On Mar 21, 2015 1:22 AM, Michael Omohundro
momohund1...@yahoo.com.dmarc.invalid wrote:
Does anyone know how to create a python script to generate an aspect
raster from the input elevation in a digital elevation model?
This topic is out of the normal scope of the mailing list. It's hard to say
On 03/31/2015 10:00 AM, Ian D wrote:
Hi
I have a list that I am splitting into pairs of values. But the list is dynamic
in size. It could have 4 values or 6 or more.
I originally split the list into pairs, by using a new list and keep a pair in
the old list by just popping 2 values. But if
Thanks I will look into these. The data going in is a list like
this:['broadcast', 'd8on', 'broadcast', 'd11on']
With the output beng something like this.
lst_0 = ['broadcast', 'd8on']
lst_0 = ['broadcast', 'd11on']
I have managed to use a dictionary as advised in a post on StackOverflow; not
Ian D wrote:
Thanks I will look into these. The data going in is a list like
this:['broadcast', 'd8on', 'broadcast', 'd11on']
With the output beng something like this.
lst_0 = ['broadcast', 'd8on']
lst_0 = ['broadcast', 'd11on']
Is that a typo, did you mean
lst_1 = ['broadcast',
Ok Thanks a lot. And sadly not a typo, my bad logic overwriting values!
To: tutor@python.org
From: __pete...@web.de
Date: Tue, 31 Mar 2015 17:50:01 +0200
Subject: Re: [Tutor] Dynamic naming of lists
Ian D wrote:
Thanks I will look into these. The
The following behavior has me stumped:
Python 2.7.8 (default, Jun 30 2014, 16:03:49) [MSC v.1500 32 bit
(Intel)] on win32
Type copyright, credits or license() for more information.
L = ['#ROI:roi_0', '#TXT:text_0', '#1:one^two^three']
for i, item in enumerate(L):
subitems =
On Tue, Mar 31, 2015 at 3:23 PM, boB Stepp robertvst...@gmail.com wrote:
The following behavior has me stumped:
Python 2.7.8 (default, Jun 30 2014, 16:03:49) [MSC v.1500 32 bit
(Intel)] on win32
Type copyright, credits or license() for more information.
L = ['#ROI:roi_0', '#TXT:text_0',
On Tue, Mar 31, 2015 at 3:28 PM, Zachary Ware
zachary.ware+py...@gmail.com wrote:
On Tue, Mar 31, 2015 at 3:23 PM, boB Stepp robertvst...@gmail.com wrote:
The following behavior has me stumped:
Python 2.7.8 (default, Jun 30 2014, 16:03:49) [MSC v.1500 32 bit
(Intel)] on win32
Type copyright,
On Tue, Mar 31, 2015 at 3:37 PM, boB Stepp robertvst...@gmail.com wrote:
On Tue, Mar 31, 2015 at 3:28 PM, Zachary Ware
zachary.ware+py...@gmail.com wrote:
On Tue, Mar 31, 2015 at 3:23 PM, boB Stepp robertvst...@gmail.com wrote:
The following behavior has me stumped:
Python 2.7.8 (default,
On Tue, Mar 31, 2015 at 3:32 PM, Dave Angel da...@davea.name wrote:
On 03/31/2015 04:23 PM, boB Stepp wrote:
The following behavior has me stumped:
Python 2.7.8 (default, Jun 30 2014, 16:03:49) [MSC v.1500 32 bit
(Intel)] on win32
Type copyright, credits or license() for more information.
Hi
I have a list that I am splitting into pairs of values. But the list is dynamic
in size. It could have 4 values or 6 or more.
I originally split the list into pairs, by using a new list and keep a pair in
the old list by just popping 2 values. But if the list is longer than 4 values.
I
On 31/03/2015 21:49, boB Stepp wrote:
On Tue, Mar 31, 2015 at 3:42 PM, Zachary Ware
zachary.ware+py...@gmail.com wrote:
Also, not that since you aren't using the index for anything, you
don't need to use enumerate() to iterate over the list. Just do for
item in L:. Of course, if you actually
On 03/31/2015 04:23 PM, boB Stepp wrote:
The following behavior has me stumped:
Python 2.7.8 (default, Jun 30 2014, 16:03:49) [MSC v.1500 32 bit
(Intel)] on win32
Type copyright, credits or license() for more information.
L = ['#ROI:roi_0', '#TXT:text_0', '#1:one^two^three']
for i, item in
On Tue, Mar 31, 2015 at 3:42 PM, Zachary Ware
zachary.ware+py...@gmail.com wrote:
Also, not that since you aren't using the index for anything, you
don't need to use enumerate() to iterate over the list. Just do for
item in L:. Of course, if you actually use the index in the real
code that
Ian -
Note that if all your keys are named 'broadcast', the dictionary approach
is probably not going to work. You'll end up with something like:
{ 'broadcast': 'last_value_in_the_list' }
On Tue, Mar 31, 2015 at 10:56 AM, Ian D dux...@hotmail.com wrote:
Ok Thanks a lot. And sadly not a
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