Re: PIL : How to write array to image ???
On Oct 4, 9:47 am, Martin wrote:
> On Oct 3, 11:56 pm, Peter Otten <__pete...@web.de> wrote:
>
>
>
>
>
> > Martin wrote:
> > > Dear group
>
> > > I'm trying to use PIL to write an array (a NumPy array to be exact) to
> > > an image.
> > > Peace of cake, but it comes out looking strange.
>
> > > I use the below mini code, that I wrote for the purpose. The print of
> > > a looks like expected:
>
> > > [[ 200. 200. 200. ..., 0. 0. 0.]
> > > [ 200. 200. 200. ..., 0. 0. 0.]
> > > [ 200. 200. 200. ..., 0. 0. 0.]
> > > ...,
> > > [ 0. 0. 0. ..., 200. 200. 200.]
> > > [ 0. 0. 0. ..., 200. 200. 200.]
> > > [ 0. 0. 0. ..., 200. 200. 200.]]
>
> > > But the image looks nothing like that.
>
> > > Please see the images on:
> > >http://hvidberg.net/Martin/temp/quat_col.png
> > >http://hvidberg.net/Martin/temp/quat_bw.png
>
> > > or run the code to see them locally.
>
> > > Please – what do I do wrong in the PIL part ???
>
> > > :-? Martin
>
> > > import numpy as np
> > > from PIL import Image
> > > from PIL import ImageOps
>
> > > maxcol = 100
> > > maxrow = 100
>
> > > a = np.zeros((maxcol,maxrow),float)
>
> > > for i in range(maxcol):
> > > for j in range(maxrow):
> > > if (i<(maxcol/2) and j<(maxrow/2)) or (i>=(maxcol/2) and j>=
> > > (maxrow/2)):
> > > a[i,j] = 200
> > > else:
> > > a[i,j] = 0
>
> > > print a
>
> > > pilImage = Image.fromarray(a,'RGB')
> > > pilImage.save('quat_col.png')
> > > pilImage = ImageOps.grayscale(pilImage)
> > > pilImage.save('quat_bw.png')
>
> > The PIL seems to copy the array contents directly from memory without any
> > conversions or sanity check. In your example The float values determine the
> > gray value of 8 consecutive pixels.
>
> > If you want a[i,j] to become the color of the pixel (i, j) you have to use
> > an array with a memory layout that is compatible to the Image.
> > Here are a few examples:
>
> > >>> import numpy
> > >>> from PIL import Image
> > >>> a = numpy.zeros((100, 100), numpy.uint8)
> > >>> a[:50, :50] = a[50:, 50:] = 255
> > >>> Image.fromarray(a).save("tmp1.png")
> > >>> b = numpy.zeros((100, 100, 3), numpy.uint8)
> > >>> b[:50, :50, :] = b[50:, 50:, :] = [255, 0, 0]
> > >>> Image.fromarray(b).save("tmp2.png")
> > >>> c = numpy.zeros((100, 100), numpy.uint32)
> > >>> c[:50, :50] = c[50:, 50:] = 0xff808000
> > >>> Image.fromarray(c, "RGBA").save("tmp3.png")
>
> > Peter
>
> Thanks All - That helped a lot...
>
> The working code ended with:
>
> imga = np.zeros((imgL.shape[1],imgL.shape[0]),np.uint8)
> for ro in range(imgL.shape[1]):
> for co in range(imgL.shape[0]):
> imga[ro,co] = imgL[ro,co]
> Image.fromarray(imga).save('_a'+str(lev)+'.png')
Without knowing how big your image is (can't remember if you said!).
Perhaps rather than looping in the way you might in C for example, the
numpy where might be quicker if you have a big image. Just a
thought...
--
http://mail.python.org/mailman/listinfo/python-list
Re: PIL : How to write array to image ???
On Oct 5, 5:14 pm, Martin wrote:
> On Oct 4, 10:16 pm, "Mart." wrote:
>
>
>
>
>
> > On Oct 4, 9:47 am, Martin wrote:
>
> > > On Oct 3, 11:56 pm, Peter Otten <__pete...@web.de> wrote:
>
> > > > Martin wrote:
> > > > > Dear group
>
> > > > > I'm trying to use PIL to write an array (a NumPy array to be exact) to
> > > > > an image.
> > > > > Peace of cake, but it comes out looking strange.
>
> > > > > I use the below mini code, that I wrote for the purpose. The print of
> > > > > a looks like expected:
>
> > > > > [[ 200. 200. 200. ..., 0. 0. 0.]
> > > > > [ 200. 200. 200. ..., 0. 0. 0.]
> > > > > [ 200. 200. 200. ..., 0. 0. 0.]
> > > > > ...,
> > > > > [ 0. 0. 0. ..., 200. 200. 200.]
> > > > > [ 0. 0. 0. ..., 200. 200. 200.]
> > > > > [ 0. 0. 0. ..., 200. 200. 200.]]
>
> > > > > But the image looks nothing like that.
>
> > > > > Please see the images on:
> > > > >http://hvidberg.net/Martin/temp/quat_col.png
> > > > >http://hvidberg.net/Martin/temp/quat_bw.png
>
> > > > > or run the code to see them locally.
>
> > > > > Please – what do I do wrong in the PIL part ???
