Python function to convert seconds into minutes, hours, and days
Question: Write a program that asks the user to enter a number of seconds, and works as follows:
There are 60 seconds in a minute. If the number of seconds
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These functions are fairly compact and only use standard Python 2.6 and later.
def ddhhmmss(seconds): """Convert seconds to a time string "[[[DD:]HH:]MM:]SS". """ dhms = '' for scale in 86400, 3600, 60: result, seconds = divmod(seconds, scale) if dhms != '' or result > 0: dhms += '{0:02d}:'.format(result) dhms += '{0:02d}'.format(seconds) return dhms def seconds(dhms): """Convert a time string "[[[DD:]HH:]MM:]SS" to seconds. """ components = [int(i) for i in dhms.split(':')] pad = 4 - len(components) if pad < 0: raise ValueError('Too many components to match [[[DD:]HH:]MM:]SS') components = [0] * pad + components return sum(i * j for i, j in zip((86400, 3600, 60, 1), components))And here are tests to go with them. I'm using the pytest package as a simple way to test exceptions.
import ddhhmmss import pytest def test_ddhhmmss(): assert ddhhmmss.ddhhmmss(0) == '00' assert ddhhmmss.ddhhmmss(2) == '02' assert ddhhmmss.ddhhmmss(12 * 60) == '12:00' assert ddhhmmss.ddhhmmss(3600) == '01:00:00' assert ddhhmmss.ddhhmmss(10 * 86400) == '10:00:00:00' assert ddhhmmss.ddhhmmss(86400 + 5 * 3600 + 30 * 60 + 1) == '01:05:30:01' assert ddhhmmss.ddhhmmss(365 * 86400) == '365:00:00:00' def test_seconds(): assert ddhhmmss.seconds('00') == 0 assert ddhhmmss.seconds('02') == 2 assert ddhhmmss.seconds('12:00') == 12 * 60 assert ddhhmmss.seconds('01:00:00') == 3600 assert ddhhmmss.seconds('1:0:0') == 3600 assert ddhhmmss.seconds('3600') == 3600 assert ddhhmmss.seconds('60:0') == 3600 assert ddhhmmss.seconds('10:00:00:00') == 10 * 86400 assert ddhhmmss.seconds('1:05:30:01') == 86400 + 5 * 3600 + 30 * 60 + 1 assert ddhhmmss.seconds('365:00:00:00') == 365 * 86400 def test_seconds_raises(): with pytest.raises(ValueError): ddhhmmss.seconds('') with pytest.raises(ValueError): ddhhmmss.seconds('foo') with pytest.raises(ValueError): ddhhmmss.seconds('1:00:00:00:00')讨论(0) -
Patching as well Ralph Bolton's answer. Moving to a class and moving tulp of tulp (intervals) to dictionary. Adding an optional rounded function depending of granularity (enable by default). Ready to translation using gettext (default is disable). This is intend to be load from an module. This is for python3 (tested 3.6 - 3.8)
import gettext import locale from itertools import chain mylocale = locale.getdefaultlocale() # see --> https://stackoverflow.com/a/10174657/11869956 thx #localedir = os.path.join(os.path.dirname(__file__), 'locales') # or python > 3.4: try: localedir = pathlib.Path(__file__).parent/'locales' lang_translations = gettext.translation('utils', localedir, languages=[mylocale[0]]) lang_translations.install() _ = lang_translations.gettext except Exception as exc: print('Error: unexcept error while initializing translation:', file=sys.stderr) print(f'Error: {exc}', file=sys.stderr) print(f'Error: localedir={localedir}, languages={mylocale[0]}', file=sys.stderr) print('Error: translation has been disabled.', file=sys.stderr) _ = gettext.gettextHere is the class:
class FormatTimestamp: """Convert seconds to, optional rounded, time depending of granularity's degrees. inspired by https://stackoverflow.com/a/24542445/11869956""" def __init__(self): # For now i haven't found a way to do it better # TODO: optimize ?!? ;) self.intervals = { # 'years' : 31556952, # https://www.calculateme.com/time/years/to-seconds/ # https://www.calculateme.com/time/months/to-seconds/ -> 2629746 seconds # But it's outputing some strange result : # So 3 seconds less (2629743) : 4 weeks, 2 days, 10 hours, 29 minutes and 3 seconds # than after 3 more seconds : 1 month ?!? # Google give me 2628000 seconds # So 3 seconds less (2627997): 4 weeks, 2 days, 9 hours, 59 minutes and 57 seconds # Strange as well # So for the moment latest is week ... #'months' : 2419200, # 60 * 60 * 24 * 7 * 4 'weeks' : 604800, # 60 * 60 * 24 * 7 'days' : 86400, # 60 * 60 * 24 'hours' : 3600, # 60 * 60 'minutes' : 60, 'seconds' : 1 } self.nextkey = { 'seconds' : 'minutes', 'minutes' : 'hours', 'hours' : 'days', 'days' : 'weeks', 'weeks' : 'weeks', #'months' : 