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Raphael Collet authored
The faulty code modifies a dictionary in-place to apply a formatting
function to each value.  In this case, the formatting is `browse`, and
the issue occurs when `ids` contain duplicates:

    for id in ids:
        result[id] = format(result.get(id, default))

Fix it by returning a new dictionary based on `result`.
bd4565e2
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