python 2.7 - How to sort defaultdict(list) such that all lists are sorted based on one key? -

to amazing programmers,

i know how can sort defaultdict(list) collection such sort order of 1 list (e.g. defaultdict(list)[list1]) applied remaining lists well. perhaps short description/example more useful. if there better way of doing ears.

example problem: had .csv file consisting of many columns (different data types) , headerline. using defaultdict(list) import .csv file using:

data = defaultdict(list) reader = csv.dictreader(open(filepath, 'r')) (k,v) in row.items():     data[k].append(v) 

now left defaultdict(list) named 'data' of structure:

data = [('vara', <list of n time.struct_time items>),         ('varb', <list of n other data type items>)',         ('varc', <list of n other data type items>)'] 

each list (vara, varb, varc) has exact same number of items. assuming vara not ordered, how order data entries based on vara.

i know

sortorder = [i[0] in sorted(enumerate(data['vara']), key=lambda x:x[1])] 

returns list correct order of indexes. there way apply order other lists.

note: read somewhere that

data[varb] = data[varb][sortorder]  data[varc] = data[varc][sortorder]  

would give me correct answer, given sortorder array (its not, above yields list). problem how can i, without using numpy, convert sortorder list array.

i think would:

• pair list vara numbers 1 n, so:
  • [ b, a, c ] becomes [ (1,b), (2,a), (3,c) ]
• sort new list value, so:
  • [ (2,a), (1,b), (3,c) ]
• unpair them , way numbers end up, ordering other lists:
  • [2,1,3] how make list order: take second item, first item, third item.
• for each row, take items in order new list

untested example code:

sort_holder = [(index, item) (index, item) in enumerate(data['vara'])] sort_holder = sorted(sort_holder, key = lambda pair: pair[1]) new_order = [pair[0] pair in sort_holder]  var, row in data.items():     new_list = []     index in new_order:         new_list.append(row[index])     data[var] = new_list