'months', #'years' : 'years' # stop here } self.translate = { 'weeks' : _('weeks'), 'days' : _('days'), 'hours' : _('hours'), 'minutes' : _('minutes'), 'seconds' : _('seconds'), ## Single 'week' : _('week'), 'day' : _('day'), 'hour' : _('hour'), 'minute' : _('minute'), 'second' : _('second'), ' and' : _('and'), ',' : _(','), # This is for compatibility '' : '\0' # same here BUT we CANNOT pass empty string to gettext # or we get : warning: Empty msgid. It is reserved by GNU gettext: # gettext("") returns the header entry with # meta information, not the empty string. # Thx to --> https://stackoverflow.com/a/30852705/11869956 - saved my day } def convert(self, seconds, granularity=2, rounded=True, translate=False): """Proceed the conversion""" def _format(result): """Return the formatted result TODO : numpy / google docstrings""" start = 1 length = len(result) none = 0 next_item = False for item in reversed(result[:]): if item['value']: # if we have more than one item if length - none > 1: # This is the first 'real' item if start == 1: item['punctuation'] = '' next_item = True elif next_item: # This is the second 'real' item # Happened 'and' to key name item['punctuation'] = ' and' next_item = False # If there is more than two 'real' item # than happened ',' elif 2 < start: item['punctuation'] = ',' else: item['punctuation'] = '' else: item['punctuation'] = '' start += 1 else: none += 1 return [ { 'value' : mydict['value'], 'name' : mydict['name_strip'], 'punctuation' : mydict['punctuation'] } for mydict in result \ if mydict['value'] is not None ] def _rstrip(value, name): """Rstrip 's' name depending of value""" if value == 1: name = name.rstrip('s') return name # Make sure granularity is an integer if not isinstance(granularity, int): raise ValueError(f'Granularity should be an integer: {granularity}') # For seconds only don't need to compute if seconds < 0: return 'any time now.' elif seconds < 60: return 'less than a minute.' result = [] for name, count in self.intervals.items(): value = seconds // count if value: seconds -= value * count name_strip = _rstrip(value, name) # save as dict: value, name_strip (eventually strip), name (for reference), value in seconds # and count (for reference) result.append({ 'value' : value, 'name_strip' : name_strip, 'name' : name, 'seconds' : value * count, 'count' : count }) else: if len(result) > 0: # We strip the name as second == 0 name_strip = name.rstrip('s') # adding None to key 'value' but keep other value # in case when need to add seconds when we will # recompute every thing result.append({ 'value' : None, 'name_strip' : name_strip, 'name' : name, 'seconds' : 0, 'count' : count }) # Get the length of the list length = len(result) # Don't need to compute everything / every time if length < granularity or not rounded: if translate: return ' '.join('{0} {1}{2}'.format(item['value'], _(self.translate[item['name']]), _(self.translate[item['punctuation']])) \ for item in _format(result)) else: return ' '.join('{0} {1}{2}'.format(item['value'], item['name'], item['punctuation']) \ for item in _format(result)) start = length - 1 # Reverse list so the firsts elements # could be not selected depending on granularity. # And we can delete item after we had his seconds to next # item in the current list (result) for item in reversed(result[:]): if granularity <= start <= length - 1: # So we have to round current_index = result.index(item) next_index = current_index - 1 # skip item value == None # if the seconds of current item is superior # to the half seconds of the next item: round if item['value'] and item['seconds'] > result[next_index]['count'] // 2: # +1 to the next item (in seconds: depending on item count) result[next_index]['seconds'] += result[next_index]['count'] # Remove item which is not selected del result[current_index] start -= 1 # Ok now recalculate everything # Reverse as well for item in reversed(result[:]): # Check if seconds is superior or equal to the next item # but not from 'result' list but from 'self.intervals' dict # Make sure it's not None if item['value']: next_item_name = self.nextkey[item['name']] # This mean we are at weeks if item['name'] == next_item_name: # Just recalcul item['value'] = item['seconds'] // item['count'] item['name_strip'] = _rstrip(item['value'], item['name']) # Stop