On Mon, Sep 16, 2013 at 12:27 PM, Suyash Bhatt bhatt.suy...@gmail.com wrote:
thanks for the response..
what if i have another list
*e = [*'*my1name1is1','**my2name2is1','xyz','abc']*
*and in the list d, I want only the elements which are present in e..*
*
Python 2.7.3 (default, Apr 10 2013,
On Sep 16, 2013 12:50 PM, Dhananjay Nene dhananjay.n...@gmail.com wrote:
On Mon, Sep 16, 2013 at 12:27 PM, Suyash Bhatt bhatt.suy...@gmail.com
wrote:
thanks for the response..
what if i have another list
*e = [*'*my1name1is1','**my2name2is1','xyz','abc']*
*and in the list d, I want
Hi all,
i need help in finding the most optimized way of making a list using
itertools.
Suppose I have 3 lists:
*a=[‘my1’,’my2’]
*
**
*b=[‘name1’,’name2’]*
*c=[‘is1’]*
I want to iter through all and form another list with the following
appended strings:
d = [''.join(x) for x in itertools.product(a, b, c)]
--
Dhruv Baldawa
(http://www.dhruvb.com)
On Fri, Sep 13, 2013 at 4:53 PM, Suyash Bhatt bhatt.suy...@gmail.comwrote:
Hi all,
i need help in finding the most optimized way of making a list using
itertools.
Suppose I have 3 lists:
On Fri, Sep 13, 2013 at 5:01 PM, Dhruv Baldawa dhruvbald...@gmail.comwrote:
d = [''.join(x) for x in itertools.product(a, b, c)]
Actually, using itertools.imap would ensure that the elements aren't
computed till necessary. So...
d = itertools.imap(''.join, itertools.product(a, b, c))
- d
