For loop heading depending on if statement (Python)

Question

I want to write a conditional for loop, where if an array is of length 1, use one for loop heading, and if this len(array) > 1 and == len(array2), use a different for loop heading, and if neither of these conditions are true, quit with an error of my choice. The real issue is that I don't want to have this if statement, and then the for loops, when the for loops actually are IDENTICAL, except for the heading, and rather long, so doubling the code seems like a waste.

Is there a nice way to do this where I only have to have the meat of the for loop written once?

Note: xarray and tarray are multi-d numpy arrays ie) xarray = array([[1,2,3],[4,5,6]])

The snippit of code looks as such:

if len(tarray) > 1 and len(xarray) == len(tarray):
        for x,ts in zip(xarray,tarray):
           #stuff
if len(tarray) == 1:
        for x in xarray:
           #same stuff as above for loop
else:
        print 'Dimension Mismatch -- Quitting:'
        quit()

It would really help if you had gave more info about what "meat of the for loop" does. I suspect, because you're using numpy, some sort of broadcasting might the the right tool for this situation. — Bi Rico
– Bi Rico, Commented Apr 5, 2013 at 19:19
the meat of the for loop does a trigonometric interpolation on the arrays in xarray (x values) corresponding to the t values in tarray. And essentially I want to be able to say "hey, xarray has 5 chunks of x data, and tarray only has one, use that same tarray value for all 5 xarrays" or "hey, xarray has n chunks, and so does tarray, so lets use xarray[i] with tarray[i] from i = 0,1,...,n" — KevinShaffer
– KevinShaffer, Commented Apr 5, 2013 at 19:21

Aswin Murugesh · Accepted Answer · 2013-04-05 19:22:00Z

4

If the contents of the for loop are the same, then you can call a function from both the for loops, containing the common code.

And the if statements, checking is necessary if you want the conditions to be met.

answered Apr 5, 2013 at 19:22

Aswin Murugesh

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2 Comments

KevinShaffer Over a year ago

Good answer, however I want to keep this function free if possible, as it's just a file that i will be putting into a package to call as needed.

Aswin Murugesh Over a year ago

so the for loop should be a file output statement only?

vivek · Accepted Answer · 2013-04-05 19:49:10Z

2

This should work:

if len(tarray) >= 1:
    res = zip(tarray, xarray) if len(tarray) == len(xarray) else xarray
else:
    # Error message

for each in res:
    # Do some stuff

answered Apr 5, 2013 at 19:49

vivek

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Comments

user545424 · Accepted Answer · 2013-04-05 19:45:28Z

If I understand your problem correctly, before the for loop you should be able to do something like:

try:
    xx, tt = np.broadcast_arrays(xarray,tarray)
 except ValueError:
    # raised if two arrays cannot be broadcast together

for x, t in zip(xx,tt):
    # do stuff

So, for example:

>>> xarray = np.arange(6).reshape((2,3))
>>> tarray = np.atleast_2d(np.arange(3))
>>> xarray
array([[0, 1, 2],
       [3, 4, 5]])
>>> tarray
array([[0, 1, 2]])
>>> xx, tt = np.broadcast_arrays(xarray,tarray)
>>> xx
array([[0, 1, 2],
       [3, 4, 5]])
>>> tt
array([[0, 1, 2],
       [0, 1, 2]])
>>> tarray = np.arange(6,12).reshape((2,3))
>>> tarray
array([[ 6,  7,  8],
       [ 9, 10, 11]])
>>> xx, tt = np.broadcast_arrays(xarray,tarray)
>>> tt
array([[ 6,  7,  8],
       [ 9, 10, 11]])

I would also suggest doing some checking beforehand that both the xarray and tarray are 2D.

Community · Accepted Answer · 2017-05-23 10:32:11Z

0

In order to change the for loop header in run time you can use exec.

codeToRun = "";

if len(tarray) > 1 and len(xarray) == len(tarray):
        codeToRun += """for x,ts in zip(xarray,tarray): """
elif len(tarray) == 1:
        codeToRun += """for x in xarray:"""
else:
        codeToRun += """print 'Dimension Mismatch -- Quitting:'
        quit()"""

exec codeToRun + """# rest of code """

see How do I execute a string containing Python code in Python? and consider to use eval.

edited May 23, 2017 at 10:32

CommunityBot

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answered Apr 5, 2013 at 20:04

0x90

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