>
> > > > > :-? Martin
>
> > > > > import numpy as np
> > > > > from PIL import Image
> > > > > from PIL import ImageOps
>
> > > > > maxcol = 100
> > > > > maxrow = 100
>
> > > > > a = np.zeros((maxcol,maxrow),float)
>
> > > > > for i in range(maxcol):
> > > > > for j in range(maxrow):
> > > > > if (i<(maxcol/2) and j<(maxrow/2)) or (i>=(maxcol/2) and j>=
> > > > > (maxrow/2)):
> > > > > a[i,j] = 200
> > > > > else:
> > > > > a[i,j] = 0
>
> > > > > print a
>
> > > > > pilImage = Image.fromarray(a,'RGB')
> > > > > pilImage.save('quat_col.png')
> > > > > pilImage = ImageOps.grayscale(pilImage)
> > > > > pilImage.save('quat_bw.png')
>
> > > > The PIL seems to copy the array contents directly from memory without
> > > > any
> > > > conversions or sanity check. In your example The float values determine
> > > > the
> > > > gray value of 8 consecutive pixels.
>
> > > > If you want a[i,j] to become the color of the pixel (i, j) you have to
> > > > use
> > > > an array with a memory layout that is compatible to the Image.
> > > > Here are a few examples:
>
> > > > >>> import numpy
> > > > >>> from PIL import Image
> > > > >>> a = numpy.zeros((100, 100), numpy.uint8)
> > > > >>> a[:50, :50] = a[50:, 50:] = 255
> > > > >>> Image.fromarray(a).save("tmp1.png")
> > > > >>> b = numpy.zeros((100, 100, 3), numpy.uint8)
> > > > >>> b[:50, :50, :] = b[50:, 50:, :] = [255, 0, 0]
> > > > >>> Image.fromarray(b).save("tmp2.png")
> > > > >>> c = numpy.zeros((100, 100), numpy.uint32)
> > > > >>> c[:50, :50] = c[50:, 50:] = 0xff808000
> > > > >>> Image.fromarray(c, "RGBA").save("tmp3.png")
>
> > > > Peter
>
> > > Thanks All - That helped a lot...
>
> > > The working code ended with:
>
> > > imga = np.zeros((imgL.shape[1],imgL.shape[0]),np.uint8)
> > > for ro in range(imgL.shape[1]):
> > > for co in range(imgL.shape[0]):
> > > imga[ro,co] = imgL[ro,co]
> > > Image.fromarray(imga).save('_a'+str(lev)+'.png')
>
> > Without knowing how big your image is (can't remember if you said!).
> > Perhaps rather than looping in the way you might in C for example, the
> > numpy where might be quicker if you have a big image. Just a
> > thought...
>
> And a good thought too... I use to teach ansi C at Univ. many ears
> ago, and it's still one of my favorits. But I prefere Python for this
> type of work. Python have a much faster develop time, is much easier
> to write (fewer linees, simpler syntax) and is much more 'hot' in
> these days. Finally my favorit GIS and RS applications support Python
> out-of-the-box.
>
> :-) Martin
I wasn't advocating doing it in c just that you could speed up your
program considerably by not looping and using some of the numpy array
options.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Help with arrays
On Aug 25, 11:50 pm, Stephen Fairchild wrote: > Philip Semanchuk wrote: > > > On Aug 25, 2009, at 6:14 PM, Gleb Belov wrote: > > >> Hello! I'm working on an exercise wherein I have to write a Guess The > >> Number game, but it's the computer who's guessing MY number. I can get > >> it to work, but there's one obvious problem: the computer generates > >> random numbers until one of them corresponds to my number, but it will > >> often generate one number (eg. 4) numerous times, meaning it doesn't > >> know that this number is invalid. What I mean is, it will sometimes > >> use 37 tries to guess a number out of 1 - 9, which makes no sense, > >> since it should only take 9 tries, at most. I was trying to find a way > >> to make a dynamic list of all the numbers the computer generates in > >> the loop and then make it re-generate the number if the previous > >> number is present in the list, so it doesn't keep on generating 4 (as > >> an example). I don't know if that makes sense... Basically, we humans > >> know that once something is incorrect, there's no point in trying to > >> use it as the answer next time, because we already know it's > >> incorrect. How do I go about coding this in Python? I'm still quite > >> new to the language so any help will be appreciated... > > > One cheap way to do it (not necessarily efficient) is to make a list > > of your possible guesses (e.g. range(1,10)), use random.shuffle() to > > put them in random order and then run through the guesses one at a time. > > import random > import time > > l = range(1, 10) > > while l: > print l.pop(random.randint(0, len(l) - 1)) > time.sleep(2) > > -- > Stephen Fairchild Perhaps generate all of the possible moves at the outset in some form of lookup table that you can refer to? I think it is a similar thing to how computers play chess, I think it is called "hash tables" -- http://mail.python.org/mailman/listinfo/python-list
Re: Help with arrays
