CODEWITHSUNDEEP
pythonpython-3.xordereddict
Alex Ivanov
Alex Ivanov

Reputation: 51

How can I update the value of key in OrderedDict in Python 3?

Here's a minimal reproducible example:

from collections import OrderedDict

d = OrderedDict([('foo', 123),
                 ('bar', 456)])

So I want to check if there's a foo key in d and if there's then I'd like to rewrite it as a single value of a list for a new hardcoded key:

print(d)

ordereddict([('bar', 456), ('newCoolHardcodedKey', [ordereddict([('foo', 123)])])])

Upvotes: 0

Views: 205

Answers (1)

joanis
joanis

Reputation: 12221

You can use a generating expression (like a list comprehension, but returns an iterator instead of storing the temporary list in memory) to do this:

d = OrderedDict(
    (
        ("newCoolHardcodedKey", OrderedDict([item])) if item[0] == "foo" else item
        for item in d.items()
    )
)

print(d)

OrderedDict([('newCoolHardcodedKey', OrderedDict([('foo', 123)])), ('bar', 456)])

The dict being ordered, the new element is where foo was.

If you need the new element to go to the end, it might be easiest to test if d["foo"] exists, and if so append the new ordered dict with its hard-coded key and delete the original entry for foo:

if "foo" in d:
    d["newCoolHardcodedKey"] = OrderedDict([("foo", d["foo"])])
    del d["foo"]

print(d)

OrderedDict([('bar', 456), ('newCoolHardcodedKey', OrderedDict([('foo', 123)]))])

Performance considerations

If d is large in your real application, the second solution is much better since it changes d in place instead of making a copy.

Upvotes: 1

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