Sort a dictionary containing child dictionary, by using value of a key in the child dictionary

Question

Suppose that I have a dictionary containing dictionary like this example:

a_dict = {
  'abcd': {'name': 'John', 'height': 180},
  'asdf': {'name': 'Lex', 'height': 160},
  'csxf': {'name': 'Amber', 'height': 193},
}

How do I sort the dictionary using value of the key in child dictionary, i.e: value of the 'height' key?

Does this answer your question? [How do I sort a list of dictionaries by a value of the dictionary?](https://stackoverflow.com/questions/72899/how-do-i-sort-a-list-of-dictionaries-by-a-value-of-the-dictionary) As well as [How do I sort a dictionary by value?](https://stackoverflow.com/q/613183/4518341) — wjandrea, Jul 09 '21 at 01:59
No, I've been thru that and, those were not nested dictionary @wjandrea — ewin.str, Jul 09 '21 at 02:20

score 1 · Accepted Answer · edited Jul 09 '21 at 02:47

1

You can try something like this

dict(sorted(a_dict.items(), key=lambda item: item[1]["height"]))

edited Jul 09 '21 at 02:47

wjandrea

answered Jul 09 '21 at 01:59

Djoby

1

`TypeError: tuple indices must be integers or slices, not str` You just forgot `item[1]` – wjandrea Jul 09 '21 at 02:01
@wjandrea still produce error: '<' not supported between instances of 'dict' and 'dict' – ewin.str Jul 09 '21 at 02:22
1

@ewin Oops, that was unclear. I meant `item[1]["height"]` – wjandrea Jul 09 '21 at 02:23
@wjandrea Thanks, it works! Now what was the indice [1] in item[1] refer to actually? – ewin.str Jul 09 '21 at 02:27
@ewin Think about it, what does `dict.items()` give you? What's the second element in each item? – wjandrea Jul 09 '21 at 02:52
@wjandrea Got it, thx. Im an idiot. – ewin.str Jul 09 '21 at 07:52

1 Answers1