On Aug 26, 3:02 am, Dave Angel wrote: > Stephen Fairchild wrote: > > Philip Semanchuk wrote: > > >> On Aug 25, 2009, at 6:14 PM, Gleb Belov wrote: > > >>> Hello! I'm working on an exercise wherein I have to write a Guess The > >>> Number game, but it's the computer who's guessing MY number. I can get > >>> it to work, but there's one obvious problem: the computer generates > >>> random numbers until one of them corresponds to my number, but it will > >>> often generate one number (eg. 4) numerous times, meaning it doesn't > >>> know that this number is invalid. What I mean is, it will sometimes > >>> use 37 tries to guess a number out of 1 - 9, which makes no sense, > >>> since it should only take 9 tries, at most. I was trying to find a way > >>> to make a dynamic list of all the numbers the computer generates in > >>> the loop and then make it re-generate the number if the previous > >>> number is present in the list, so it doesn't keep on generating 4 (as > >>> an example). I don't know if that makes sense... Basically, we humans > >>> know that once something is incorrect, there's no point in trying to > >>> use it as the answer next time, because we already know it's > >>> incorrect. How do I go about coding this in Python? I'm still quite > >>> new to the language so any help will be appreciated... > > >> One cheap way to do it (not necessarily efficient) is to make a list > >> of your possible guesses (e.g. range(1,10)), use random.shuffle() to > >> put them in random order and then run through the guesses one at a time. > > > import random > > import time > > > l = range(1, 10) > > > while l: > > print l.pop(random.randint(0, len(l) - 1)) > > time.sleep(2) > > While both of these will work well, I'd point out that a direct > translation of your question is to use a set. Define an empty set, and > each time you try a number unsuccessfully, add it to the set. Then just use >while x in myset: >x = newguess() > > to find the next guess. This approach isn't as efficient, but it's a > useful paradigm to understand. > > A separate question is whether the human is supposed to tell the > computer whether the guess is high or low. If so, you can eliminate > many numbers on each guess. For example, suppose the solution is 8. A > guess of 5 would say "too low." Then you'd cross off now only 5 but 1-4 > as well. > > With this change the best solution changes from a random shuffle to a > binary search. > > DaveA That's a good point if you can define whether the computer has guessed too low/high, then you could use the previous guess to set a bounds and rescale the next random guess. min + (random_number * max) although I think random.randomint does this for you! -- http://mail.python.org/mailman/listinfo/python-list
Re: regexp help
On Aug 27, 7:15 pm, Bakes wrote: > If I were using the code: > > (?P[0-9]+) > > to get an integer between 0 and 9, how would I allow it to register > negative integers as well? -? -- http://mail.python.org/mailman/listinfo/python-list
Re: Extracting patterns after matching a regex
On Sep 8, 2:15 pm, MRAB wrote: > Martin wrote: > > Hi, > > > I need to extract a string after a matching a regular expression. For > > example I have the string... > > > s = "FTPHOST: e4ftl01u.ecs.nasa.gov" > > > and once I match "FTPHOST" I would like to extract > > "e4ftl01u.ecs.nasa.gov". I am not sure as to the best approach to the > > problem, I had been trying to match the string using something like > > this: > > > m = re.findall(r"FTPHOST", s) > > > But I couldn't then work out how to return the "e4ftl01u.ecs.nasa.gov" > > part. Perhaps I need to find the string and then split it? I had some > > help with a similar problem, but now I don't seem to be able to > > transfer that to this problem! > > > Thanks in advance for the help, > > m = re.search(r"FTPHOST: (.*)", s) > print m.group(1) so the .* means to match everything after the regex? That doesn't help in this case as the string is placed amongst others for example. MEDIATYPE: FtpPull\r\n', 'MEDIAFORMAT: FILEFORMAT\r\n', 'FTPHOST: e4ftl01u.ecs.nasa.gov\r\n', 'FTPDIR: /PullDir/0301872638CySfQB\r\n', 'Ftp Pull Download Links: \r\n', -- http://mail.python.org/mailman/listinfo/python-list
Re: Extracting patterns after matching a regex
On Sep 8, 2:21 pm, "Mark Tolonen" wrote: > "Martin" wrote in message > > news:5941d8f1-27c0-47d9-8221-d21f07200...@j39g2000yqh.googlegroups.com... > > > > > Hi, > > > I need to extract a string after a matching a regular expression. For > > example I have the string... > > > s = "FTPHOST: e4ftl01u.ecs.nasa.gov" > > > and once I match "FTPHOST" I would like to extract > > "e4ftl01u.ecs.nasa.gov". I am not sure as to the best approach to the > > problem, I had been trying to match the string using something like > > this: > > > m = re.findall(r"FTPHOST", s) > > > But I couldn't then work out how to return the "e4ftl01u.ecs.nasa.gov" > > part. Perhaps I need to find the string and then split it? I had some > > help with a similar problem, but now I don't seem to be able to > > transfer that to this problem! > > In regular expressions, you match the entire string you are interested in, > and parenthesize the parts that you want to parse out of that string. The > group() method is used to get the whole string with group(0), and each of > the parenthesized parts with group(n). An example: > > >>> s = "FTPHOST: e4ftl01u.ecs.nasa.gov" > >>> import re > >>> re.search(r'FTPHOST: (.*)',s).group(0) > > 'FTPHOST: e4ftl01u.ecs.nasa.gov'>>> re.search(r'FTPHOST: (.*)',s).group(1) > > 'e4ftl01u.ecs.nasa.gov' > > -Mark I see what you mean regarding the groups. Because my string is nested in amongst others e.g. MEDIATYPE: FtpPull\r\n', 'MEDIAFORMAT: FILEFORMAT\r\n', 'FTPHOST: e4ftl01u.ecs.nasa.gov\r\n', 'FTPDIR: /PullDir/0301872638CySfQB\r\n', 'Ftp Pull Download Links: \r\n', I get the information that follows as well. So is the only way to then parse the new string? I am trying to construct something that is fairly robust, so not sure just printing before the \r is the best solution. Thanks -- http://mail.python.org/mailman/listinfo/python-list
