I love this [Tutor] list. There are always new tricks that change the
way I write code. And it makes me like Python more every day.
I keep a script file with "notes" on the things I learn here and I refer
to these notes frequently. Here are the notes I made for this thread:
""" iterate/map/modulus/(zip) """
a = (1, 2, 3, 4, 5)
b = ('r', 'g', 'b')
na = len(a)
nb = len(b)
print
print "==================================="
# zip
A = ''.join([str(i) for i in a]) # "12345"
B = ''.join(b) # "rgb"
print zip(A, B), " - zip (inadequate)"
print
print "==================================="
# brute force
m = 0
n = 0
t = []
##x = 10
##for i in range(x):
for i in range(max(na, nb)):
if m == na:
m = 0
if n == nb:
n = 0
t.append((a[m], b[n]))
m += 1
n += 1
print t, "- brute force"
print
print "==================================="
# itertools
from itertools import izip, cycle
print list(izip(a, cycle(b))), "- itertools"
print
print "==================================="
# map
print map(None,a,b*(na/nb+1))[:na], "- map"
print
print "==================================="
# modulus
print [(a[i], b[i%nb]) for i in range(na)], "- modulus"
print
print "-----------------------------------"
##x = max(na, nb)
x = 10
print [(a[i%na], b[i%nb]) for i in range(x)], "- modulus (extended)"
print
print "==================================="
This mailing list is great. Thanks to all the experienced Python coders
who offer various solutions to the questions, and to all the beginners
who ask them.
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Reed O'Brien
Sent: Sunday, June 17, 2007 8:59 AM
To: Iyer; tutor@python.org
Subject: Re: [Tutor] iterating over a sequence question..
On Jun 17, 2007, at 3:44 AM, John Fouhy wrote:
> On 17/06/07, Iyer <[EMAIL PROTECTED]> wrote:
>>
>> say, if I have a list l = [1,2,3,5]
>>
>> and another tuple t = ('r', 'g', 'b')
>>
>> Suppose I iterate over list l, and t at the same time, if I use
>> the zip
>> function as in zip(l,t) , I will not be able to cover elements 3
>> and 5 in
>> list l
>>
>>>>> l = [1,2,3,5]
>>>>> t = ('r', 'g', 'b')
>>>>> for i in zip(l,t):
>> ... print i
>> ...
>> (1, 'r')
>> (2, 'g')
>> (3, 'b')
>>
>> is there an elegant way in python to print all the elements in l,
>> while
>> looping over list t, if len(t) != len(l) as to get the output:
>>
>> (1, 'r')
>> (2, 'g')
>> (3, 'b')
>> (5, 'r')
>
> Check out the itertools module. I don't have the ability to test this
> right now, but try something like:
>
> import itertools
> lst = [1,2,3,5]
> t = ('r', 'g', 'b')
>
> itertools.izip(lst, itertools.cycle(t))
>
> --
> John.
>
+1 for John's solution
usage:
>>> [x for x in itertools.izip(lst, itertools.cycle(t)]
>>> [(1, 'r'), (2, 'g'), (3, 'b'), (5, 'r')]
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