python dictionary sorted based on time

Question

I have a dictionary such as below.

d = {
    '0:0:7': '19734',
    '0:0:0': '4278',
    '0:0:21': '19959',
    '0:0:14': '9445',
    '0:0:28': '14205',
    '0:0:35': '3254'
}

Now I want to sort it by keys with time priority.

Answer 1

Dictionaries are not sorted, if you want to print it out or iterate through it in sorted order, you should convert it to a list first:

e.g.:

sorted_dict = sorted(d.items(), key=parseTime)
#or
for t in sorted(d, key=parseTime):
    pass

def parseTime(s):
    return tuple(int(x) for x in s.split(':'))

Note that this will mean you can not use the d['0:0:7'] syntax for sorted_dict though.

Passing a 'key' argument to sorted tells python how to compare the items in your list, standard string comparison will not work to sort by time.

Answer 2

Dictionaries in python have no guarantees on order. There is collections.OrderedDict, which retains insertion order, but if you want to work through the keys of a standard dictionary in order you can just do:

for k in sorted(d):

In your case, the problem is that your time strings won't sort correctly. You need to include the additional zeroes needed to make them do so, e.g. "00:00:07", or interpret them as actual time objects, which will sort correctly. This function may be useful:

def padded(s, c=":"):
    return c.join("{0:02d}".format(int(i)) for i in s.split(c))

You can use this as a key for sorted if you really want to retain the current format in your output:

for k in sorted(d, key=padded):

Answer 3

Have a look at the collections.OrderedDict module