Re: Extracting patterns after matching a regex
On Sep 8, 2:16 pm, "Andreas Tawn" wrote: > > Hi, > > > I need to extract a string after a matching a regular expression. For > > example I have the string... > > > s = "FTPHOST: e4ftl01u.ecs.nasa.gov" > > > and once I match "FTPHOST" I would like to extract > > "e4ftl01u.ecs.nasa.gov". I am not sure as to the best approach to the > > problem, I had been trying to match the string using something like > > this: > > > m = re.findall(r"FTPHOST", s) > > > But I couldn't then work out how to return the "e4ftl01u.ecs.nasa.gov" > > part. Perhaps I need to find the string and then split it? I had some > > help with a similar problem, but now I don't seem to be able to > > transfer that to this problem! > > > Thanks in advance for the help, > > > Martin > > No need for regex. > > s = "FTPHOST: e4ftl01u.ecs.nasa.gov" > If "FTPHOST" in s: > return s[9:] > > Cheers, > > Drea Sorry perhaps I didn't make it clear enough, so apologies. I only presented the example s = "FTPHOST: e4ftl01u.ecs.nasa.gov" as I thought this easily encompassed the problem. The solution presented works fine for this i.e. re.search(r'FTPHOST: (.*)',s).group(1). But when I used this on the actual file I am trying to parse I realised it is slightly more complicated as this also pulls out other information, for example it prints e4ftl01u.ecs.nasa.gov\r\n', 'FTPDIR: /PullDir/0301872638CySfQB\r\n', 'Ftp Pull Download Links: \r\n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/ 0301872638CySfQB\r\n', 'Down load ZIP file of packaged order:\r\n', etc. So I need to find a way to stop it before the \r slicing the string wouldn't work in this scenario as I can envisage a situation where the string lenght increases and I would prefer not to keep having to change the string. Many thanks -- http://mail.python.org/mailman/listinfo/python-list
Re: Extracting patterns after matching a regex
On Sep 8, 3:53 pm, MRAB wrote:
> Mart. wrote:
> > On Sep 8, 3:14 pm, "Andreas Tawn" wrote:
> >>>>> Hi,
> >>>>> I need to extract a string after a matching a regular expression. For
> >>>>> example I have the string...
> >>>>> s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
> >>>>> and once I match "FTPHOST" I would like to extract
> >>>>> "e4ftl01u.ecs.nasa.gov". I am not sure as to the best approach to the
> >>>>> problem, I had been trying to match the string using something like
> >>>>> this:
> >>>>> m = re.findall(r"FTPHOST", s)
> >>>>> But I couldn't then work out how to return the "e4ftl01u.ecs.nasa.gov"
> >>>>> part. Perhaps I need to find the string and then split it? I had some
> >>>>> help with a similar problem, but now I don't seem to be able to
> >>>>> transfer that to this problem!
> >>>>> Thanks in advance for the help,
> >>>>> Martin
> >>>> No need for regex.
> >>>> s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
> >>>> If "FTPHOST" in s:
> >>>> return s[9:]
> >>>> Cheers,
> >>>> Drea
> >>> Sorry perhaps I didn't make it clear enough, so apologies. I only
> >>> presented the example s = "FTPHOST: e4ftl01u.ecs.nasa.gov" as I
> >>> thought this easily encompassed the problem. The solution presented
> >>> works fine for this i.e. re.search(r'FTPHOST: (.*)',s).group(1). But
> >>> when I used this on the actual file I am trying to parse I realised it
> >>> is slightly more complicated as this also pulls out other information,
> >>> for example it prints
> >>> e4ftl01u.ecs.nasa.gov\r\n', 'FTPDIR: /PullDir/0301872638CySfQB\r\n',
> >>> 'Ftp Pull Download Links: \r\n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/
> >>> 0301872638CySfQB\r\n', 'Down load ZIP file of packaged order:\r\n',
> >>> etc. So I need to find a way to stop it before the \r
> >>> slicing the string wouldn't work in this scenario as I can envisage a
> >>> situation where the string lenght increases and I would prefer not to
> >>> keep having to change the string.
> >> If, as Terry suggested, you do have a tuple of strings and the first
> >> element has FTPHOST, then s[0].split(":")[1].strip() will work.
>
> > It is an email which contains information before and after the main
> > section I am interested in, namely...
>
> > FINISHED: 09/07/2009 08:42:31
>
> > MEDIATYPE: FtpPull
> > MEDIAFORMAT: FILEFORMAT
> > FTPHOST: e4ftl01u.ecs.nasa.gov
> > FTPDIR: /PullDir/0301872638CySfQB
> > Ftp Pull Download Links:
> >ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB
> > Down load ZIP file of packaged order:
> >ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB.zip
> > FTPEXPR: 09/12/2009 08:42:31
> > MEDIA 1 of 1
> > MEDIAID:
>
> > I have been doing this to turn the email into a string
>
> > email = sys.argv[1]
> > f = open(email, 'r')
> > s = str(f.readlines())
>
> To me that seems a strange thing to do. You could just read the entire
> file as a string:
>
> f = open(email, 'r')
> s = f.read()
>
> > so FTPHOST isn't the first element, it is just part of a larger
> > string. When I turn the email into a string it looks like...
