按标签对嵌套的python字典进行排序

Question

I'm trying to find a smart way to sort the following data structure by std:

{'4555':{'std':5656, 'var': 5664}, '5667':{'std':5656, 'var': 5664}}

Ideally like to have a sorted dictionary (bad I know), or a list of sorted tuples, but I don't know how to get the 'std' part in my lambda expression. I'm trying the following, but how should I get at the 'stdev' bit in a smart manner? Which I want to go give a list of tuples (each tuple contains index such as [(4555, 5656), (5667, 5656)] .

sorted_list = sorted(sd_dict.items(), key=lambda x:x['std'])

Answer 1

Since sd_dict.items() returns a list of tuples, you no longer can access the elements as if it was a dictionary in the key function. Instead, the key function gets a two-element tuple with the first element being the key and the second element being the value. So to get the std value, you need to access it like this:

lambda x: x[1]['std']

But since in your example both values are identical, you don't actually change anything:

>>> list(sorted(sd_dict.items(), key=lambda x: x[1]['std']))
[('5667', {'var': 5664, 'std': 5656}), ('4555', {'var': 5664, 'std': 5656})]

And if you just want a pair of the outer dictionary key and the std value, then you should use a list comprehension first to transform the dictionary values:

>>> lst = [(key, value['std']) for key, value in sd_dict.items()]
>>> lst.sort(key=lambda x: x[1])
>>> lst
[('5667', 5656), ('4555', 5656)]

Or maybe you want to include an int conversion, and also sort by the key too:

>>> lst = [(int(key), value['std']) for key, value in sd_dict.items()]
>>> lst.sort(key=lambda x: (x[1], x[0]))
>>> lst
[(4555, 5656), (5667, 5656)]

Answer 2

Each element in your input list is a tuple of (k, dictionary) items, so you need to index into that to get to the std key in the dictionary value:

sorted(sd_dict.items(), key=lambda i: i[1]['std'])

If you wanted a tuple of just the key and the std value from the dictionary, you need to pick those out; it doesn't matter if you do so before or after sorting, just adjust your sort key accordingly

sorted([(int(k), v['std']) for k, v in sd_dict.items()], key=lambda i: i[1])

or

[(int(k), v['std']) for k, v in sorted(sd_dict.items()], key=lambda i: i[1]['std'])

However, extracting the std values just once instead of both for sorting and for extraction is going to be faster.