variable name you are assigning the array to. Something like below would work

{i[1]: i[0].shape for i in [(a, "a"), (b, "b"), (c, "c") , (d, "d")]}

answered Mar 28, 2018 at 18:03

David

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1 Comment

Eric Over a year ago

Or: {name: locals()[name].shape for name in "abcd"}

Matt Ellis · Accepted Answer · 2018-03-28 18:03:34Z

4

You can't have a numpy array as a key. Try using something like:

{i:arr.shape for i,arr in enumerate([a,b,c,d])}

answered Mar 28, 2018 at 18:03

Matt Ellis

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Comments

sacuL · Accepted Answer · 2018-03-28 19:05:35Z

3

Another solution (just to give you options, as other people have already posted good ones):

{name: arr.shape for arr,name in zip([a,b,c,d],'abcd')}

edited Mar 28, 2018 at 19:05

answered Mar 28, 2018 at 18:11

sacuL

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Comments

Hultner · Accepted Answer · 2018-03-28 18:02:35Z

1

You’re trying to use a numpy.ndarray as the key for your shape. This won’t work as the object isn’t hashable and you probably want to use something else as your key.

answered Mar 28, 2018 at 18:02

Hultner

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1 Comment

Dino Alessi Over a year ago

But how can I tell the dict to take "a", "b" ... as keys?

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