Sorting python dictionary based on keys which is alphanumeric values-CodePudding

I am new to python. In Python, I want to sort list of dictionary items first based on 'year' key and later with key which is alphanumeric ie P1, P2, P3.

Code snippet is as follows:

[
    {"P9": 0, "year": 2023},
    {"P13": 0, "year": 2023},
    {"P10": 0, "year": 2023},
    {"P11": 121200, "year": 2022},
    {"P12": 0, "year": 2023},
]

Which first sorted by year in acending order becomes:

[
    {"P11": 121200, "year": 2022},
    {"P9": 0, "year": 2023},
    {"P13": 0, "year": 2023},
    {"P10": 0, "year": 2023},
    {"P12": 0, "year": 2023},
]

Again later sorted by period key ie, P1,P2,P3 etc becomes:

[
    {"P11": 121200, "year": 2022},
    {"P9": 0, "year": 2023},
    {"P10": 0, "year": 2023},
    {"P12": 0, "year": 2023},
    {"P13": 0, "year": 2023},
]

Above is the expected output, Now I don't know how do I sort it like this

CodePudding user response：

The sort function do all you need

You can pass a key function (a callable) for identifying the correct order. In your case:

unordered = [{'P9': 0, 'year': 2023}, {'P13': 0, 'year': 2023}, {'P10': 0, 'year': 2023}, {'P11': 121200, 'year': 2022}, {'P12': 0, 'year': 2023}]

#by year

sorted(unordered, key=lambda x: x['year'])
# out
[{'P11': 121200, 'year': 2022},
 {'P9': 0, 'year': 2023},
 {'P13': 0, 'year': 2023},
 {'P10': 0, 'year': 2023},
 {'P12': 0, 'year': 2023}]

For the period, the solution is a little bit tricky, you must identify how to select the P** key. We don't know how these dictionaries are built but doing some assumptions maybe something like:

sorted(unordered, key=lambda x: [lbl for lbl in x if lbl.startswith('P')][0] )

but here there are lacks of information to be more precise

CodePudding user response：

You can use sorted function, and specify a key function with it. But as you want to use two separate data points as keys, you can use tuple to build a key.

Because when tuples are compared, it first compares the very first entity in it, and if the first one matches, then it goes to check the second entity.

Example:

>>> (1,3) > (1, 4)
False

>>> (1, 4) < (2,2)
True

>>> (1, 4, 1) < (2, 1)
True

So, here's how you can build your key function:

>>> unordered = [{'P9': 0, 'year': 2023}, {'P13': 0, 'year': 2023}, {'P10': 0, 'year': 2023}, {'P11': 121200, 'year': 2022}, {'P12': 0, 'year': 2023}]
>>> ordered = sorted(
    unordered,
    key=lambda d: (d['year'], next(int(k[1:]) for k in d.keys() if k.startswith('P')))
)

>>> ordered
[{'P11': 121200, 'year': 2022}, {'P9': 0, 'year': 2023}, {'P10': 0, 'year': 2023}, {'P12': 0, 'year': 2023}, {'P13': 0, 'year': 2023}]

Here, I'm just assuming that you'll always have a key in the dictionary that starts with P, and the remaining part of it is an integer.

CodePudding user response：

For sorting first base years then sorting base first key of each dict you can catch number and convert to int then sorting like below:

>>> sorted(lst_dct, key=lambda x: (x['year'], int(list(x.keys())[0][1:])))
[{'P11': 121200, 'year': 2022},
 {'P9': 0, 'year': 2023},
 {'P10': 0, 'year': 2023},
 {'P12': 0, 'year': 2023},
 {'P13': 0, 'year': 2023}]