Python-按字典中元组中的某个元素对列表进行排序

Question

data = [{'info': ('orange', 400000, 'apple'), 'photo': None}, {'info': ('grape', 485000, 'watermelon'), 'photo': None}]

I want to sort data by the 2nd element (400000, 485000) in the tuple in the dictionary. How do I do this?

I followed another answer and my closest attempt was data.sort(key=lambda tup: tup[1]) , but that produces this error:

KeyError: 1

Answer 1

Use the inplace list.sort method with a key that indexes info first , and then the data in the tuple:

data.sort(key=lambda x: x['info'][1])

Or, the not in-place sorted function:

data = sorted(data, key=lambda x: x['info'][1])

Between the two, list.sort is faster, because it sorts the data in-place, whereas sorted has to create and return a copy of the data in data .

A couple of other things you should think about;

1. What if your dictionary does not have info as a key?
2. What if the second element in the info tuple isn't even numeric?

Do you want sort to error out when it hits invalid data? Or would you rather introduce some means of handling these on a case-by-case basis? The key would need to change accordingly.

Answer 2

You can sort via key :

data = [{'info': ('orange', 400000, 'apple'), 'photo': None}, {'info': ('grape', 485000, 'watermelon'), 'photo': None}]
new_data = sorted(data, key=lambda x:x.get('info', [0, 0])[1])

Answer 3

For each dict item in the list you want to sort, you want to take the item's value keyed by 'info', which is a list, and sort on the second item (addressed as [1], counting from zero.

So: data.sort(key=lambda item: item['info'][1])