🔒 How can I sort a dictionary from the values in the key?

[young jacinth]

Hey, so I have a dictionary
{'Century Park': [1442.28, 490376.02, 2], 'Mill Woods': [1450.75, 482257.75, 4], 'Strathcona': [1447.04, 549875.98, 2], 'Clareview': [1230.12, 452684.16, 6], 'Downtown': [1436.14, 586575.18, 6], 'McKernan': [1367.38, 560626.48, 4], 'Bonnie Doon': [1473.92, 505376.79, 6], 'Southgate': [1642.81, 605456.01, 2]}

How would I sort this numerically based on the second value in the list?

[grizzled badger]

@young jacinth

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[nova elm]

!d sorted

[grizzled badger]

sorted

sorted(iterable, /, *, key=None, reverse=False)```
Return a new sorted list from the items in *iterable*.

Has two optional arguments which must be specified as keyword arguments.

*key* specifies a function of one argument that is used to extract a comparison key from each element in *iterable* (for example, `key=str.lower`). The default value is `None` (compare the elements directly)...

[nova elm]

you'd need to deconstruct and reconstruct the dict

for example

!e ```py
d = {3: "c", 1: "a", 2: "b"}

e = {
k: d[k]
for k in sorted(d)
}

print(e)

[grizzled badger]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

{1: 'a', 2: 'b', 3: 'c'}

[nova elm]

you make key a function that transforms each element into whatever, and then it numerically sorts them all

[young jacinth]

gotcha ok thanks ill try that out

!closr

[grizzled badger]

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!close

[young jacinth]

!close

[grizzled badger]

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