How would you translate this from Perl to Python?
I\'ve got a Perl function which takes a timestamp and returns either the unchanged timestamp (if it\'s never seen it before) or otherwise, it appends some letters to make it uni
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Well, sorry to say, but you can't just do a direct translation from Perl to Python (including bit-for-bit Perlisms) and expect the outcome to be prettier. It won't be, it will be considerably uglier.
If you want the prettiness of Python you will need to use Python idioms instead.
Now for the question at hand:
from string import uppercase class Uniquifier(object): def __init__(self): self.last_timestamp = None self.last_suffix = 0 def uniquify(self, timestamp): if timestamp == self.last_timestamp: timestamp = '%s%s' % (timestamp, uppercase[self.last_suffix]) self.last_suffix += 1 else: self.last_suffix = 0 self.timestamp = timestamp return timestamp uniquifier = Uniquifier() uniquifier.uniquify(a_timestamp)Prettier? Maybe. More readable? Probably.
Edit (re comments): Yes this fails after Z, and I am altogether unhappy with this solution. So I won't fix it, but might offer something better, like using a number instead:
timestamp = '%s%s' % (timestamp, self.last_suffix)If it were me, I would do this:
import uuid def uniquify(timestamp): return '%s-%s' % (timestamp, uuid.uuid4())And just be happy.
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I just tested this up to 1000 against the original perl implementation and diff returns the same results for both. The suffix code is tricky -- this is not a base 36 counter. Hasen J's solution - though it produces a unique timestamp - isn't quite the same since it goes from 'Z' to 'BA', when it should instead go to 'AA' to match the perl ++ operator.
#!/usr/bin/python class uniqify: def __init__(self): self.last_timestamp = -1 self.next_suffix = 'A' return def suffix(self): s = self.next_suffix letters = [l for l in self.next_suffix] if letters[-1] == 'Z': letters.reverse() nonz = None for i in range(len(letters)): if letters[i] != 'Z': nonz = i break if nonz is not None: letters[nonz] = chr(ord(letters[nonz]) + 1) for i in range(0, nonz): letters[i] = 'A' else: letters = ['A'] * (len(letters) + 1) letters.reverse() else: letters[-1] = chr(ord(letters[-1]) + 1) self.next_suffix = ''.join(letters) return s def reset(self): self.next_suffix = 'A' return def __call__(self, timestamp): if timestamp == self.last_timestamp: timestamp_str = '%s%s' % (timestamp, self.suffix()) else: self.last_timestamp = timestamp self.reset() timestamp_str = '%s' % timestamp return timestamp_str uniqify = uniqify() if __name__ == '__main__': for n in range(1000): print uniqify(1) for n in range(1000): print uniqify(2)讨论(0) -
Look at this answer for a robust method to convert a number to an alphanumeric id
The code I present doesn't go from 'Z' to 'AA', instead goes to 'BA', but I suppose that doesn't matter, it still produces a unique id
from string import uppercase as up import itertools def to_base(q, alphabet): if q < 0: raise ValueError( "must supply a positive integer" ) l = len(alphabet) converted = [] while q != 0: q, r = divmod(q, l) converted.insert(0, alphabet[r]) return "".join(converted) or alphabet[0] class TimestampUniqifier( object ): def __init__(self): self.last = '' self.counter = itertools.count() def __call__( self, str ): if str == self.last: suf = self.counter.next() return str + to_base( suf, up ) else: self.last = str self.counter = itertools.count() return str timestamp_uniqify = TimestampUniqifier()usage:
timestamp_uniqify('1') '1' timestamp_uniqify('1') '1A' timestamp_uniqify('1') '1B' timestamp_uniqify('1') '1C' timestamp_uniqify('2') '2' timestamp_uniqify('3') '3' timestamp_uniqify('3') '3A' timestamp_uniqify('3') '3B'You can call it maaaany times and it will still produce good results:
for i in range(100): print timestamp_uniqify('4') 4 4A 4B 4C 4D 4E 4F 4G 4H 4I 4J 4K 4L 4M 4N 4O 4P 4Q 4R 4S 4T 4U 4V 4W 4X 4Y 4Z 4BA 4BB 4BC 4BD 4BE 4BF 4BG 4BH 4BI 4BJ 4BK 4BL 4BM 4BN 4BO 4BP 4BQ 4BR 4BS 4BT 4BU 4BV 4BW 4BX 4BY 4BZ 4CA 4CB 4CC 4CD 4CE 4CF 4CG 4CH 4CI 4CJ 4CK 4CL 4CM 4CN 4CO 4CP 4CQ 4CR 4CS 4CT 4CU 4CV 4CW 4CX 4CY 4CZ 4DA 4DB 4DC 4DD 4DE 4DF 4DG 4DH 4DI 4DJ 4DK 4DL 4DM 4DN 4DO 4DP 4DQ 4DR 4DS 4DT 4DU讨论(0) -
Does the suffix have to be letters like that?
