将两个循环合二为一

Question

I have three different loops in my code but I am wondering if there is a way to combine two of them where it only loops through once. My code is below:

for x in groups:
    A bunch of code that produces the following...
    frames[x]
    highcs[x]

I then loop through two loops below. These are the ones I'm trying to combine.

for x, frame in frames.items():
    more unrelated code

for x, highc in highcs.items():

The final two loops are really just creating excel workbooks using xlsxwriter. The first loop creates the workbook and inputs dataframes from frames[x] onto the worksheets. Then highc just goes back over the same sheets and adds new dataframes.

Answer 1

frames and highcs are the same length, you can use zip() to iterate them at once:

for (f,frame), (h, highc) in zip(frames.items(), highcs.items()):
    # do your work here

zip()

This function returns a list of tuples, where the i-th tuple contains the i-th element from each of the argument sequences or iterables.The returned list is truncated in length to the length of the shortest argument sequence.

Answer 2

Use izip from itertools

from itertools import izip
for (x, frame), (x, highc) in izip(frames.items(), highcs.items()):
   ...

https://docs.python.org/2/library/itertools.html#itertools.izip

Make an iterator that aggregates elements from each of the iterables. Like zip() except that it returns an iterator instead of a list. Used forlock-step iteration over several iterables at a time.

izip is more efficient than zip because izip returns an iterator instead of a list. To make it even more efficient, replace .items() with .iteritems()

Answer 3

Not sure what you are doing, but maybe iterate over the groups again?

for group in groups:
    frame = frames[group]
    highc = highcs[group]
    ...

Answer 4

Dunno how pythonic this approach is, but I can think of two quick and dirty ways to do it.

Are they guaranteed to be keyed with the same keys? If so, you could try:

for key in frames.keys():
    doSomething(frames[key])
    doSomethingElse(highcs[key])

Or, if not, do they have the same length? You could:

for i in range(len(frames.keys())):
    frameKey = frames.keys[i]
    highcsKey = highcs.keys[i]
    doSomething(frames[frameKey])
    doSomethingElse(highcs[highcsKey])

Even if they aren't the same length, you could (and probably should) test the lengths of the keys() lists for both dictionaries anyway, and only index into the longer one once you pass the bound of the shorter one.