Python sorting 2D Array of tuples in python

Question

I'm new to python and have an array, where each element is an array of tuples. I need to sort the tuples in the inner array by the first element in the tuple. My array looks something like this:

[[(u'10:30', u'11:20'), (u'08:30', u'09:20')], [(u'14:30', u'15:50'), (u'10:00', u'11:20'), (u'10:30', u'11:20'), (u'13:00', u'14:20')], [(u'10:30', u'11:20'), (u'08:30', u'09:20')], [(u'14:30', u'15:50'), (u'10:00', u'11:20'), (u'13:00', u'14:20')], [(u'10:30', u'11:20'), (u'08:30', u'09:20')]]

I tried doing something like

for index, elm in array:
    array[index] = sorted(elm, key=lambda x: x[0])

But I get the ValueError: need more than 0 values to unpack. Not sure if my syntax is off or something

The expected result is

  [[ (u'08:30', u'09:20'), (u'10:30', u'11:20')], [ (u'10:00', u'11:20'), (u'10:30', u'11:20'), (u'13:00', u'14:20'), (u'14:30', u'15:50')], [(u'08:30', u'09:20'),(u'10:30', u'11:20') ], [ (u'10:00', u'11:20'), (u'13:00', u'14:20'), (u'14:30', u'15:50'),] [ (u'08:30', u'09:20'), (u'10:30', u'11:20')]]

falsetru · Accepted Answer · 2016-01-07 06:44:55Z

4

for statement is missing enumerate:

for index, elm in enumerate(array):

But, if you want to sort the list in-place, you don't need to use index. Use list.sort instead:

for elm in array:                                                              
    elm.sort(key=lambda x: x[0])

edited Jan 7, 2016 at 6:44

answered Jan 7, 2016 at 6:39

falsetru

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5 Comments

falsetru Over a year ago

@The6thSense, Strings like 10:30 cannot be converted to int.

The6thSense Over a year ago

Sorry did not see the actual data now I have a little doubt should not it be sorted according to time than normal strings ? It provides the correct output but is string comparsion enough here ?

falsetru Over a year ago

@The6thSense, Strings are sorted according to lexicographical order. '10:30' < '11:45'

PM 2Ring Over a year ago

@The6thSense: Normal (lexicographic) string sorting will sort these time strings correctly because they are zero-padded. If we had times like 2:30 then there would be problems because eg, '2:30' > '10:30' is True.

The6thSense Over a year ago

@PM2Ring hmm yeah that makes sense :).

TigerhawkT3 · Accepted Answer · 2016-01-07 06:37:45Z

1

When you do for index, elm in array:, you're iterating over pairs of objects. On the first loop iteration, index is (u'10:30', u'11:20') and elm is (u'08:30', u'09:20'), which isn't what you want (I get TypeError when I enter your posted code, not ValueError). You can get what you planned with the enumerate() function:

for index, elm in enumerate(array):
    array[index] = sorted(elm, key=lambda x: x[0])

answered Jan 7, 2016 at 6:37

TigerhawkT3

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2 Comments

TigerhawkT3 Over a year ago

@The6thSense - There's a leading zero, so that's not necessary. If there was no leading zero, a simple int() call wouldn't be enough to parse these time strings. Adding a leading zero with string formatting would be easier.

PM 2Ring Over a year ago

FWIW, using the sorted function when the result is going to replace the original list isn't very efficient. That's because sorted copies the source list to a new list and does an in-place sort of that new list. So in this situation it's much better to simply do an in-place sort directly, using the .sort() method.

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