ValueError：太多值无法解包

Question

I am trying to sort dictionaries in MongoDB. However, I get the value error "too many values to unpack" because I think it's implying that there are too many values in each dictionary (there are 16 values in each one). This is my code:

FortyMinute.find().sort(['Rank', 1])

Anyone know how to get around this?

EDIT: Full traceback

Traceback (most recent call last):
  File "main.py", line 33, in <module>
    main(sys.argv[1:])
  File "main.py", line 21, in main
    fm.readFortyMinute(args[0])
  File "/Users/Yih-Jen/Documents/Rowing Project/FortyMinute.py", line 71, in readFortyMinute
    writeFortyMinute(FortyMinData)
  File "/Users/Yih-Jen/Documents/Rowing Project/FortyMinute.py", line 104, in writeFortyMinute
    FortyMinute.find().sort(['Rank', 1])        
  File "/Users/Yih-Jen/anaconda/lib/python2.7/site-packages/pymongo/cursor.py", line 692, in sort
    self.__ordering = helpers._index_document(keys)
  File "/Users/Yih-Jen/anaconda/lib/python2.7/site-packages/pymongo/helpers.py", line 65, in       _index_document
    for (key, value) in index_list:
ValueError: too many values to unpack

Answer 1

You pass the arguments and values in unpacked as so:

FortyMinute.find().sort('Rank', 1)

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It is only when you're passing multiple sort parameters that you group arguments and values using lists, and then too you must surround all your parameters with a tuple as so:

 FortyMinute.find().sort([(Rank', 1), ('Date', 1)]) 

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Pro-tip: Even the Cursor.sort documentation linked below recommends using pymongo.DESCENDING and pymongo.ASCENDING instead of 1 and -1; in general, you should use descriptive variable names instead of magic constants in your code as so:

 FortyMinute.find().sort('Rank',pymongo.DESCENDING) 

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Finally, if you are so inclined, you can sort the list using Python's built-in as the another answerer mentioned; but even thought sorted accepts iterators and not just sequences it might be more inefficient and nonstandard:

 sorted(FortyMinute.find(), key=key_function) 

where you might define key_function to return the Rank column of a record.

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Link to the official documentation

Answer 2

If you want mong/pymongo to sort:

FortyMinute.find().sort('Rank', 1)

If you want to sort using multiple fields:

FortyMinute.find().sort([('Rank': 1,), ('other', -1,)])

You also have constants to make it more clear what you're doing:

FortyMinute.find().sort('Rank',pymongo.DESCENDING)

If you want to sort in python first you have to return the result and use a sorting method in python:

sorted(FortyMinute.find(), key=<some key...>)