[Tutor] Help with a Conversion

[Peter Otten]

S. P. Molnar wrote:

> I have just started attempting programming in Python and am using Spyder
> with Python 3.5.2 on a Linux platform. (I first started programing in
> Fortran II using punched paper tape.  Yes, am a rather elderly . .  .).
> 
> I have bumbled through, what I foolishly thought was a simple problem, a
> short program to change frequency to wavelength for a plot of
> ultraviolet spectra.  I have attached a pdf of the program.
> 
> During my attempt at programming I have printed results at various
> stages.  Printing wavelength = [row[0] for row in data] gives me 25000
> as the first frequency in the wavelength list (the corresponding
> wavelength is 400).
> 
> To change the frequency to wave length I did the following:
> 
> 
> p=1/1e7
> wave_length = p*np.array(frequency)
> 
> (The relationship between wavelength and frequency is: wavelength =
> 1.0e7/frequency, where 1e7 is the speed of light)
> 
> 
> Apparently whhat I have managed to do is divide each element of the
> frequency list by 1/1e7.
> 
> What I want to do is divide 1e7 by each element of the freqquency list.
> 
> How di I do this?

Since you are using numpy anyway I'd put the frequencies into a numpy.array 
as soon as possible:

>>> import numpy
>>> frequencies = numpy.array([25000, 1250, 400])

Because of numpy's "broadcasting" you can mix skalars and vectors as you 
already tried -- and with the right formula, lamda = c / nu, you get the 
correct result:

>>> speed_of_light = 1e7
>>> wavelengths = speed_of_light / frequencies
>>> wavelengths
array([   400.,   8000.,  25000.])

The equivalent list comprehension in plain Python looks like this:

>>> frequencies = [25000, 1250, 400]
>>> wavelengths = [speed_of_light/freq for freq in frequencies]
>>> wavelengths
[400.0, 8000.0, 25000.0]

> Please keep in mind that many, many hyears ago I learned the ole
> arithmetic

That hasn't changed and is honoured by numpy; you were probably confused by 
the new tool ;)

> and an not trying to start a flame war.

> Thanks in advance for the assistance tha I am sure will be most helpful.