>
> > 'FINISHED: 09/07/2009 08:42:31\r\n', '\r\n', 'MEDIATYPE: FtpPull\r\n',
> > 'MEDIAFORMAT: FILEFORMAT\r\n', 'FTPHOST: e4ftl01u.ecs.nasa.gov\r\n',
> > 'FTPDIR: /PullDir/0301872638CySfQB\r\n', 'Ftp Pull Download Links: \r
> > \n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB\r\n', 'Down
> > load ZIP file of packaged order:\r\n',
>
> > So not sure splitting it like you suggested works in this case.
>
>
Within the file are a list of files, e.g.
TOTAL FILES: 2
FILENAME: MOD13A2.A2007033.h17v08.005.2007101023605.hdf
FILESIZE: 11028908
FILENAME: MOD13A2.A2007033.h17v08.005.2007101023605.hdf.xml
FILESIZE: 18975
and what i want to do is get the ftp address from the file and collect
these files to pull down from the web e.g.
MOD13A2.A2007033.h17v08.005.2007101023605.hdf
MOD13A2.A2007033.h17v08.005.2007101023605.hdf.xml
Thus far I have
#!/usr/bin/env python
import sys
import re
import urllib
email = sys.argv[1]
f = open(email, 'r')
s = str(f.readlines())
m = re.findall(r"MOD\.\.h..v..\.005\..\
\", s)
ftphost = re.search(r'FTPHOST: (.*?)\\r',s).group(1)
ftpdir = re.search(r'FTPDIR: (.*?)\\r',s).group(1)
url = 'ftp://' + ftphost + ftpdir
for i in xrange(len(m)):
print i, ':', len(m)
file1 = m[i][:-4] # remove xml bit.
file2 = m[i]
urllib.urlretrieve(url, file1)
urllib.urlretrieve(url, file2)
which works, clearly my match for the MOD13A2* files isn't ideal I
guess, but they will always occupt those dimensions, so it should
work. Any suggestions on how to improve this are appreciated.
Thanks.
--
http://mail.python.org/mailman/listinfo/python-list
Re: Extracting patterns after matching a regex
On Sep 8, 3:14 pm, "Andreas Tawn" wrote:
> > > > Hi,
>
> > > > I need to extract a string after a matching a regular expression. For
> > > > example I have the string...
>
> > > > s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
>
> > > > and once I match "FTPHOST" I would like to extract
> > > > "e4ftl01u.ecs.nasa.gov". I am not sure as to the best approach to the
> > > > problem, I had been trying to match the string using something like
> > > > this:
>
> > > > m = re.findall(r"FTPHOST", s)
>
> > > > But I couldn't then work out how to return the "e4ftl01u.ecs.nasa.gov"
> > > > part. Perhaps I need to find the string and then split it? I had some
> > > > help with a similar problem, but now I don't seem to be able to
> > > > transfer that to this problem!
>
> > > > Thanks in advance for the help,
>
> > > > Martin
>
> > > No need for regex.
>
> > > s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
> > > If "FTPHOST" in s:
> > > return s[9:]
>
> > > Cheers,
>
> > > Drea
>
> > Sorry perhaps I didn't make it clear enough, so apologies. I only
> > presented the example s = "FTPHOST: e4ftl01u.ecs.nasa.gov" as I
> > thought this easily encompassed the problem. The solution presented
> > works fine for this i.e. re.search(r'FTPHOST: (.*)',s).group(1). But
> > when I used this on the actual file I am trying to parse I realised it
> > is slightly more complicated as this also pulls out other information,
> > for example it prints
>
> > e4ftl01u.ecs.nasa.gov\r\n', 'FTPDIR: /PullDir/0301872638CySfQB\r\n',
> > 'Ftp Pull Download Links: \r\n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/
> > 0301872638CySfQB\r\n', 'Down load ZIP file of packaged order:\r\n',
>
> > etc. So I need to find a way to stop it before the \r
>
> > slicing the string wouldn't work in this scenario as I can envisage a
> > situation where the string lenght increases and I would prefer not to
> > keep having to change the string.
>
> If, as Terry suggested, you do have a tuple of strings and the first element
> has FTPHOST, then s[0].split(":")[1].strip() will work.
It is an email which contains information before and after the main
section I am interested in, namely...
FINISHED: 09/07/2009 08:42:31
MEDIATYPE: FtpPull
MEDIAFORMAT: FILEFORMAT
FTPHOST: e4ftl01u.ecs.nasa.gov
FTPDIR: /PullDir/0301872638CySfQB
Ftp Pull Download Links:
ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB
Down load ZIP file of packaged order:
ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB.zip
FTPEXPR: 09/12/2009 08:42:31
MEDIA 1 of 1
MEDIAID:
I have been doing this to turn the email into a string
email = sys.argv[1]
f = open(email, 'r')
s = str(f.readlines())
so FTPHOST isn't the first element, it is just part of a larger
string. When I turn the email into a string it looks like...
'FINISHED: 09/07/2009 08:42:31\r\n', '\r\n', 'MEDIATYPE: FtpPull\r\n',
'MEDIAFORMAT: FILEFORMAT\r\n', 'FTPHOST: e4ftl01u.ecs.nasa.gov\r\n',
'FTPDIR: /PullDir/0301872638CySfQB\r\n', 'Ftp Pull Download Links: \r
\n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB\r\n', 'Down
load ZIP file of packaged order:\r\n',
So not sure splitting it like you suggested works in this case.
