[Python] Python print first row of list

I'm sure there a simple solution to this question but so far it has eluded me.

I simply want to print out the first row the below list where [0] is the None value and [1] is the time stamp. Where ss is the list name.

None 2018-05-21 00:15:23.154428
None 2018-05-21 00:15:25.154428
None 2018-05-21 00:28:49.154627
None 2018-05-21 00:30:21.154642
None 2018-05-21 00:30:23.154642
None 2018-05-21 00:43:17.154838
None 2018-05-21 00:43:19.154838
None 2018-05-21 00:45:19.154857
None 2018-05-21 00:45:21.154857
None 2018-05-21 00:45:23.154858
None 2018-05-21 01:43:17.155843
None 2018-05-21 01:43:19.155843
None 2018-05-21 01:43:21.155843
None 2018-05-21 02:28:17.156551
None 2018-05-21 02:28:19.156552
None 2018-05-21 02:28:21.156552
None 2018-05-21 02:28:23.156553
None 2018-05-21 02:28:25.156553
None 2018-05-21 02:28:27.156553
None 2018-05-21 02:30:17.156571
None 2018-05-21 02:30:19.156571
None 2018-05-21 02:30:21.156571
None 2018-05-21 02:58:17.157040
None 2018-05-21 03:13:17.157249
None 2018-05-21 03:43:17.157726
None 2018-05-21 03:58:21.157929
None 2018-05-21 03:58:23.157930
None 2018-05-21 04:00:17.158008
None 2018-05-21 04:00:19.158009
None 2018-05-21 04:00:21.158009
None 2018-05-21 04:58:17.158840
None 2018-05-21 04:58:19.158841
None 2018-05-21 05:13:17.159086
None 2018-05-21 05:13:19.159086


Here are several things I've attempted.

#print ss[:2]

#print next(ss)

#for item in ss[0:1]:
# print item

#for i in ss :
# print(i[0])
#print(ss(zip(*0))[0])

#print ss [0:1]

#for row[0] for row in ss:
print row

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