maintaining list order

[michalmonday]

You could make use of "max_but_less_than" max_lt function to make the code shorter

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from bisect import bisect_left   def max_lt(seq, val):     # http://code.activestate.com/recipes/415233-getting-minmax-in-a-sequence-greaterless-than-some/#c8     return seq[bisect_left(seq, val) - 1]   thres = [0,3000,6000] nested_list = [[3], [4, 400], [3, 4222]]   substituted = [[max_lt(thres, val) for val in seq] for seq in nested_list]   print(substituted)

Output:
[[0], [0, 0], [0, 3000]]

[3Pinter]

Interesting! Thanks for that suggestion, going to have look at it!

[perfringo]

Now my clumsy brain has processed it and as a result I refactored your original code a little bit:

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comparelist = [0, 3000, 6000] elements = [[3], [4, 400], [3,4222]] result = []   for element in elements:     result.append([])                            # add sublist for every list in elements         for num in element:                                for value in reversed(comparelist):                  if value < num:                     result[-1].append(value)    # append first match and break                     break

I assume that 'comparelist' is always sorted and starting from end is required to return closest match (in last case 3000 instead of 0)

However, there is some ambiguity: what should be returned when condition fully or partially not met? (for example comparelist = [4500, 5000, 6000])

[3Pinter]

@perfringo,

Well pointed out the ambiguity. Eventhough chances that will occur is small, why not take the effort to handle it the way I would want it to.

So the possibilties:
1. Negative values in element lists.
Right now they aren't included in the resultlist.
It should return the first item in the comparelist. So -1000 (element list) should return first in compare list (nearest)

2. Comparelist has values larger than the largest value in elementlist.
It should return the smallest (so nearest) in the comparelist So in your suggestion that would be 4500.

This is what I think:

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#comparelist = [4500, 5000, 6000] comparelist = [0, 3000, 6000] elements = [[-1,3], [4, 400], [3,4222]] result = []    for element in elements:     result.append([])     for num in element:         if all(value > num for value in comparelist):             result[-1].append(comparelist[0])         else:             for value in reversed(comparelist):              if num > value:                 result[-1].append(value)                 break   print(result)