Thanks
--
http://mail.python.org/mailman/listinfo/python-list
Re: Extracting patterns after matching a regex
On Sep 8, 4:33 pm, MRAB wrote:
> Mart. wrote:
> > On Sep 8, 3:53 pm, MRAB wrote:
> >> Mart. wrote:
> >>> On Sep 8, 3:14 pm, "Andreas Tawn" wrote:
> >>>>>>> Hi,
> >>>>>>> I need to extract a string after a matching a regular expression. For
> >>>>>>> example I have the string...
> >>>>>>> s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
> >>>>>>> and once I match "FTPHOST" I would like to extract
> >>>>>>> "e4ftl01u.ecs.nasa.gov". I am not sure as to the best approach to the
> >>>>>>> problem, I had been trying to match the string using something like
> >>>>>>> this:
> >>>>>>> m = re.findall(r"FTPHOST", s)
> >>>>>>> But I couldn't then work out how to return the "e4ftl01u.ecs.nasa.gov"
> >>>>>>> part. Perhaps I need to find the string and then split it? I had some
> >>>>>>> help with a similar problem, but now I don't seem to be able to
> >>>>>>> transfer that to this problem!
> >>>>>>> Thanks in advance for the help,
> >>>>>>> Martin
> >>>>>> No need for regex.
> >>>>>> s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
> >>>>>> If "FTPHOST" in s:
> >>>>>> return s[9:]
> >>>>>> Cheers,
> >>>>>> Drea
> >>>>> Sorry perhaps I didn't make it clear enough, so apologies. I only
> >>>>> presented the example s = "FTPHOST: e4ftl01u.ecs.nasa.gov" as I
> >>>>> thought this easily encompassed the problem. The solution presented
> >>>>> works fine for this i.e. re.search(r'FTPHOST: (.*)',s).group(1). But
> >>>>> when I used this on the actual file I am trying to parse I realised it
> >>>>> is slightly more complicated as this also pulls out other information,
> >>>>> for example it prints
> >>>>> e4ftl01u.ecs.nasa.gov\r\n', 'FTPDIR: /PullDir/0301872638CySfQB\r\n',
> >>>>> 'Ftp Pull Download Links: \r\n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/
> >>>>> 0301872638CySfQB\r\n', 'Down load ZIP file of packaged order:\r\n',
> >>>>> etc. So I need to find a way to stop it before the \r
> >>>>> slicing the string wouldn't work in this scenario as I can envisage a
> >>>>> situation where the string lenght increases and I would prefer not to
> >>>>> keep having to change the string.
> >>>> If, as Terry suggested, you do have a tuple of strings and the first
> >>>> element has FTPHOST, then s[0].split(":")[1].strip() will work.
> >>> It is an email which contains information before and after the main
> >>> section I am interested in, namely...
> >>> FINISHED: 09/07/2009 08:42:31
> >>> MEDIATYPE: FtpPull
> >>> MEDIAFORMAT: FILEFORMAT
> >>> FTPHOST: e4ftl01u.ecs.nasa.gov
> >>> FTPDIR: /PullDir/0301872638CySfQB
> >>> Ftp Pull Download Links:
> >>>ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB
> >>> Down load ZIP file of packaged order:
> >>>ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB.zip
> >>> FTPEXPR: 09/12/2009 08:42:31
> >>> MEDIA 1 of 1
> >>> MEDIAID:
> >>> I have been doing this to turn the email into a string
> >>> email = sys.argv[1]
> >>> f = open(email, 'r')
> >>> s = str(f.readlines())
> >> To me that seems a strange thing to do. You could just read the entire
> >> file as a string:
>
> >> f = open(email, 'r')
> >> s = f.read()
>
> >>> so FTPHOST isn't the first element, it is just part of a larger
> >>> string. When I turn the email into a string it looks like...
> >>> 'FINISHED: 09/07/2009 08:42:31\r\n', '\r\n', 'MEDIATYPE: FtpPull\r\n',
> >>> 'MEDIAFORMAT: FILEFORMAT\r\n', 'FTPHOST: e4ftl01u.ecs.nasa.gov\r\n',
> >>> 'FTPDIR: /PullDir/0301872638CySfQB\r\n', 'Ftp Pull Download Links: \r
> >>> \n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB\r\n', 'Down
> >>> load ZIP file of packaged order:\r\n',
>
Re: Extracting patterns after matching a regex
On Sep 8, 3:21 pm, nn wrote: > On Sep 8, 9:55 am, "Mart." wrote: > > > > > On Sep 8, 2:16 pm, "Andreas Tawn" wrote: > > > > > Hi, > > > > > I need to extract a string after a matching a regular expression. For > > > > example I have the string... > > > > > s = "FTPHOST: e4ftl01u.ecs.nasa.gov" > > > > > and once I match "FTPHOST" I would like to extract > > > > "e4ftl01u.ecs.nasa.gov". I am not sure as to the best approach to the > > > > problem, I had been trying to match the string using something like > > > > this: > > > > > m = re.findall(r"FTPHOST", s) > > > > > But I couldn't then work out how to return the "e4ftl01u.ecs.nasa.gov" > > > > part. Perhaps I need to find the string and then split it? I had some > > > > help with a similar problem, but now I don't seem to be able to > > > > transfer that to this problem! > > > > > Thanks in advance for the help, > > > > > Martin > > > > No need for regex. > > > > s = "FTPHOST: e4ftl01u.ecs.nasa.gov" > > > If "FTPHOST" in s: > > > return s[9:] > > > > Cheers, > > > > Drea > > > Sorry perhaps I didn't make it clear enough, so apologies. I only > > presented the example s = "FTPHOST: e4ftl01u.ecs.nasa.gov" as I > > thought this easily encompassed the problem. The solution presented > > works fine for this i.e. re.search(r'FTPHOST: (.