How about using list.index() and storing month names in a list? You may want to measure performance your self and conclude.
Regards, Ashish Vyas -----Original Message----- From: python-list-bounces+ntb837=motorola....@python.org [mailto:python-list-bounces+ntb837=motorola....@python.org] On Behalf Of wiso Sent: Wednesday, January 06, 2010 4:34 PM To: python-list@python.org Subject: Convert month name to month number faster I'm optimizing the inner most loop of my script. I need to convert month name to month number. I'm using python 2.6 on linux x64. month_dict = {"Jan":1,"Feb":2,"Mar":3,"Apr":4, "May":5, "Jun":6, "Jul":7,"Aug":8,"Sep":9,"Oct":10,"Nov":11,"Dec":12} def to_dict(name): return month_dict[name] def to_if(name): if name == "Jan": return 1 elif name == "Feb": return 2 elif name == "Mar": return 3 elif name == "Apr": return 4 elif name == "May": return 5 elif name == "Jun": return 6 elif name == "Jul": return 7 elif name == "Aug": return 8 elif name == "Sep": return 9 elif name == "Oct": return 10 elif name == "Nov": return 11 elif name == "Dec": return 12 else: raise ValueError import random l = [random.choice(month_dict.keys()) for _ in range(1000000)] from time import time t = time(); xxx=map(to_dict,l); print time() - t # 0.5 t = time(); xxx=map(to_if,l); print time() - t # 1.0 is there a faster solution? Maybe something with str.translate? The problem is a little different because I don't read random data, but sorted data. For example: l = [x for x in ("Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec" ) for _ in range(1000)] # ["Jan","Jan", ..., "Feb", "Feb", ...] so maybe the to_if approach will be faster if I write the case in the best order. Look: l = ["Jan"] * 1000000 # to_if is in the best order for "Jan" t = time(); xxx=map(to_dict,l); print time() - t # 0.5 t = time(); xxx=map(to_if,l); print time() - t # 0.5 -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