to weeks to stay 'right' elif item['seconds'] >= self.intervals[next_item_name]: # First make sure we have the 'next item' # found via --> https://stackoverflow.com/q/26447309/11869956 # maybe there is a faster way to do it ? - TODO if any(search_item['name'] == next_item_name for search_item in result): next_item_index = result.index(item) - 1 # Append to result[next_item_index]['seconds'] += item['seconds'] # recalculate value result[next_item_index]['value'] = result[next_item_index]['seconds'] // \ result[next_item_index]['count'] # strip or not result[next_item_index]['name_strip'] = _rstrip(result[next_item_index]['value'], result[next_item_index]['name']) else: # Creating next_item_index = result.index(item) - 1 # get count next_item_count = self.intervals[next_item_name] # convert seconds next_item_value = item['seconds'] // next_item_count # strip 's' or not next_item_name_strip = _rstrip(next_item_value, next_item_name) # added to dict next_item = { 'value' : next_item_value, 'name_strip' : next_item_name_strip, 'name' : next_item_name, 'seconds' : item['seconds'], 'count' : next_item_count } # insert to the list result.insert(next_item_index, next_item) # Remove current item del result[result.index(item)] else: # for current item recalculate # keys 'value' and 'name_strip' item['value'] = item['seconds'] // item['count'] item['name_strip'] = _rstrip(item['value'], item['name']) if translate: return ' '.join('{0} {1}{2}'.format(item['value'], _(self.translate[item['name']]), _(self.translate[item['punctuation']])) \ for item in _format(result)) else: return ' '.join('{0} {1}{2}'.format(item['value'], item['name'], item['punctuation']) \ for item in _format(result))To use it:
myformater = FormatTimestamp() myconverter = myformater.convert(seconds)granularity = 1 - 5, rounded = True / False, translate = True / False
Some test to show difference:
myformater = FormatTimestamp() for firstrange in [131440, 563440, 604780, 2419180, 113478160]: print(f'#### Seconds : {firstrange} ####') print('\tFull - function: {0}'.format(display_time(firstrange, granularity=5))) print('\tFull - class: {0}'.format(myformater.convert(firstrange, granularity=5))) for secondrange in range(1, 6, 1): print('\tGranularity this answer ({0}): {1}'.format(secondrange, myformater.convert(firstrange, granularity=secondrange, translate=False))) print('\tGranularity Bolton\'s answer ({0}): {1}'.format(secondrange, display_time(firstrange, granularity=secondrange))) print()Seconds : 131440
Seconds : 563440Full - function: 1 day, 12 hours, 30 minutes, 40 seconds Full - class: 1 day, 12 hours, 30 minutes and 40 seconds Granularity this answer (1): 2 days Granularity Bolton's answer (1): 1 day Granularity this answer (2): 1 day and 13 hours Granularity Bolton's answer (2): 1 day, 12 hours Granularity this answer (3): 1 day, 12 hours and 31 minutes Granularity Bolton's answer (3): 1 day, 12 hours, 30 minutes Granularity this answer (4): 1 day, 12 hours, 30 minutes and 40 seconds Granularity Bolton's answer (4): 1 day, 12 hours, 30 minutes, 40 seconds Granularity this answer (5): 1 day, 12 hours, 30 minutes and 40 seconds Granularity Bolton's answer (5): 1 day, 12 hours, 30 minutes, 40 seconds
Seconds : 604780Full - function: 6 days, 12 hours, 30 minutes, 40 seconds Full - class: 6 days, 12 hours, 30 minutes and 40 seconds Granularity this answer (1): 1 week Granularity Bolton's answer (1): 6 days Granularity this answer (2): 6 days and 13 hours Granularity Bolton's answer (2): 6 days, 12 hours Granularity this answer (3): 6 days, 12 hours and 31 minutes Granularity Bolton's answer (3): 6 days, 12 hours, 30 minutes Granularity this answer (4): 6 days, 12 hours, 30 minutes and 40 seconds Granularity Bolton's answer (4): 6 days, 12 hours, 30 minutes, 40 seconds Granularity this answer (5): 6 days, 12 hours, 30 minutes and 40 seconds Granularity Bolton's answer (5): 6 days, 12 hours, 30 minutes, 40 seconds
Seconds : 2419180Full - function: 6 days, 23 hours, 59 minutes, 40 seconds Full - class: 6 days, 23 hours, 59 minutes and 40 seconds Granularity this answer (1): 1 week Granularity Bolton's answer (1): 6 days Granularity this answer (2): 1 week Granularity Bolton's answer (2): 6 days, 23 hours Granularity this answer (3): 1 week Granularity Bolton's answer (3): 6 days, 23 hours, 59 minutes Granularity this answer (4): 6 days, 23 hours, 59 minutes and 40 seconds Granularity Bolton's answer (4): 6 days, 23 hours, 59 minutes, 40 seconds Granularity this answer (5): 6 days, 23 hours, 59 minutes and 40 seconds Granularity Bolton's answer (5): 6 days, 23 hours, 59 minutes, 40 seconds