*)',s).group(1). But > > when I used this on the actual file I am trying to parse I realised it > > is slightly more complicated as this also pulls out other information, > > for example it prints > > > e4ftl01u.ecs.nasa.gov\r\n', 'FTPDIR: /PullDir/0301872638CySfQB\r\n', > > 'Ftp Pull Download Links: \r\n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/ > > 0301872638CySfQB\r\n', 'Down load ZIP file of packaged order:\r\n', > > > etc. So I need to find a way to stop it before the \r > > > slicing the string wouldn't work in this scenario as I can envisage a > > situation where the string lenght increases and I would prefer not to > > keep having to change the string. > > > Many thanks > > It is not clear from your post what the input is really like. But just > guessing this might work: > > >>> print s > > 'MEDIATYPE: FtpPull\r\n', 'MEDIAFORMAT: FILEFORMAT\r\n','FTPHOST: > e4ftl01u.ecs.nasa.gov\r\n', 'FTPDIR: /PullDir/0301872638CySfQB\r > \n','Ftp Pull Download Links: \r\n' > > >>> re.search(r'FTPHOST: (.*?)\\r',s).group(1) > > 'e4ftl01u.ecs.nasa.gov' Hi, That does work. So the \ escapes the \r, does this tell it to stop when it reaches the "\r"? Thanks -- http://mail.python.org/mailman/listinfo/python-list
Re: Extracting patterns after matching a regex
On Sep 8, 4:33 pm, MRAB wrote:
> Mart. wrote:
> > On Sep 8, 3:53 pm, MRAB wrote:
> >> Mart. wrote:
> >>> On Sep 8, 3:14 pm, "Andreas Tawn" wrote:
> >>>>>>> Hi,
> >>>>>>> I need to extract a string after a matching a regular expression. For
> >>>>>>> example I have the string...
> >>>>>>> s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
> >>>>>>> and once I match "FTPHOST" I would like to extract
> >>>>>>> "e4ftl01u.ecs.nasa.gov". I am not sure as to the best approach to the
> >>>>>>> problem, I had been trying to match the string using something like
> >>>>>>> this:
> >>>>>>> m = re.findall(r"FTPHOST", s)
> >>>>>>> But I couldn't then work out how to return the "e4ftl01u.ecs.nasa.gov"
> >>>>>>> part. Perhaps I need to find the string and then split it? I had some
> >>>>>>> help with a similar problem, but now I don't seem to be able to
> >>>>>>> transfer that to this problem!
> >>>>>>> Thanks in advance for the help,
> >>>>>>> Martin
> >>>>>> No need for regex.
> >>>>>> s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
> >>>>>> If "FTPHOST" in s:
> >>>>>> return s[9:]
> >>>>>> Cheers,
> >>>>>> Drea
> >>>>> Sorry perhaps I didn't make it clear enough, so apologies. I only
> >>>>> presented the example s = "FTPHOST: e4ftl01u.ecs.nasa.gov" as I
> >>>>> thought this easily encompassed the problem. The solution presented
> >>>>> works fine for this i.e. re.search(r'FTPHOST: (.*)',s).group(1). But
> >>>>> when I used this on the actual file I am trying to parse I realised it
> >>>>> is slightly more complicated as this also pulls out other information,
> >>>>> for example it prints
> >>>>> e4ftl01u.ecs.nasa.gov\r\n', 'FTPDIR: /PullDir/0301872638CySfQB\r\n',
> >>>>> 'Ftp Pull Download Links: \r\n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/
> >>>>> 0301872638CySfQB\r\n', 'Down load ZIP file of packaged order:\r\n',
> >>>>> etc. So I need to find a way to stop it before the \r
> >>>>> slicing the string wouldn't work in this scenario as I can envisage a
> >>>>> situation where the string lenght increases and I would prefer not to
> >>>>> keep having to change the string.
> >>>> If, as Terry suggested, you do have a tuple of strings and the first
> >>>> element has FTPHOST, then s[0].split(":")[1].strip() will work.
> >>> It is an email which contains information before and after the main
> >>> section I am interested in, namely...
> >>> FINISHED: 09/07/2009 08:42:31
> >>> MEDIATYPE: FtpPull
> >>> MEDIAFORMAT: FILEFORMAT
> >>> FTPHOST: e4ftl01u.ecs.nasa.gov
> >>> FTPDIR: /PullDir/0301872638CySfQB
> >>> Ftp Pull Download Links:
> >>>ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB
> >>> Down load ZIP file of packaged order:
> >>>ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB.zip
> >>> FTPEXPR: 09/12/2009 08:42:31
> >>> MEDIA 1 of 1
> >>> MEDIAID:
> >>> I have been doing this to turn the email into a string
> >>> email = sys.argv[1]
> >>> f = open(email, 'r')
> >>> s = str(f.readlines())
> >> To me that seems a strange thing to do. You could just read the entire
> >> file as a string:
>
> >> f = open(email, 'r')
> >> s = f.read()
>
> >>> so FTPHOST isn't the first element, it is just part of a larger
> >>> string. When I turn the email into a string it looks like...
