Re: Managing Google Groups headaches
Dennis Lee Bieber writes: > [NNTP] clients provide full-fledged editors and conversely full-fledged editors provide NNTP clients -- https://mail.python.org/mailman/listinfo/python-list
Re: cascading python executions only if return code is 0
Roy Smith writes: > Or how to deal with languages where 26 letters isn't enough. English! that is, imvho English is in sore need of some more letters[*] and of diacriticals too g [*] unable to quantify! -- https://mail.python.org/mailman/listinfo/python-list
Re: list comprehension return a list and sum over in loop
KK Sasa writes: > Hi there, > > The list comprehension is results = [d2(t[k]) for k in > xrange(1000)], where d2 is a function returning a list, say > [x1,x2,x3,x4] for one example. So "results" is a list consisting of > 1000 lists, each of length four. Here, what I want to get is the sum > of 1000 lists, and then the result is a list of length four. Is > there any efficient way to do this? Because I found it is slow in my > case. I tried sum(d2(t[k]) for k in xrange(1000)), but it returned > error: TypeError: unsupported operand type(s) for +: 'int' and > 'list'. Thanks. Why didn't you follow Mark Lawrence's advice? In your problem, results is a list of N sublists, each containing exactly four numerical values, Let's try with N=2 In [36]: results = [d2(t[k]) for k in range(2)] In [37]: print results [[1, 2, 3, 4], [5, 6, 7, 8]] Let's try the obvious method to sum In [38]: [sum(el) for el in results] Out[38]: [10, 26] not what you're looking for, but what if we had In [39]: [sum(el) for el in zip(*results)] Out[39]: [6, 8, 10, 12] correct. BTW, as you're using the scientific stack the answer of Christian Gollwitzer is the more appropriate. -- https://mail.python.org/mailman/listinfo/python-list
Re: Explanation of list reference
dave em writes:
> He is asking a question I am having trouble answering which is how a
> variable containing a value differs from a variable containing a
> list or more specifically a list reference.
s/list/mutable object/
# Mr Bond and Mr Tont are two different ob^H^H persons
james_bond = SecretAgent()
james_tont = SecretAgent()
# in some circles, Mr Bond is know as agent 007
agent_007 = james_bond
# Mr Bond, aka 007, is sent to the Caribbeans to crush Spectre
agent_007.move_to('Barbados')
print agent_007.location
print james_bond.location
# Mr Bond, alas, retires and his place in the Mi5 is taken, alas, by Mr Tont
agent_007 = james_tont
# Mr Tont, aka 007, is sent to Hong Kong to, to, whatever...
agent_007.move_to('Hong Kong')
print agent_007.location
print james_bond.location
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Re: [OFF-TOPIC] How do I find a mentor when no one I work with knows what they are doing?
> """ “I’m going to put wings on a [bleep] tank”. class FairchildA10(... -- https://mail.python.org/mailman/listinfo/python-list
Re: 1-0.95
Pedro Izecksohn writes:
> pedro@microboard:~$ /usr/bin/python3
> Python 3.3.2+ (default, Feb 28 2014, 00:52:16)
> [GCC 4.8.1] on linux
> Type "help", "copyright", "credits" or "license" for more information.
1-0.95
> 0.050044
>
> How to get 0.05 as result?