Seconds : 113478160Full - function: 3 weeks, 6 days, 23 hours, 59 minutes, 40 seconds Full - class: 3 weeks, 6 days, 23 hours, 59 minutes and 40 seconds Granularity this answer (1): 4 weeks Granularity Bolton's answer (1): 3 weeks Granularity this answer (2): 4 weeks Granularity Bolton's answer (2): 3 weeks, 6 days Granularity this answer (3): 4 weeks Granularity Bolton's answer (3): 3 weeks, 6 days, 23 hours Granularity this answer (4): 4 weeks Granularity Bolton's answer (4): 3 weeks, 6 days, 23 hours, 59 minutes Granularity this answer (5): 3 weeks, 6 days, 23 hours, 59 minutes and 40 seconds Granularity Bolton's answer (5): 3 weeks, 6 days, 23 hours, 59 minutes, 40 secondsFull - function: 187 weeks, 4 days, 9 hours, 42 minutes, 40 seconds Full - class: 187 weeks, 4 days, 9 hours, 42 minutes and 40 seconds Granularity this answer (1): 188 weeks Granularity Bolton's answer (1): 187 weeks Granularity this answer (2): 187 weeks and 4 days Granularity Bolton's answer (2): 187 weeks, 4 days Granularity this answer (3): 187 weeks, 4 days and 10 hours Granularity Bolton's answer (3): 187 weeks, 4 days, 9 hours Granularity this answer (4): 187 weeks, 4 days, 9 hours and 43 minutes Granularity Bolton's answer (4): 187 weeks, 4 days, 9 hours, 42 minutes Granularity this answer (5): 187 weeks, 4 days, 9 hours, 42 minutes and 40 seconds Granularity Bolton's answer (5): 187 weeks, 4 days, 9 hours, 42 minutes, 40 secondsI have a french translation ready. But it's fast to do the translation ... just few words. Hope this could help as the other answer help me a lot.
讨论(0) -
Patching Mr.B's answer (sorry, not enough rep. to comment), we can return variable granularity based on the amount of time. For example, we don't say "1 week, 5 seconds", we just say "1 week":
def display_time(seconds, granularity=2): result = [] for name, count in intervals: value = seconds // count if value: seconds -= value * count if value == 1: name = name.rstrip('s') result.append("{} {}".format(value, name)) else: # Add a blank if we're in the middle of other values if len(result) > 0: result.append(None) return ', '.join([x for x in result[:granularity] if x is not None])Some sample input:
for diff in [5, 67, 3600, 3605, 3667, 24*60*60, 24*60*60+5, 24*60*60+57, 24*60*60+3600, 24*60*60+3667, 2*24*60*60, 2*24*60*60+5*60*60, 7*24*60*60, 7*24*60*60 + 24*60*60]: print "For %d seconds: %s" % (diff, display_time(diff, 2))...returns this output:
For 5 seconds: 5 seconds For 67 seconds: 1 minute, 7 seconds For 3600 seconds: 1 hour For 3605 seconds: 1 hour For 3667 seconds: 1 hour, 1 minute For 86400 seconds: 1 day For 86405 seconds: 1 day For 86457 seconds: 1 day For 90000 seconds: 1 day, 1 hour For 90067 seconds: 1 day, 1 hour For 172800 seconds: 2 days For 190800 seconds: 2 days, 5 hours For 604800 seconds: 1 week For 691200 seconds: 1 week, 1 day讨论(0) -
At first glance, I figured divmod would be faster since it's a single statement and a built-in function, but timeit seems to show otherwise. Consider this little example I came up with when I was trying to figure out the fastest method for use in a loop that continuously runs in a gobject idle_add splitting a seconds counter into a human readable time for updating a progress bar label.
import timeit def test1(x,y, dropy): while x > 0: y -= dropy x -= 1 # the test minutes = (y-x) / 60 seconds = (y-x) % 60.0 def test2(x,y, dropy): while x > 0: y -= dropy x -= 1 # the test minutes, seconds = divmod((y-x), 60) x = 55 # litte number, also number of tests y = 10000 # make y > x by factor of drop dropy = 7 # y is reduced this much each iteration, for variation print "division and modulus:", timeit.timeit( lambda: test1(x,y,dropy) ) print "divmod function:", timeit.timeit( lambda: test2(x,y,dropy) )The built-in divmod function seems incredibly slower compared to using the simple division and modulus.
division and modulus: 12.5737669468 divmod function: 17.2861430645讨论(0)
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