from itertools import count def unique(timestamp): if timestamp in unique.ts.keys(): return timestamp + '.' + str(unique.ts[timestamp].next()) else: unique.ts[timestamp] = count() return timestamp unique.ts = {}You can define a different count if you want the letters back.
This isn't the same as your perl code, though.
- It keeps a dict around so if you have lots of unique timestamps then you'll use lots of memory.
- It handles out of order calls, which the original doesn't (i.e. u(1), u(2), u(1)).
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Looking at the problem it seems like a good fit for a coroutine (Python 2.5 or higher). Here's some code that will roughly produce the same result:
def uniqify(): seen = {} val = (yield None) while True: if val in seen: idxa, idxb = seen[val] idxb += 1 else: idxa, idxb = (len(seen)+1, ord('a')) seen[val] = (idxa, idxb) uniq = "%s%s" % (idxa, chr(idxb)) val = (yield uniq)And here's how you use it:
>>> u = send.uniqify() >>> u.next() #need this to start the generator >>> u.send(1) '1a' >>> u.send(1) '1b' >>> u.send(1) '1c' >>> u.send(2) '2a' >>> u.send(2) '2b' >>> u.send(1) #you can go back to previous values '1d' >>> u.send('stringy') #you can send it anything that can be used as a dict key '3a'讨论(0) -
This is my first time answering, and I used globals, but it seemed the simplest way to me.
from string import uppercase last_ts = None letters = None def increment(letters): if not letters: return "A" last_letter = letters[-1] if last_letter == "Z": return increment(letters[:-1]) + "A" return letters[:-1] + uppercase[uppercase.index(last_letter) + 1] def uniquify(timestamp): global last_ts, letters if timestamp == last_ts: letters = increment(letters) return timestamp + letters last_ts = timestamp letters = None return timestamp print uniquify("1") print uniquify('1') print uniquify("1") print uniquify("2") for each in range(100): print uniquify("2") 1 1A 1B 2 2A 2B 2C 2D 2E 2F 2G 2H 2I 2J 2K 2L 2M 2N 2O 2P 2Q 2R 2S 2T 2U 2V 2W 2X 2Y 2Z 2AA 2AB 2AC 2AD 2AE 2AF 2AG 2AH 2AI 2AJ 2AK 2AL 2AM 2AN 2AO 2AP 2AQ 2AR 2AS 2AT 2AU 2AV 2AW 2AX 2AY 2AZ 2BA 2BB 2BC 2BD 2BE 2BF 2BG 2BH 2BI 2BJ 2BK 2BL 2BM 2BN 2BO 2BP 2BQ 2BR 2BS 2BT 2BU 2BV 2BW 2BX 2BY 2BZ 2CA 2CB 2CC 2CD 2CE 2CF 2CG 2CH 2CI 2CJ 2CK 2CL 2CM 2CN 2CO 2CP 2CQ 2CR 2CS 2CT 2CU 2CV讨论(0)
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