> >>> 'FINISHED: 09/07/2009 08:42:31\r\n', '\r\n', 'MEDIATYPE: FtpPull\r\n',
> >>> 'MEDIAFORMAT: FILEFORMAT\r\n', 'FTPHOST: e4ftl01u.ecs.nasa.gov\r\n',
> >>> 'FTPDIR: /PullDir/0301872638CySfQB\r\n', 'Ftp Pull Download Links: \r
> >>> \n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB\r\n', 'Down
> >>> load ZIP file of packaged order:\r\n',
>
Re: Extracting patterns after matching a regex
On Sep 9, 4:58 pm, Al Fansome wrote:
> Mart. wrote:
> > On Sep 8, 4:33 pm, MRAB wrote:
> >>Mart. wrote:
> >>> On Sep 8, 3:53 pm, MRAB wrote:
> >>>>Mart. wrote:
> >>>>> On Sep 8, 3:14 pm, "Andreas Tawn" wrote:
> >>>>>>>>> Hi,
> >>>>>>>>> I need to extract a string after a matching a regular expression.
> >>>>>>>>> For
> >>>>>>>>> example I have the string...
> >>>>>>>>> s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
> >>>>>>>>> and once I match "FTPHOST" I would like to extract
> >>>>>>>>> "e4ftl01u.ecs.nasa.gov". I am not sure as to the best approach to
> >>>>>>>>> the
> >>>>>>>>> problem, I had been trying to match the string using something like
> >>>>>>>>> this:
> >>>>>>>>> m = re.findall(r"FTPHOST", s)
> >>>>>>>>> But I couldn't then work out how to return the
> >>>>>>>>> "e4ftl01u.ecs.nasa.gov"
> >>>>>>>>> part. Perhaps I need to find the string and then split it? I had
> >>>>>>>>> some
> >>>>>>>>> help with a similar problem, but now I don't seem to be able to
> >>>>>>>>> transfer that to this problem!
> >>>>>>>>> Thanks in advance for the help,
> >>>>>>>>> Martin
> >>>>>>>> No need for regex.
> >>>>>>>> s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
> >>>>>>>> If "FTPHOST" in s:
> >>>>>>>> return s[9:]
> >>>>>>>> Cheers,
> >>>>>>>> Drea
> >>>>>>> Sorry perhaps I didn't make it clear enough, so apologies. I only
> >>>>>>> presented the example s = "FTPHOST: e4ftl01u.ecs.nasa.gov" as I
> >>>>>>> thought this easily encompassed the problem. The solution presented
> >>>>>>> works fine for this i.e. re.search(r'FTPHOST: (.*)',s).group(1). But
> >>>>>>> when I used this on the actual file I am trying to parse I realised it
> >>>>>>> is slightly more complicated as this also pulls out other information,
> >>>>>>> for example it prints
> >>>>>>> e4ftl01u.ecs.nasa.gov\r\n', 'FTPDIR: /PullDir/0301872638CySfQB\r\n',
> >>>>>>> 'Ftp Pull Download Links: \r\n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/
> >>>>>>> 0301872638CySfQB\r\n', 'Down load ZIP file of packaged order:\r\n',
> >>>>>>> etc. So I need to find a way to stop it before the \r
> >>>>>>> slicing the string wouldn't work in this scenario as I can envisage a
> >>>>>>> situation where the string lenght increases and I would prefer not to
> >>>>>>> keep having to change the string.
> >>>>>> If, as Terry suggested, you do have a tuple of strings and the first
> >>>>>> element has FTPHOST, then s[0].split(":")[1].strip() will work.
> >>>>> It is an email which contains information before and after the main
> >>>>> section I am interested in, namely...
> >>>>> FINISHED: 09/07/2009 08:42:31
> >>>>> MEDIATYPE: FtpPull
> >>>>> MEDIAFORMAT: FILEFORMAT
> >>>>> FTPHOST: e4ftl01u.ecs.nasa.gov
> >>>>> FTPDIR: /PullDir/0301872638CySfQB
> >>>>> Ftp Pull Download Links:
> >>>>>ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB
> >>>>> Down load ZIP file of packaged order:
> >>>>>ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB.zip
> >>>>> FTPEXPR: 09/12/2009 08:42:31
> >>>>> MEDIA 1 of 1
> >>>>> MEDIAID:
> >>>>> I have been doing this to turn the email into a string
> >>>>> email = sys.argv[1]
> >>>>> f = open(email, 'r')
> >>>>> s = str(f.readlines())
> >>>> To me that seems a strange thing to do. You could just read the entire
> >>>> file as a string:
> >>&
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On Sep 11, 7:22 pm, C Barr Leigh wrote: > I'm trying to get started with plotting maps in python. I need to read > "shape files" (.shp) and make maps. There seem to be many efforts but > none is complete? I'm looking for suggestions and troubleshooting. > > The basemap package is obviously at an impressive stage and comes with > some data:http://www.scipy.org/Cookbook/Matplotlib/Maps > > but it cannot read shapefiles when their coordinates are not in > geographic projection. min are in a lambert, so the readshapefile > fails. > > Apparently there is a utility that can convert a .shp file to lat/lon > coordinates, but it fails for me. “You can convert the shapefile to > geographic - coordinates using the shpproj utility from the shapelib > tools - (http://shapelib.maptools.org/shapelib-tools.html)" > For me, this gives: > “unable to process projection, exiting...” > > Has anyone overcome these problems? > > Thanks! > Chris Not sure - but perhaps the GDAL? -- http://mail.python.org/mailman/listinfo/python-list
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