On 01/18/2014 07:20 PM, Pierre Dagenais wrote:
Hello,
I wish to fill a list called years[] with a hundred lists called
year1900[], year1901[], year1902[], ..., year1999[]. That is too much
typing of course. Any way of doing this in a loop? I've tried stuff like
("year" + str(1900)) = [0,0] but nothing works.
Any solution?
Thank you,
PierreD.
I think Danny & Wiktor's solutions are the right ones. Danny's is faster a
simpler (because of direct array indexing), but does not allow giving the true
year "names" (actually numbers). Wiktor more correctly matches you expectation
bt is a bit slower because we're searching individual years in a dict.
Here is an alternative, which should about as slow (since it is also searching
in a dict), and give true (string) names, at the cost of some complication in
code. The trick is to make a plain object (unfortunately we need a class in
python for that: it's a "fantom" class) and then directly set years on it as
plain attributes. Unfortunately, I don't remember of any nice way to traverse an
object's attributes: we have to traverse its __dict__ explicitely.
======================================
class Years: pass # fantom class
years = Years()
# set years as attributes:
start_year = 1900
n_years = 3 # 3 years only as example
for i in range (n_years):
# use setattr(obj, name, value) :
name = "year" + str(start_year + i)
value = [i] # give meaningful value maybe ;-)
setattr(years, name, value)
# access individual year:
print(years.year1901)
# traverse years:
for name in years.__dict__:
value = years.__dict__[name]
# unordered, unfortunately
print("%s : %s" % (name, value))
""" output (by me: order is undefined):
[1]
year1900 : [0]
year1901 : [1]
year1902 : [2]
"""
======================================
denis
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