什么是在循环中在两个函数之间交替的pythonic方法

Question

It seems like a simple question to me, but a search yielded nothing useful.

I have code like below:

    for key, v in enumerate(ranges):

        ### Used to switch between voltage steps
        second_iter = 0

        for step in v:

            ### Set voltage sources to differential voltages
            if second_iter == 0:
                self.tf.institf.dcm.vsource1(step)
                second_iter = 1
            elif second_iter == 1:
                self.tf.institf.dcm.vsource2(step)

            self.measure_something()

v is a 2 element list. ranges is a list of v's. I want to cycle through each v, but use the second element in a different function each time.

Is there a pythonic way to write the code above?

Edit: To clarify, I want to do some separate code afterwards which measures something. So I can't set both functions at the same time.

Answer 1

Your code is very little intuitive... Is this not enough to do that, without abusing loops?

vsource1 = self.tf.institf.dcm.vsource1
vsource2 = self.tf.institf.dcm.vsource2

for key, (v1, v2) in enumerate(ranges):

    vsource1(v1)
    # Do some other stuff

    vsource2(v2)
    # Do some other stuff  

Alternatively, if the "stuff" is the same you can indeed use a loop like this to avoid repeating yourself:

vsource1 = self.tf.institf.dcm.vsource1
vsource2 = self.tf.institf.dcm.vsource2

funcs = (vsource1, vsource2)

for key, steps in enumerate(ranges):
    for func, step in zip(funcs, steps):
        func(step)
        # Do some other stuff

Answer 2

I would do it like this:

for key, v in enumerate(range(0, 100)):
     [fun_call1() if key % 2 else fun_call2()]

Answer 3

This reduces the amount of code and reduces the need of conditional evaluation to 1 for each iteration.

for key, v in enumerate(ranges):

    if key % 2 == 0:
        self.tf.institf.dcm.vsource1(v[0])
    else:
        self.tf.institf.dcm.vsource2(v[1])

Answer 4

If v is a 2 element list, I'd write it like this:

for key, (step1, step2) in enumerate(ranges):
    self.tf.institf.dcm.vsource1(step1)
    self.measure_something()

    self.tf.institf.dcm.vsource2(step2)
    self.measure_something()

Pythonic doesn't always mean to use the most advanced functions to generate the shortest code. Because of the fixed length of v in your example, I'd say that the solution above is the easiest to read and therefore pythonic.

Answer 5

Simple indexing into second list can help avoid loop and second_iter variable

for key, v in enumerate(ranges):
    self.tf.institf.dcm.vsource1(v[0])
    self.measure_something()

    self.tf.institf.dcm.vsource2(v[1])
    self.measure_something()

Answer 6

You could use itertools.cycle and zip :

from itertools import cycle

vsources = [self.tf.institf.dcm.vsource1, self.tf.institf.dcm.vsource2]


for key, v in enumerate(ranges):
    for (step, vsource) in zip(v, cycle(vsources)):
        ### Set voltage sources to differential voltages
        vsource(step)

Edit: I realise now that there is only two values in each v , so no need to use cycle . Delgan's answer seems to cover this well, so I'll just leave this as-is, and defer to his answer.

Answer 7

Not strictly pythonic, but you can use the remainder operator to switch between the two functions:

funcs = [func1, func2]

for i, voltage in enumerate(ranges):
    for step in voltage:
        func = funcs[i % 2]
        func(step)

OR

This one's not DRY, but probably makes better use of the cache by grouping all calls to each function. Slice ranges into the two arrays you need, then call the function on the subarray:

range1 = ranges[0::2]
range2 = ranges[1::2]

for step in range1: func1(step)
for step in range2: func2(step)

Inlining the slices:

for step in ranges[0::2]: func1(step)
for step in ranges[1::2]: func2(step)