Sorting a list of tuples based on the first items

Question

How to sort a list of tuples based on the first value i.e, in a dictionary we can use sorted(a.keys()).

How to do it for a list of tuples?

If these are the tuple values

t = [('2010-09-11', 'somedata', somedata),
     ('2010-06-11', 'somedata', somedata),
     ('2010-09-12', 'somedata', somedata)]

tuples should be sorted based on dates in the first field.

Answer 1

Usually, just sorted(t) works, as tuples are sorted by lexicographical order. If you really want to ignore everything after the first item (instead of sorting tuples with the same first element by the following elements), you can supply a key that picks out the first element. The simplest way would be operator.itemgetter:

import operator
...
for item in sorted(t, key=operator.itemgetter(0)):
    ...

Of course if you want to sort the list in-place, you can use t.sort(key=operator.itemgetter(0)) instead.

Answer 2

Or you can use something like this to be sure that list of tuples sorted by dates:

from datetime import datetime
initData = [('2010-09-11','somedata',1), ('2010-06-11','somedata',2), ('2010-09-12','somedata',3)]
outData = sorted(initData , key=lambda x: datetime.strptime(x[0],"%Y-%m-%d"))

Answer 3

If '2010-09-11' is year-month-day , you do:

somedata = 'jyhghg'
t = [('2010-09-11','somedata',somedata),
     ('2010-06-11','somedata',somedata),
     ('2010-09-12','somedata',somedata),
     ('2010-08-12','somedata',somedata)]

from operator import itemgetter
t.sort(key = itemgetter(0))
print t

result

[('2010-06-11', 'somedata', 'jyhghg'),
 ('2010-08-12', 'somedata', 'jyhghg'),
 ('2010-09-11', 'somedata', 'jyhghg'),
 ('2010-09-12', 'somedata', 'jyhghg')]

.

If '2010-09-11' is year-day-month, you do:

from time import strptime,strftime

somedata = 'jyhghg'
t = [('2010-09-11','somedata',somedata),
     ('2010-06-11','somedata',somedata),
     ('2010-09-12','somedata',somedata),
     ('2010-08-12','somedata',somedata)]

t.sort(key = lambda x: strftime('%Y%m%d',strptime(x[0],'%Y-%d-%m')))
print t

result

[('2010-06-11', 'somedata', 'jyhghg'),
 ('2010-09-11', 'somedata', 'jyhghg'),
 ('2010-08-12', 'somedata', 'jyhghg'),
 ('2010-09-12', 'somedata', 'jyhghg')]

.

Edit 1

Reading the answer of Artsiom Rudzenka in which he uses strptime() alone, I realized that strptime() produces a struct_time object that is sorted by nature . Such an object has attributes tm_year, tm_mon, tm_mday, tm_hour, tm_min, tm_sec, tm_wday, tm_yday, tm_isdst that are accessible through common dot-notation access (toto.tm_mon for exemple), but also through index-notation access (toto[1] for exemple) , because the attributes of a struc_time object are registered in this order tm_year, tm_mon, tm_mday, tm_hour, tm_min, tm_sec, tm_wday, tm_yday, tm_isdst . The struct_time data type has a named tuple's interface .

Since a struct_time object is ordered by nature, it isn't necessary to apply strftime() to obtain a date string having year-month-day in this order: this order is already present in the struct_time object.

Then , I correct my code for the case in which 11 in '2010-06-11' is the month : I eliminate strftime()

from time import strptime

somedata = 'jyhghg'
t = [('2010-09-11','somedata',somedata),
     ('2010-06-11','somedata',somedata),
     ('2010-09-12','somedata',somedata),
     ('2010-08-12','somedata',somedata)]

t.sort(key = lambda x: strptime(x[0],'%Y-%d-%m'))
print t

Edit 2

Taking Kirk Strauser's info in consideration:

import re

regx = re.compile('(\d{4})-(\d\d)-(\d\d)')

somedata = 'jyhghg'
t = [('2010-09-11','somedata',somedata),
     ('2010-06-11','somedata',somedata),
     ('2010-09-12','somedata',somedata),
     ('2010-08-12','somedata',somedata)]

t.sort(key = lambda x: regx.match(x[0]).group(1,3,2))
print t

Answer 4

You can use the very simple

t.sort()

see: How does Python sort a list of tuples?