print("%4.2f"%(1-0.95))
i.e., you can change how a result is displayed, not its internal
representation
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Re: Exploring Python for next desktop GUI Project
>> 2. Python 2 or 3? Which will serve me better in the future? > > Long term (7 years), [Python] 3. I have STRONG suicidal intent and no access to treatment, should I better learn Python 2? -- https://mail.python.org/mailman/listinfo/python-list
Re: Exploring Python for next desktop GUI Project
Roy Smith writes: > In article <87mwbtjg9r@pascolo.net>, pec...@pascolo.net wrote: > >> >> 2. Python 2 or 3? Which will serve me better in the future? >> > >> > Long term (7 years), [Python] 3. >> >> I have STRONG suicidal intent and no access to treatment, >> should I better learn Python 2? > > In that case, go with PHP. ah, well said! but I went with PHP already and look at me now! -- https://mail.python.org/mailman/listinfo/python-list
Re: How to look up historical time zones by date and location
Ian Kelly writes: > Some experimentation determines that the timedelta between Shanghai > and Urumqi Urumqi is on the way for the level of popularity that Piraeus enjoyed in the good ol'days -- per crimini come l'umanita'. MMAX, in IPI+IHC -- https://mail.python.org/mailman/listinfo/python-list
Re: How to look up historical time zones by date and location
Joel Goldstick writes: > Local Mean Time is time based on the actually astronomical position > of the sun. It is defined as 12 noon when the sun is at its high > point, directly south in the sky. This is the time you get when you > read a sundial! a sundial measures the Apparent Time, where the duration between two consecutive noons is different from 24*3600 s due to the variations in the Sun apparent angular velocity, that are originated by the Earth's orbit eccentricity and by the inclination of the Earth's rotation axis these variation are small but cumulative, so the AT can be fast or slow with respect to the LMT by as much as ±15' don't know if the same considerations apply also in Urumqi -- https://mail.python.org/mailman/listinfo/python-list
Re: Matplotlib Contour Plots
Jamie Mitchell writes:
> You were right Christian I wanted a shape (2,150).
>
> Thank you Rustom and Steven your suggestion has worked.
>
> Unfortunately the data doesn't plot as I imagined.
>
> What I would like is:
>
> X-axis - hs_con_sw
> Y-axis - te_con_sw
> Z-axis - Frequency
>
> What I would like is for the Z-axis to contour the frequency or
> amount of times that the X-axis data and Y-axis data meet at a
> particular point or bin.
>
> Does anyone know what function or graph could best show this?
in my understanding, you have 3 arrays of data that describe 3D data
points, and you want to draw a 2D contour plot...
in this case you have to interpolate the z-values on a regular grid,
that's very easy if you already know what to do ;-)
here I assume that data is in a .csv file
% cat a.csv
0 ≤ x ≤ 10, 0 ≤ y ≤ 10, z = cos(sqrt((x-5)**2_(y-5)**2))
1.922065,5.827944,-0.998953
7.582322,0.559370,0.411861
5.001753,3.279957,-0.148694
...
of course my z's are different from yours, but this shouldn't be a
real problem --- and here it is my *tested* solution (tested on python
2.7, that is), please feel free to adapt to your needs
hth, ciao
g
% cat contour.py
from numpy import loadtxt, linspace
from matplotlib.mlab import griddata
import matplotlib.pyplot as pl
# open 'a.csv', specify the delimiter, specify how many header rows,
# slurp the data
temp_array = loadtxt(open('a.csv'),delimiter=',',skiprows=1)
# the shape of temp_array is (N,3), we want its transpose
temp_array = temp_array.transpose()
# now the shape is (3,N) and we can do "unpack and assignment:
x, y, z = temp_array
# now the tricky part,
# 1: create two arrays with 101 (arbitrary number) equispaced values
# between 0 and 10 --- that is the ranges of data x and data y
xi = linspace(0,10,101)
yi = linspace(0,10,101)
# 2: create, by interpolation, the 2D array that contourf so eagerly
# awaited!
print griddata.__doc__
zi = griddata(x,y,z, xi,yi)
# eventually, lets plot the stuff...
# see http://matplotlib.org/examples/pylab_examples/griddata_demo.html
# for further details and ideas
pl.contour (xi,yi,zi,11,linewidths=1,colors='black')
pl.contourf(xi,yi,zi); pl.colorbar()
# optional
pl.gca().set_aspect('equal', 'box')
pl.show()
% python contour.py
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Re: Rosetta: Sequence of non-squares
breamore...@gmail.com writes: > A problem at the moment is that although the gmane side works, you > can't get onto the website gmane ex-maintainer got fed up of people complaining through lawyers he gave to new maintainers the gmane spools and the infrastructure to harvest the mailing lists, but not the web site's code while setting up a nntp server was a simple matter the new maintainers are still struggling to build a new web site -- https://mail.python.org/mailman/listinfo/python-list
